Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically. 3. SOLUTION: When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ). Create a table of values using x-values in each interval. (− , 2.5): (2.5, ): The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ). x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9 x f (x) 3 −2 5 246 8 2553 10 6991 12 15,541 eSolutions Manual - Powered by Cognero Page 1 1-4 Extrema and Average Rates of Change
Transcript
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 1
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 2
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 3
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 4
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 5
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 6
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 7
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 8
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 9
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 10
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 11
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 12
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 13
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 14
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 15
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
eSolutions Manual - Powered by Cognero Page 16
1-4 Extrema and Average Rates of Change
Use the graph of each function to estimate intervals to the nearest 0.5 unit on which the function is increasing, decreasing, or constant. Support the answer numerically.
3.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is decreasing on (− , 2.5) and
increasing on (2.5, ). Create a table of values using x-values in each interval.
(− , 2.5):
(2.5, ):
The tables support the conjecture that the graph of f is decreasing on (− , 2.5) and increasing on (2.5, ).
x f (x) −12 25,933 −10 13,011 −8 5641 −5 1006 −3 166 0 1 2 −9
x f (x) 3 −2 5 246 8 2553
10 6991 12 15,541
6.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is increasing on (− , −6),
decreasing on (−6, −3), decreasing on (−3, 0), and increasing on (0, ). Create a table of values using x-values in each interval.
(− , −6):
(−6, −3):
(−3, 0):
(0, ):
The tables support the conjecture that the graph of f is increasing on (− , −6), decreasing on (−6, −3), decreasing
on (−3, 0), and increasing on (0, ).
x f (x) −15 −18.75 −10 −14.29 −8 −12.8 −7 −12.25
x f (x) –5.5 –12.1 −5 −12.5 –4.5 –13.5 −4 −16 –3.5 –24.5
x f (x) –2.5 12.5 −2 4 –1.5 1.5 −1 0.5 –0.5 0.10
x f (x) 2 0.8 3 1.5 5 3.13 8 5.82
10 7.69
9.
SOLUTION:
When the graph is viewed from left to right, we can estimate that the graph of f is constant on (− , −5), increasing
on (−5, −3.5), and decreasing on (−3.5, ). Create a table of values using x-values in each interval.
(− , −5):
(−5, −3.5):
(−3.5, ):
The tables support the conjecture that the graph of f is constant on (− , −5), increasing on (−5, −3.5), and
decreasing on (−3.5, ).
x f (x) –12 –4 −10 −4 −8 −4 −7 −4 −6 −4
x f (x) –4.75 10.688 −4.5 11.25 –4.25 11.688 −4 12
–3.75 12.188
x f (x) −3 12 −1 6 0 0 2 −18 5 −60
Estimate and classify the extrema for the graph of each function. Support the answers numerically.
12.
SOLUTION:
Sample answer: It appears that f (x) has relative maxima at x = −0.5 and x = 0.75 and relative minima at x = 0 and x
= 2.25. It also appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−0.5) > f (−1) and f (−0.5) > f (0), there is a relative maximum in the interval (−1, 0). The approximate value of this relative maximum is 0.28. Likewise, because f (0.75) > f (0.5) and f (0.75) > f (1), there is a relative maximum in the interval (0.5, 1). The approximate value of this relative maximum is 0.41. Because f (0) < f (−0.5) and f (0) < f (0.5), there is a relative minimum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. Likewise, because f (2.25) < f (2) and f (2.25) < f (2.5), there is a relative minimumin the interval (2, 2.5). The approximate value of this relative minimum is –9.1. f(−100) < f (−0.5) and f (100) > f (2.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 ·
1010
−1 −2 −0.5 0.28
0 0 0.5 0.34 0.75 0.41
1 0 2 –8
2.25 –9.1 2.5 –7.03 100 9.7 ·
109
15.
SOLUTION: Sample answer: It appears that f (x) has an absolute minimum at x = –3.75, a relative maximum at x = 0, and a
relative minimum at x = 3.5. It also appears that = and = , so we conjecture that this
function has no absolute maximum. Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−3.75) < f (−4) and f (−3.75) < f (−3.5), there is an absolute minimum in the interval (−4, −3.5). The approximate value of the absolute minimum is –1332.37. Likewise, because f (3.5) < f (3) and f (3.5) < f (4), there is a relative minimum in the interval (3, 4). The approximate value of this relative minimum is –1034.36. Because f (0) > f (−0.5) and f (0) > f (0.5), there is a relative maximum in the interval (−0.5, 0.5). The approximate value of this relative minimum is 0. f(−100) > f (−3.5) and f (100) > f (3.5), which supports our conjecture that the function has no absolute maximum.
x f (x) −100 1 · 10
12
−4 −1216 −3.75 –1332.37 −3.5 –1291.61 −0.5 –0.86
0 0 0.5 –0.86 3 –810
3.5 –1034.36 4 –832
100 1 · 1012
18.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.5 and a relative minimum at x = 1.5. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.5) > f (−2) and f (−1.5) > f (−1), there is a relative maximum in the interval (−2, −1). The approximate value of this relative maximum is 5.7. Because f (1.5) < f (1) and f (1.5) < f (2), there is a relative minimum in the interval (1, 2). The approximate value of this relative minimum is –5.7. f(−100) < f (−1.5) and f (100) > f (1.5), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−2 −32 −1.5 5.70 −1 2 1 −2
1.5 −5.70 2 32
100 1 · 1014
21.
SOLUTION: Sample answer: It appears that f (x) has a relative maximum at x = –1.25 and a relative minimum at x = 1.25. It also
appears that and , so we conjecture that this function has no absolute extrema.
Make a table of values. Choose x-values on either side of the estimated x-value for each extremum, as well as one very large and one very small value for x.
Because f (−1.25) > f (−1.5) and f (−1.25) > f (−1), there is a relative maximum in the interval (−1.5, −1). The approximate value of this relative maximum is –1.96. Because f (1.25) < f (1) and f (1.25) < f (1.5), there is a relative minimum in the interval (1, 1.5). The approximate value of this relative minimum is –8.04. f(−100) < f (−1.25) and f (100) > f (1.25), which supports our conjecture that the function has no absolute extrema.
x f (x) −100 −1 · 10
14
−1.5 −8.59 –1.25 –1.96 −1 −2 1 −8
1.25 −8.04 1.5 −1.41 100 1 · 1014
GRAPHING CALCULATOR Approximate to the nearest hundredth the relative or absolute extrema of each function. State the x-values where they occur.
24. f (x) = −x4 + 3x
3 – 2
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. There appears to be an absolute maximum and no relative extrema. Use the maximum tool from the CALC menu to locate the absolute maximum.
The absolute maximum is at (2.25, 6.54).
27. f (x) = −x5 + 3x
2 + x – 1
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Use the minimum and maximum tools from the CALC menu to locate the relative maximum and relative minimum.
The relative maximum is (1.11, 2.12) and the relative minimum is (−0.17, −1.08).
30. f (x) = 0.008x5 – 0.05x
4 – 0.2x
3 + 1.2x
2 – 0.7x
SOLUTION: Graph the function.
The behavior of the graph appears to be visible in the standard window. Expand the y-interval of the window to identify the relative minimum more clearly. Use the minimum and maximum tools from the CALC menu to the relative maxima and relative minima.
The relative maximums are (2.49, 1.45) and (−3.72, 14.23). The relative minimums are (0.32, −0.11) and (5.90, −6.83).
33. GEOMETRY Determine the radius and height that will maximize the volume of the drinking glass shown. Round tothe nearest hundredth of an inch, if necessary.
SOLUTION:
The formula for the surface area of a cylinder is SA = 2πr2 +2πrh. Substitute 20.5π for SA and get h in terms of r.
We need to maximize the volume of the cylinder. The formula for the volume is V = πr2h. Substitute for h.
Graph the equation and locate the maximum.
The maximum occurs when x equals about 1.8484. This is the radius of the cylinder. Substitute this answer and solve for h.
Therefore, the radius should be about 1.85 and the height about 3.70 to maximize the volume of a cylinder with a
surface area of in2.
Find the average rate of change of each function on the given interval.
36. f (x) = 3x3 – 2x
2 + 6; [2, 6]
SOLUTION:
f(x) = 3x3 – 2x
2 + 6; [2, 6]
f(6) = 3(6)3 – 2(6)
2 + 6
f(6) = 648 – 72 + 6 f(6) = 582
f(3) = 3(3)3 – 2(3)
2 + 6
f(3) = 81 – 18 + 6 f(3) = 22
39. f (x) = −2x4 – 5x
3 + 4x – 6; [−1, 5]
SOLUTION:
f(5) = −2(5)4 – 5(5)
3 + 4(5) – 6
f(5) = −1250 – 625 + 20 – 6 f(5) = –1861
f(–1) = −2(–1)4 – 5(–1)
3 + 4(–1) – 6
f(–1) = −2 + 5 – 4 – 6 f(–1) = –7
42. f (x) = ; [5, 12]
SOLUTION:
46. WEATHER The average high temperature by month in Pensacola, Florida, can be modeled by f (x) = −0.9x2 + 13x
+ 43, where x is the month and x = 1 represents January. Find the average rate of change for each time interval, andexplain what this rate represents. a. April to May b. July to November
SOLUTION: a. For April, x = 4, and for May, x = 5.
f (x) = −0.9x2 + 13x + 43
f (5) = −0.9(5)2 + 13(5) + 43
f (5) = −22.5 + 65 + 43 f (5) = 85.5
f(4) = −0.9(4)2 + 13(4) + 43
f(4) = −14.4 + 52 + 43
f(4) = 80.6
4.9° per month; The average temperature increases from early spring to mid-spring. b. For July, x = 7, and for November, x = 11.
f (x) = −0.9x2 + 13x + 43
f (11) = −0.9(11)2 + 13(11) + 43
f (11) = −108.9 + 143 + 43 f (11) = 77.1
f(7) = −0.9(7)2 + 13(7) + 43
f(7) = −44.1 + 91 + 43 f(7) = 89.9
−3.2° per month; The average temperature decreases from summer to late autumn.
47. COFFEE The world coffee consumption from 1990 to 2000 can be modeled by
f (x) = −0.004x4 + 0.077x
3 – 0.38x
2 + 0.46x + 12, where x is the year, x = 0 corresponds with 1990, and the
consumption is measured in millions of pounds. Find the average rate of change for each time interval. a. 1990 to 2000 b. 1995 to 2000
SOLUTION: a. For 1990, x = 0, and for 2000, x = 10.
f(10) = −0.004(10)4 + 0.077(10)
3 – 0.38(10)
2 + 0.46(10) + 12
f(10) = −40 + 77 – 38 + 4.6 + 12 f(10) = 15.6
f(0) = −0.004(0)4 + 0.077(0)
3 – 0.38(0)
2 + 0.46(0) + 12
f(0) = 0 + 0 – 0 + 0+ 12 f(0) = 12
The average rate of change between 1990 and 2000 was 0.36 million pounds more of coffee per year. b. For 1995, x = 5, and for 2000, x = 10.
The average rate of change between 1995 and 2000 was 0.735 million pounds more of coffee per year.
49. Use the graph to complete the following.
a. Find the average rate of change for [5, 15], [15, 20], and [25, 45]. b. Compare and contrast the nature of the speed of the object over these time intervals. c. What conclusions can you make about the magnitude of the rate of change, the steepness of the graph, and the nature of the function?
SOLUTION: a.
b. The average rate of change is the greatest along [5, 15] and the least along [25, 45]. The average rate of change is positive for all three intervals, so, on average, the object is accelerating over all three intervals. It is accelerating atthe fastest rate for interval [5, 15]. While it is very slowly accelerating for [25, 45], it is still increasing speed on average. The graph confirms that it is accelerating for the entire duration of time. c. The steeper the graph, the greater the slope, the greater the average rate of change, and the sharper increase in yfor the same change in x. Steep graph = high magnitude rate of change = rapidly increasing or decreasing; flat graph = low magnitude rate of change = minimal increasing or decreasing.
Sketch a graph of a function with each set of characteristics.54. f (x) is continuous and always increasing.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to always be increasing, f (x + 1) must be greater than f (x) for all elements x of the domain. Sample answer:
58. f (x) is continuous, increasing for x < −2 and decreasing for x > −2.
SOLUTION: For the function to be continuous, there must be no breaks, holes, or gaps in the graph. For the function to be
increasing for x < −2 and decreasing for x > −2, the graph must go up for [− , −2), reach a maximum at x = −2,
then go down for [−2, ). A quadratic or absolute value function would satisfy these conditions. Place the maximum at x = −2 and have the graph open down by multiplying by a negative. Sample answer:
