1. 1FUNCTIONS AND MODELS1.1 Four Ways to Represent a Function In exercises requiring estimations or approximations, your answers may vary slightly from the answers given here. 1. (a) The point (1, 2) is on the graph of f , so f (1) = 2.(b) When x = 2, y is about 2.8, so f (2) 2.8. (c) f (x) = 2 is equivalent to y = 2. When y = 2, we have x = 3 and x = 1. (d) Reasonable estimates for x when y = 0 are x = 2.5 and x = 0.3. (e) The domain of f consists of all x-values on the graph of f . For this function, the domain is 3 x 3, or [3, 3]. The range of f consists of all y-values on the graph of f . For this function, the range is 2 y 3, or [2, 3].(f ) As x increases from 1 to 3, y increases from 2 to 3. Thus, f is increasing on the interval [1, 3]. 3. From Figure 1 in the text, the lowest point occurs at about (t, a) = (12, 85). The highest point occurs at about (17, 115).Thus, the range of the vertical ground acceleration is 85 a 115. Written in interval notation, we get [85, 115].5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve failsthe Vertical Line Test. 7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [3, 2] and the rangeis [3, 2) [1, 3]. 9. The persons weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years. The persons weightdropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30 years saw a gradual increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems. 11. The water will cool down almost to freezing as the ice13. Of course, this graph depends strongly on themelts. Then, when the ice has melted, the water willgeographical location!slowly warm up to room temperature.15. As the price increases, the amount sold decreases.17.9

2. 10CHAPTER 1 FUNCTIONS AND MODELS19. (a)(b) From the graph, we estimate the number of cell-phone subscribers worldwide to be about 92 million in 1995 and 485 million in 1999.21. f (x) = 3x2 x + 2.f (2) = 3(2)2 2 + 2 = 12 2 + 2 = 12. f (2) = 3(2)2 (2) + 2 = 12 + 2 + 2 = 16. f (a) = 3a2 a + 2. f (a) = 3(a)2 (a) + 2 = 3a2 + a + 2. f (a + 1) = 3(a + 1)2 (a + 1) + 2 = 3(a2 + 2a + 1) a 1 + 2 = 3a2 + 6a + 3 a + 1 = 3a2 + 5a + 4. 2f (a) = 2 f (a) = 2(3a2 a + 2) = 6a2 2a + 4. f (2a) = 3(2a)2 (2a) + 2 = 3(4a2 ) 2a + 2 = 12a2 2a + 2. f (a2 ) = 3(a2 )2 (a2 ) + 2 = 3(a4 ) a2 + 2 = 3a4 a2 + 2. [f (a)]2 = 3a2 a + 22= 3a2 a + 2 3a2 a + 2= 9a4 3a3 + 6a2 3a3 + a2 2a + 6a2 2a + 4 = 9a4 6a3 + 13a2 4a + 4.f (a + h) = 3(a + h)2 (a + h) + 2 = 3(a2 + 2ah + h2 ) a h + 2 = 3a2 + 6ah + 3h2 a h + 2. 23. f (x) = 4 + 3x x2 , so f (3 + h) = 4 + 3(3 + h) (3 + h)2 = 4 + 9 + 3h (9 + 6h + h2 ) = 4 3h h2 ,and(4 3h h2 ) 4 h(3 h) f (3 + h) f (3) = = = 3 h. h h h1 ax 1 f (x) f (a) x a = xa = a x = 1(x a) = 1 = 25. xa xa xa xa(x a) xa(x a) ax 27. f (x) = x/(3x 1) is dened for all x except when 0 = 3x 1isx R | x 6=1 3= ,1 31 3,. x = 1 , so the domain 3 29. f (t) = t + 3 t is dened when t 0. These values of t give real number results for t, whereas any value of t gives a real number result for 3 t. The domain is [0, ). 31. h(x) = 1 4 x2 5x is dened when x2 5x > 0x(x 5) > 0. Note that x2 5x 6= 0 since that would result indivision by zero. The expression x(x 5) is positive if x < 0 or x > 5. (See Appendix A for methods for solving inequalities.) Thus, the domain is (, 0) (5, ). 3. SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION33. f (x) = 5 is dened for all real numbers, so the domain is R, or (, ).The graph of f is a horizontal line with y-intercept 5.35. f (t) = t2 6t is dened for all real numbers, so the domain is R, or(, ). The graph of f is a parabola opening upward since the coefcient of t2 is positive. To nd the t-intercepts, let y = 0 and solve for t.0 = t2 6t = t(t 6)t = 0 and t = 6. The t-coordinate of thevertex is halfway between the t-intercepts, that is, at t = 3. Sincef (3) = 32 6 3 = 9, the vertex is (3, 9). x 5 is dened when x 5 0 or x 5, so the domain is [5, ). Since y = x 5 y 2 = x 5 x = y 2 + 5, we see that g is the37. g(x) =top half of a parabola.39. G(x) =3x + |x| . Since |x| = x 3x + x x G(x) = 3x x xx xif x 0 if x < 0 4x x = 2x if x < 0 x if x > 0, we haveif x > 0 = if x < 04if x > 02if x < 0Note that G is not dened for x = 0. The domain is (, 0) (0, ). 41. f (x) =x+2if x < 01xif x 043. f (x) =The domain is R.x + 2 if x 1 x2if x > 1Note that for x = 1, both x + 2 and x2 are equal to 1. The domain is R.45. Recall that the slope m of a line between the two points (x1 , y1 ) and (x2 , y2 ) is m =connecting those two points is y y1 = m(x x1 ). The slope of this line segment is y (3) = 5 (x 1). The function is f (x) = 5 x 2 211 , 21 x 5.y2 y1 and an equation of the line x2 x1 7 (3) 5 = , so an equation is 51 211 4. 12CHAPTER 1 FUNCTIONS AND MODELS47. We need to solve the given equation for y. x + (y 1)2 = 0 (y 1)2 = x y 1 = x x. The expression with the positive radical represents the top half of the parabola, and the one with the negative radical represents the bottom half. Hence, we want f (x) = 1 x. Note that the domain is x 0. y =149. For 0 x 3, the graph is the line with slope 1 and y-intercept 3, that is, y = x + 3. For 3 < x 5, the graph is the linewith slope 2 passing through (3, 0); that is, y 0 = 2(x 3), or y = 2x 6. So the function is f (x) =x + 3 if 0 x 3 2x 6 if 3 < x 551. Let the length and width of the rectangle be L and W . Then the perimeter is 2L + 2W = 20 and the area is A = LW .Solving the rst equation for W in terms of L gives W =20 2L = 10 L. Thus, A(L) = L(10 L) = 10L L2 . Since 2lengths are positive, the domain of A is 0 < L < 10. If we further restrict L to be larger than W , then 5 < L < 10 would be the domain. 53. Let the length of a side of the equilateral triangle be x. Then by the Pythagorean Theorem, the height y of the triangle satisesy2 +2 1 2x= x2 , so that y 2 = x2 1 x2 = 3 x2 and y = 4 4A = 1 (base)(height), we obtain A(x) = 1 (x) 2 23 x 2=3 2 x.3 2 x , 4Using the formula for the area A of a triangle,with domain x > 0.55. Let each side of the base of the box have length x, and let the height of the box be h. Since the volume is 2, we know that2 = hx2 , so that h = 2/x2 , and the surface area is S = x2 + 4xh. Thus, S(x) = x2 + 4x(2/x2 ) = x2 + (8/x), with domain x > 0. 57. The height of the box is x and the length and width are L = 20 2x, W = 12 2x. Then V = LW x and soV (x) = (20 2x)(12 2x)(x) = 4(10 x)(6 x)(x) = 4x(60 16x + x2 ) = 4x3 64x2 + 240x. The sides L, W , and x must be positive. Thus, L > 0 20 2x > 0 x < 10; W > 0 12 2x > 0 x < 6; and x > 0. Combining these restrictions gives us the domain 0 < x < 6. 59. (a)(b) On $14,000, tax is assessed on $4000, and 10%($4000) = $400. On $26,000, tax is assessed on $16,000, and 10%($10,000) + 15%($6000) = $1000 + $900 = $1900.(c) As in part (b), there is $1000 tax assessed on $20,000 of income, so the graph of T is a line segment from (10,000, 0) to (20,000, 1000). The tax on $30,000 is $2500, so the graph of T for x > 20,000 is the ray with initial point (20,000, 1000) that passes through (30,000, 2500). 5. SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS1361. f is an odd function because its graph is symmetric about the origin. g is an even function because its graph is symmetric withrespect to the y-axis. 63. (a) Because an even function is symmetric with respect to the y-axis, and the point (5, 3) is on the graph of this even function,the point (5, 3) must also be on its graph. (b) Because an odd function is symmetric with respect to the origin, and the point (5, 3) is on the graph of this odd function, the point (5, 3) must also be on its graph. 65. f (x) =x . x2 + 1f (x) =67. f (x) =x x x = 2 = 2 = f(x). (x)2 + 1 x +1 x +1x x x , so f(x) = = . x+1 x + 1 x1Since this is neither f (x) nor f (x), the function f is neither even nor odd.So f is an odd function.69. f (x) = 1 + 3x2 x4 .f (x) = 1 + 3(x)2 (x)4 = 1 + 3x2 x4 = f (x). So f is an even function.1.2 Mathematical Models: A Catalog of Essential Functions 1. (a) f (x) =(b) g(x) = 5 x is a root function with n = 5. 1 x2 is an algebraic function because it is a root of a polynomial.(c) h(x) = x9 + x4 is a polynomial of degree 9. (d) r(x) =x2 + 1 is a rational function because it is a ratio of polynomials. x3 + x(e) s(x) = tan 2x is a trigonometric function. (f ) t(x) = log10 x is a logarithmic function. 3. We notice from the gure that g and h are even functions (symmetric with respect to the y-axis) and that f is an odd function(symmetric with respect to the origin). So (b) y = x5 must be f . Since g is atter than h near the origin, we must have (c) y = x8 matched with g and (a) y = x2 matched with h. 6. 14CHAPTER 1 FUNCTIONS AND MODELS5. (a) An equation for the family of linear functions with slope 2is y = f (x) = 2x + b, where b is the y-intercept.(b) f (2) = 1 means that the point (2, 1) is on the graph of f . We can use the point-slope form of a line to obtain an equation for the family of linear functions through the point (2, 1). y 1 = m(x 2), which is equivalent to y = mx + (1 2m) in slope-intercept form.(c) To belong to both families, an equation must have slope m = 2, so the equation in part (b), y = mx + (1 2m), becomes y = 2x 3. It is the only function that belongs to both families. 7. All members of the family of linear functions f(x) = c x have graphsthat are lines with slope 1. The y-intercept is c.9. Since f (1) = f (0) = f(2) = 0, f has zeros of 1, 0, and 2, so an equation for f is f (x) = a[x (1)](x 0)(x 2),or f (x) = ax(x + 1)(x 2). Because f (1) = 6, well substitute 1 for x and 6 for f (x). 6 = a(1)(2)(1) 2a = 6 a = 3, so an equation for f is f (x) = 3x(x + 1)(x 2). 11. (a) D = 200, so c = 0.0417D(a + 1) = 0.0417(200)(a + 1) = 8.34a + 8.34. The slope is 8.34, which represents thechange in mg of the dosage for a child for each change of 1 year in age. (b) For a newborn, a = 0, so c = 8.34 mg. 13. (a)(b) The slope of9 5means that F increases9 5degrees for each increaseof 1 C. (Equivalently, F increases by 9 when C increases by 5 and F decreases by 9 when C decreases by 5.) The F -intercept of 32 is the Fahrenheit temperature corresponding to a Celsius temperature of 0. 7. SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS15. (a) Using N in place of x and T in place of y, we nd the slope to beequation is T 80 = 1 (N 173) T 80 = 1 N 6 6 (b) The slope of1 6173 610 1 T2 T1 80 70 = = . So a linear = N2 N1 173 113 60 6 T = 1N + 6307 6307 6= 51.16 .means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricketchirps per minute. Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of 1 F. (c) When N = 150, the temperature is given approximately by T = 1 (150) + 6 17. (a) We are given307 6= 76.16 F 76 F.4.34 change in pressure = = 0.434. Using P for pressure and d for depth with the point 10 feet change in depth 10(d, P ) = (0, 15), we have the slope-intercept form of the line, P = 0.434d + 15. (b) When P = 100, then 100 = 0.434d + 15 0.434d = 85 d =85 0.434 195.85 feet. Thus, the pressure is100 lb/in2 at a depth of approximately 196 feet. 19. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the formf (x) = a cos(bx) + c seems appropriate. (b) The data appear to be decreasing in a linear fashion. A model of the form f (x) = mx + b seems appropriate. Some values are given to many decimal places. These are the results given by several computer algebra systems rounding is left to the reader.(b) Using the points (4000, 14.1) and (60,000, 8.2), we obtain21. (a)8.2 14.1 (x 4000) or, equivalently, 60,000 4000 y 0.000105357x + 14.521429.y 14.1 =A linear model does seem appropriate.(c) Using a computing device, we obtain the least squares regression line y = 0.0000997855x + 13.950764.The following commands and screens illustrate how to nd the least squares regression line on a TI-83 Plus. Enter the data into list one (L1) and list two (L2). PressFind the regession line and store it in Y1 . Pressto enter the editor..15 8. 16CHAPTER 1 FUNCTIONS AND MODELSNote from the last gure that the regression line has been stored in Y1 and that Plot1 has been turned on (Plot1 is highlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressing .pressingNow pressor byto produce a graph of the data and the regressionline. Note that choice 9 of the ZOOM menu automatically selects a window that displays all of the data. (d) When x = 25,000, y 11.456; or about 11.5 per 100 population. (e) When x = 80,000, y 5.968; or about a 6% chance. (f ) When x = 200,000, y is negative, so the model does not apply. 23. (a)(b)A linear model does seem appropriate.Using a computing device, we obtain the least squares regression line y = 0.089119747x 158.2403249,where x is the year and y is the height in feet.(c) When x = 2000, the model gives y 20.00 ft. Note that the actual winning height for the 2000 Olympics is less than the winning height for 1996so much for that prediction. (d) When x = 2100, y 28.91 ft. This would be an increase of 9.49 ft from 1996 to 2100. Even though there was an increase of 8.59 ft from 1900 to 1996, it is unlikely that a similar increase will occur over the next 100 years. 25.Using a computing device, we obtain the cubic function y = ax3 + bx2 + cx + d with a = 0.0012937, b = 7.06142, c = 12,823, and d = 7,743,770. When x = 1925, y 1914 (million). 9. SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS171.3 New Functions from Old Functions 1. (a) If the graph of f is shifted 3 units upward, its equation becomes y = f (x) + 3.(b) If the graph of f is shifted 3 units downward, its equation becomes y = f (x) 3. (c) If the graph of f is shifted 3 units to the right, its equation becomes y = f (x 3). (d) If the graph of f is shifted 3 units to the left, its equation becomes y = f (x + 3). (e) If the graph of f is reected about the x-axis, its equation becomes y = f (x). (f ) If the graph of f is reected about the y-axis, its equation becomes y = f (x). (g) If the graph of f is stretched vertically by a factor of 3, its equation becomes y = 3f (x). (h) If the graph of f is shrunk vertically by a factor of 3, its equation becomes y = 1 f (x). 3 3. (a) (graph 3) The graph of f is shifted 4 units to the right and has equation y = f (x 4).(b) (graph 1) The graph of f is shifted 3 units upward and has equation y = f (x) + 3. (c) (graph 4) The graph of f is shrunk vertically by a factor of 3 and has equation y = 1 f(x). 3 (d) (graph 5) The graph of f is shifted 4 units to the left and reected about the x-axis. Its equation is y = f (x + 4). (e) (graph 2) The graph of f is shifted 6 units to the left and stretched vertically by a factor of 2. Its equation is y = 2f (x + 6). 5. (a) To graph y = f (2x) we shrink the graph of fhorizontally by a factor of 2.(b) To graph y = f1 x 2we stretch the graph of fhorizontally by a factor of 2.The point (4, 1) on the graph of f corresponds to the The point (4, 1) on the graph of f corresponds to the point1 2point (2 4, 1) = (8, 1). 4, 1 = (2, 1).(c) To graph y = f (x) we reect the graph of f about the y-axis.The point (4, 1) on the graph of f corresponds to the point (1 4, 1) = (4, 1).(d) To graph y = f (x) we reect the graph of f about the y-axis, then about the x-axis.The point (4, 1) on the graph of f corresponds to the point (1 4, 1 1) = (4, 1). 10. 18CHAPTER 1 FUNCTIONS AND MODELS7. The graph of y = f (x) = 3x x2 has been shifted 4 units to the left, reected about the x-axis, and shifted downward1 unit. Thus, a function describing the graph is y=1 f (x + 4)reect about x-axisshift 4 units left 1 shift 1 unit leftThis function can be written as y = f (x + 4) 1 = 3(x + 4) (x + 4)2 1 = 3x + 12 (x2 + 8x + 16) 1 = x2 5x 4 19. y = x3 : Start with the graph of y = x3 and reectabout the x-axis. Note: Reecting about the y-axisgives the same result since substituting x for x gives us y = (x)3 = x3 .11. y = (x + 1)2 : Start with the graph of y = x2and shift 1 unit to the left.13. y = 1 + 2 cos x: Start with the graph of y = cos x, stretch vertically by a factor of 2, and then shift 1 unit upward.15. y = sin(x/2): Start with the graph of y = sin x and stretch horizontally by a factor of 2. 11. SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS x + 3 : Start with the graph of y = x and shift 3 units to the left.17. y =19. y =2 1 2 (x+ 8x) = 1 (x2 + 8x + 16) 8 = 1 (x + 4)2 8: Start with the graph of y = x2 , compress vertically by a 2 2factor of 2, shift 4 units to the left, and then shift 8 units downward.000021. y = 2/(x + 1): Start with the graph of y = 1/x, shift 1 unit to the left, and then stretch vertically by a factor of 2.23. y = |sin x|: Start with the graph of y = sin x and reect all the parts of the graph below the x-axis about the x-axis.25. This is just like the solution to Example 4 except the amplitude of the curve (the 30 N curve in Figure 9 on June 21) is14 12 = 2. So the function is L(t) = 12 + 2 sin2 365 (t 80) . March 31 is the 90th day of the year, so the model givesL(90) 12.34 h. The daylight time (5:51 AM to 6:18 PM) is 12 hours and 27 minutes, or 12.45 h. The model value differs from the actual value by12.4512.34 12.45 0.009, less than 1%.27. (a) To obtain y = f (|x|), the portion of the graph of y = f (x) to the right of the y-axis is reected about the y-axis.(b) y = sin |x|(c) y =|x|19 12. 20CHAPTER 1 FUNCTIONS AND MODELS29. f (x) = x3 + 2x2 ; g(x) = 3x2 1.D = R for both f and g.(f + g)(x) = (x3 + 2x2 ) + (3x2 1) = x3 + 5x2 1, D = R. (f g)(x) = (x3 + 2x2 ) (3x2 1) = x3 x2 + 1, D = R. (f g)(x) = (x3 + 2x2 )(3x2 1) = 3x5 + 6x4 x3 2x2 , D = R. f x3 + 2x2 (x) = , D= g 3x2 1 31. f (x) = x2 1, D = R;1 x | x 6= 3since 3x2 1 6= 0.g(x) = 2x + 1, D = R.(a) (f g)(x) = f (g(x)) = f(2x + 1) = (2x + 1)2 1 = (4x2 + 4x + 1) 1 = 4x2 + 4x, D = R.(b) (g f)(x) = g(f (x)) = g(x2 1) = 2(x2 1) + 1 = (2x2 2) + 1 = 2x2 1, D = R. (c) (f f )(x) = f(f (x)) = f (x2 1) = (x2 1)2 1 = (x4 2x2 + 1) 1 = x4 2x2 , D = R. (d) (g g)(x) = g(g(x)) = g(2x + 1) = 2(2x + 1) + 1 = (4x + 2) + 1 = 4x + 3, D = R.33. f (x) = 1 3x; g(x) = cos x.D = R for both f and g, and hence for their composites.(a) (f g)(x) = f (g(x)) = f(cos x) = 1 3 cos x.(b) (g f)(x) = g(f (x)) = g(1 3x) = cos(1 3x). (c) (f f )(x) = f(f (x)) = f (1 3x) = 1 3(1 3x) = 1 3 + 9x = 9x 2. (d) (g g)(x) = g(g(x)) = g(cos x) = cos(cos x) [Note that this is not cos x cos x.] x+1 1 , D = {x | x 6= 0}; g(x) = , D = {x | x 6= 2} x x+2 x+1 x+1 x+1 1 x+2 = = + + (a) (f g)(x) = f (g(x)) = f x+1 x+2 x+2 x+2 x+1 x+235. f (x) = x +=x2 + 2x + 1 + x2 + 4x + 4 (x + 1)(x + 1) + (x + 2)(x + 2) 2x2 + 6x + 5 = = (x + 2)(x + 1) (x + 2)(x + 1) (x + 2)(x + 1)Since g(x) is not dened for x = 2 and f (g(x)) is not dened for x = 2 and x = 1, the domain of (f g)(x) is D = {x | x 6= 2, 1}. 1 (b) (g f)(x) = g(f (x)) = g x + x1 x 1 x+ xx+ =x2 + 1 + x x2 + x + 1 x2 + x + 1 = = 2 = 2 x x + 2x + 1 (x + 1)2 x + 1 + 2x +2 x +1Since f(x) is not dened for x = 0 and g(f (x)) is not dened for x = 1, the domain of (g f )(x) is D = {x | x 6= 1, 0}. (c) (f f )(x) = f (f (x)) = f x +1 x=x+1 x+1 x+1 x=x+1 + x1 x2 +1 x=x(x) x2 + 1 + 1 x2 + 1 + x(x) x4 + x2 + x2 + 1 + x2 = 2 + 1) x(x x(x2 + 1)=x4 + 3x2 + 1 , x(x2 + 1)D = {x | x 6= 0}=x+x 1 + 2 x x +1 13. SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONSx+1 (d) (g g)(x) = g(g(x)) = g x+2x + 1 + 1(x + 2) x+1 +1 x+1+x+2 2x + 3 x+2 x+2 = = = = x+1 x + 1 + 2(x + 2) x + 1 + 2x + 4 3x + 5 +2 x+2 x+2Since g(x) is not dened for x = 2 and g(g(x)) is not dened for x = 5 , 3 the domain of (g g)(x) is D = x | x 6= 2, 5 . 3 37. (f g h)(x) = f (g(h(x))) = f (g(x 1)) = f(2(x 1)) = 2(x 1) + 1 = 2x 1 39. (f g h)(x) = f (g(h(x))) = f (g(x3 + 2)) = f [(x3 + 2)2 ]= f (x6 + 4x3 + 4) =(x6 + 4x3 + 4) 3 = x6 + 4x3 + 141. Let g(x) = x2 + 1 and f (x) = x10 . Then (f g)(x) = f (g(x)) = f(x2 + 1) = (x2 + 1)10 = F (x). 43. Let g(x) = 3 x and f (x) =45. Let g(t) = cos t and f (t) = 3 x x = F (x). . Then (f g)(x) = f (g(x)) = f ( 3 x ) = 1+x 1+ 3x t. Then (f g)(t) = f (g(t)) = f (cos t) = cos t = u(t).47. Let h(x) = x2 , g(x) = 3x , and f(x) = 1 x. Then 2(f g h)(x) = f (g(h(x))) = f (g(x2 )) = f 3x2= 1 3x = H(x). x, g(x) = sec x, and f(x) = x4 . Then 4 (f g h)(x) = f (g(h(x))) = f (g( x )) = f (sec x ) = (sec x ) = sec4 ( x ) = H(x).49. Let h(x) =51. (a) g(2) = 5, because the point (2, 5) is on the graph of g. Thus, f(g(2)) = f (5) = 4, because the point (5, 4) is on thegraph of f . (b) g(f(0)) = g(0) = 3 (c) (f g)(0) = f (g(0)) = f (3) = 0(d) (g f )(6) = g(f (6)) = g(6). This value is not dened, because there is no point on the graph of g that has x-coordinate 6. (e) (g g)(2) = g(g(2)) = g(1) = 4(f ) (f f )(4) = f (f (4)) = f (2) = 253. (a) Using the relationship distance = rate time with the radius r as the distance, we have r(t) = 60t.(b) A = r2 (A r)(t) = A(r(t)) = (60t)2 = 3600t2 . This formula gives us the extent of the rippled area(in cm ) at any time t. 255. (a) From the gure, we have a right triangle with legs 6 and d, and hypotenuse s.By the Pythagorean Theorem, d2 + 62 = s2 s = f (d) = d 2 + 36.(b) Using d = rt, we get d = (30 km/hr)(t hr) = 30t (in km). Thus, d = g(t) = 30t. (c) (f g)(t) = f (g(t)) = f (30t) =(30t)2 + 36 = 900t2 + 36. This function represents the distance between thelighthouse and the ship as a function of the time elapsed since noon.21 14. 22CHAPTER 1 FUNCTIONS AND MODELS(b)57. (a)H(t) =0 1if t < 0V (t) =if t 00if t < 0120 if t 0so V (t) = 120H(t).Starting with the formula in part (b), we replace 120 with 240 to reect the(c)different voltage. Also, because we are starting 5 units to the right of t = 0, we replace t with t 5. Thus, the formula is V (t) = 240H(t 5). 59. If f(x) = m1 x + b1 and g(x) = m2 x + b2 , then(f g)(x) = f (g(x)) = f (m2 x + b2 ) = m1 (m2 x + b2 ) + b1 = m1 m2 x + m1 b2 + b1 . So f g is a linear function with slope m1 m2 . 61. (a) By examining the variable terms in g and h, we deduce that we must square g to get the terms 4x2 and 4x in h. If we letf (x) = x2 + c, then (f g)(x) = f (g(x)) = f (2x + 1) = (2x + 1)2 + c = 4x2 + 4x + (1 + c). Since h(x) = 4x2 + 4x + 7, we must have 1 + c = 7. So c = 6 and f (x) = x2 + 6. (b) We need a function g so that f (g(x)) = 3(g(x)) + 5 = h(x). But h(x) = 3x2 + 3x + 2 = 3(x2 + x) + 2 = 3(x2 + x 1) + 5, so we see that g(x) = x2 + x 1. 63. (a) If f and g are even functions, then f (x) = f (x) and g(x) = g(x).(i) (f + g)(x) = f(x) + g(x) = f(x) + g(x) = (f + g)(x), so f + g is an even function. (ii) (f g)(x) = f (x) g(x) = f (x) g(x) = (f g)(x), so f g is an even function. (b) If f and g are odd functions, then f (x) = f (x) and g(x) = g(x).(i) (f + g)(x) = f(x) + g(x) = f (x) + [g(x)] = [f (x) + g(x)] = (f + g)(x), so f + g is an odd function.(ii) (f g)(x) = f (x) g(x) = f (x) [g(x)] = f (x) g(x) = (f g)(x), so f g is an even function. 65. We need to examine h(x).h(x) = (f g)(x) = f(g(x)) = f (g(x)) [because g is even] Because h(x) = h(x), h is an even function.= h(x) 15. SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS1.4 Graphing Calculators and Computers 1. f (x) = x3 5x2(a) [5, 5] by [5, 5](b) [0, 10] by [0, 2](c) [0, 10] by [0, 10](There is no graph shown.)The most appropriate graph is produced in viewing rectangle (c). 3. Since the graph of f (x) = 5 + 20x x2 is aparabola opening downward, an appropriate viewing rectangle should include the maximum point.5. f (x) = 4 81 x4 is dened when 81 x4 0 x4 81 |x| 3, so the domain of f is [3, 3]. Also 0 4 81 x4 4 81 = 3, so the range is [0, 3].7. The graph of f (x) = x3 225x is symmetric with respect to the origin.Since f (x) = x3 225x = x(x2 225) = x(x + 15)(x 15), there are x-intercepts at 0, 15, and 15. f (20) = 3500.9. The period of g(x) = sin(1000x) is2 1000 0.0063 and its range is[1, 1]. Since f (x) = sin2 (1000x) is the square of g, its range is [0, 1] and a viewing rectangle of [0.01, 0.01] by [0, 1.1] seems appropriate.11. The domain of y = x is x 0, so the domain of f(x) = sin x is [0, )and the range is [1, 1]. With a little trial-and-error experimentation, we nd that an Xmax of 100 illustrates the general shape of f , so an appropriate viewing rectangle is [0, 100] by [1.5, 1.5].23 16. 24CHAPTER 1 FUNCTIONS AND MODELS13. The rst term, 10 sin x, has period 2 and range [10, 10]. It will be the dominant term in any large graph ofy = 10 sin x + sin 100x, as shown in the rst gure. The second term, sin 100x, has period2 100= 50and range [1, 1].It causes the bumps in the rst gure and will be the dominant term in any small graph, as shown in the view near the origin in the second gure.15. We must solve the given equation for y to obtain equations for the upper andlower halves of the ellipse. 4x2 + 2y 2 = 1 2y 2 = 1 4x2 y= y2 =1 4x2 21 4x2 217. From the graph of y = 3x2 6x + 1and y = 0.23x 2.25 in the viewing rectangle [1, 3] by [2.5, 1.5], it isdifcult to see if the graphs intersect. If we zoom in on the fourth quadrant, we see the graphs do not intersect. 19. From the graph of f (x) = x3 9x2 4, we see that there is one solutionof the equation f (x) = 0 and it is slightly larger than 9. By zooming in or using a root or zero feature, we obtain x 9.05.21. We see that the graphs of f (x) = x2 and g(x) = sin x intersect twice. Onesolution is x = 0. The other solution of f = g is the x-coordinate of the point of intersection in the rst quadrant. Using an intersect feature or zooming in, we nd this value to be approximately 0.88. Alternatively, we could nd that value by nding the positive zero of h(x) = x2 sin x. Note: After producing the graph on a TI-83 Plus, we can nd the approximate value 0.88 by using the following keystrokes: . The 1 is just a guess for 0.88. 17. SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS2523. g(x) = x3 /10 is larger than f (x) = 10x2 whenever x > 100.We see from the graphs of y = |sin x x| and y = 0.1 that there are25.two solutions to the equation |sin x x| = 0.1: x 0.85 and x 0.85. The condition |sin x x| < 0.1 holds for any x lying between these two values, that is, 0.85 < x < 0.85. 27. (a) The root functions y =y= x, 4 x and y = 6 x(b) The root functions y = x, y = 3 x and y = 5 x(c) The root functions y = y = 4 x and y = 5 x x, y = 3 x,(d) For any n, the nth root of 0 is 0 and the nth root of 1 is 1; that is, all nth root functions pass through the points (0, 0) and (1, 1). For odd n, the domain of the nth root function is R, while for even n, it is {x R | x 0}. Graphs of even root functions look similar to that of x, while those of odd root functions resemble that of 3 x. As n increases, the graph of n x becomes steeper near 0 and atter for x > 1. 29. f (x) = x4 + cx2 + x. If c < 1.5, there are three humps: two minimum pointsand a maximum point. These humps get atter as c increases, until at c = 1.5 two of the humps disappear and there is only one minimum point. This single hump then moves to the right and approaches the origin as c increases. 31. y = xn 2x . As n increases, the maximum of thefunction moves further from the origin, and gets larger. Note, however, that regardless of n, the function approaches 0 as x . 18. 26CHAPTER 1 FUNCTIONS AND MODELS33. y 2 = cx3 + x2 . If c < 0, the loop is to the right of the origin, and if c is positive,it is to the left. In both cases, the closer c is to 0, the larger the loop is. (In the limiting case, c = 0, the loop is innite, that is, it doesnt close.) Also, the larger |c| is, the steeper the slope is on the loopless side of the origin. 35. The graphing window is 95 pixels wide and we want to start with x = 0 and end with x = 2. Since there are 94 gapsbetween pixels, the distance between pixels is20 94 .Thus, the x-values that the calculator actually plots are x = 0 +where n = 0, 1, 2, . . . , 93, 94. For y = sin 2x, the actual points plotted by the calculator are n = 0, 1, . . . , 94. For y = sin 96x, the points plotted are sin 96 2 94 n = sin 94 = sin 2 2 942 942 94 n, sin 96 n+2n2 942 94n n, sin 2 2 94for n = 0, 1, . . . , 94. But n = sin 2n + 2 [by periodicity of sine],2 942 94nn = 0, 1, . . . , 94So the y-values, and hence the points, plotted for y = sin 96x are identical to those plotted for y = sin 2x. Note: Try graphing y = sin 94x. Can you see why all the y-values are zero?1.5 Exponential Functions 1. (a) f (x) = ax , a > 0(b) R(c) (0, )(d) See Figures 4(c), 4(b), and 4(a), respectively.3. All of these graphs approach 0 as x , all of them pass through the point(0, 1), and all of them are increasing and approach as x . The larger the base, the faster the function increases for x > 0, and the faster it approaches 0 as x .Note: The notation x can be thought of as x becomes large at this point. More details on this notation are given in Chapter 2. 5. The functions with bases greater than 1 (3x and 10x ) are increasing, while thosewith bases less than 11 x 3and1 x 10are decreasing. The graph ofreection of that of 3x about the y-axis, and the graph of1 x 101 x 3is theis the reection ofthat of 10x about the y-axis. The graph of 10x increases more quickly than that of 3x for x > 0, and approaches 0 faster as x . 7. We start with the graph of y = 4x (Figure 3) and thenshift 3 units downward. This shift doesnt affect the domain, but the range of y = 4x 3 is (3, ) . There is a horizontal asymptote of y = 3. y = 4xy = 4x 3n2 94 n,for 19. SECTION 1.5 EXPONENTIAL FUNCTIONS279. We start with the graph of y = 2x (Figure 3),reect it about the y-axis, and then about the x-axis (or just rotate 180 to handle both reections) to obtain the graph of y = 2x .In each graph, y = 0 is the horizontal asymptote.y = 2xy = 2xy = 2x11. We start with the graph of y = ex (Figure 13) and reect about the y-axis to get the graph of y = ex . Then we compress thegraph vertically by a factor of 2 to obtain the graph of y = 1 ex and then reect about the x-axis to get the graph of 2 y = 1 ex . Finally, we shift the graph upward one unit to get the graph of y = 1 1 ex . 2 213. (a) To nd the equation of the graph that results from shifting the graph of y = ex 2 units downward, we subtract 2 from theoriginal function to get y = ex 2. (b) To nd the equation of the graph that results from shifting the graph of y = ex 2 units to the right, we replace x with x 2 in the original function to get y = e(x2) . (c) To nd the equation of the graph that results from reecting the graph of y = ex about the x-axis, we multiply the original function by 1 to get y = ex . (d) To nd the equation of the graph that results from reecting the graph of y = ex about the y-axis, we replace x with x in the original function to get y = ex . (e) To nd the equation of the graph that results from reecting the graph of y = ex about the x-axis and then about the y-axis, we rst multiply the original function by 1 (to get y = ex ) and then replace x with x in this equation to get y = ex . 15. (a) The denominator 1 + ex is never equal to zero because ex > 0, so the domain of f (x) = 1/(1 + ex ) is R.(b) 1 ex = 0 ex = 1 x = 0, so the domain of f (x) = 1/(1 ex ) is (, 0) (0, ). 17. Use y = Cax with the points (1, 6) and (3, 24).4 = a2 a = 2 [since a > 0] and C =19. If f (x) = 5x , then6 = Ca1 6 2C=6 aand 24 = Ca3 24 == 3. The function is f (x) = 3 2x .5x 5h 1 5h 1 5x+h 5x 5x 5h 5x f (x + h) f (x) = = = = 5x . h h h h h6 3 a a 20. 28CHAPTER 1 FUNCTIONS AND MODELS21. 2 ft = 24 in, f (24) = 242 in = 576 in = 48 ft.g(24) = 224 in = 224 /(12 5280) mi 265 mi23. The graph of g nally surpasses that of f at x 35.8.25. (a) Fifteen hours represents 5 doubling periods (one doubling period is three hours). 100 25 = 3200(b) In t hours, there will be t/3 doubling periods. The initial population is 100, so the population y at time t is y = 100 2t/3 . (c) t = 20 y = 100 220/3 10,159 (d) We graph y1 = 100 2x/3 and y2 = 50,000. The two curves intersect at x 26.9, so the population reaches 50,000 in about 26.9 hours.27. An exponential model is y = abt , where a = 3.154832569 1012and b = 1.017764706. This model gives y(1993) 5498 million and y(2010) 7417 million.From the graph, it appears that f is an odd function (f is undened for x = 0).29.To prove this, we must show that f (x) = f (x). 1 1 1/x e1/x 1 e1/(x) 1 e(1/x) e1/x 1 e f (x) = = = 1/x = 1/x 1 1 + e1/(x) 1 + e(1/x) e e +1 1 + 1/x e 1 e1/x = f (x) = 1 + e1/x so f is an odd function.1.6 Inverse Functions and Logarithms 1. (a) See Denition 1.(b) It must pass the Horizontal Line Test. 3. f is not one-to-one because 2 6= 6, but f (2) = 2.0 = f (6). 5. No horizontal line intersects the graph of f more than once. Thus, by the Horizontal Line Test, f is one-to-one. 21. SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS297. The horizontal line y = 0 (the x-axis) intersects the graph of f in more than one point. Thus, by the Horizontal Line Test,f is not one-to-one. 9. The graph of f (x) = x2 2x is a parabola with axis of symmetry x = 2 b = = 1. Pick any x-values equidistant 2a 2(1)from 1 to nd two equal function values. For example, f (0) = 0 and f (2) = 0, so f is not one-to-one. 11. g(x) = 1/x.x1 6= x2 g (x1 ) 6= g (x2 ), so g is one-to-one. 1/x1 6= 1/x2Geometric solution: The graph of g is the hyperbola shown in Figure 14 in Section 1.2. It passes the Horizontal Line Test, so g is one-to-one. 13. A football will attain every height h up to its maximum height twice: once on the way up, and again on the way down. Thus,even if t1 does not equal t2 , f (t1 ) may equal f (t2 ), so f is not 1-1. 15. Since f (2) = 9 and f is 1-1, we know that f 1 (9) = 2. Remember, if the point (2, 9) is on the graph of f, then the point(9, 2) is on the graph of f 1 . 17. First, we must determine x such that g(x) = 4. By inspection, we see that if x = 0, then g(x) = 4. Since g is 1-1 (g is anincreasing function), it has an inverse, and g1 (4) = 0. 19. We solve C =5 9 (F 32) for F : 9 C = F 32 F = 9 C + 32. This gives us a formula for the inverse function, that 5 5is, the Fahrenheit temperature F as a function of the Celsius temperature C. F 459.67 9 5C9 5C+ 32 459.67 491.67 C 273.15, the domain of the inverse function.21. f (x) = 10 3x y = 10 3x (y 0) y 2 = 10 3x 3x = 10 y 2Interchange x and y: y = 1 x2 + 3 23. y = f (x) = ex3 ln y = x325. y = f (x) = ln (x + 3)10 3 .So f 1 (x) = 1 x2 + 3 x= x + 3 = ey10 3 .10 . 3Note that the domain of f 1 is x 0. 3 ln y. Interchange x and y: y = 3 ln x. So f 1 (x) = 3 ln x. x = ey 3. Interchange x and y: y = ex 3. So f 1 (x) = ex 3. y 1 = x4 x = 4 y 1 (not since x 0). Interchange x and y: y = 4 x 1. So f 1 (x) = 4 x 1. The graph of y = 4 x 1 is just the graph of y = 4 x shifted right one unit.27. y = f (x) = x4 + 1From the graph, we see that f and f 1 are reections about the line y = x.29. Reect the graph of f about the line y = x. The points (1, 2), (1, 1),(2, 2), and (3, 3) on f are reected to (2, 1), (1, 1), (2, 2), and (3, 3) on f 1 . x = 1 y2 + 3 22. 30CHAPTER 1 FUNCTIONS AND MODELS31. (a) It is dened as the inverse of the exponential function with base a, that is, loga x = y(b) (0, )(c) R(d) See Figure 11.33. (a) log5 125 = 3 since 53 = 125.(b) log3 ay = x.1 1 1 = 3 since 33 = 3 = . 27 3 276 35. (a) log2 6 log2 15 + log2 20 = log2 ( 15 ) + log2 20[by Law 2] [by Law 1]6 = log2 ( 15 20)= log2 8, and log2 8 = 3 since 23 = 8. (b) log3 100 log3 18 log3 50 = log3100 18 log3 50 = log3= log3 ( 1 ), and log3 91 9100 1850= 2 since 32 = 1 . 9[by Law 3]37. ln 5 + 5 ln 3 = ln 5 + ln 35[by Law 1]= ln(5 35 ) = ln 1215 39. ln(1 + x2 ) +1 2 (1 + x2 ) x ln x ln sin x = ln(1 + x2 ) + ln x1/2 ln sin x = ln[(1 + x2 ) x ] ln sin x = ln sin x41. To graph these functions, we use log1.5 x =ln x ln x and log50 x = . ln 1.5 ln 50These graphs all approach as x 0+ , and they all pass through thepoint (1, 0). Also, they are all increasing, and all approach as x .The functions with larger bases increase extremely slowly, and the ones with smaller bases do so somewhat more quickly. The functions with large bases approach the y-axis more closely as x 0+ . 43. 3 ft = 36 in, so we need x such that log2 x = 3668,719,476,736 in x = 236 = 68,719,476,736. In miles, this is1 mi 1 ft 1,084,587.7 mi. 12 in 5280 ft45. (a) Shift the graph of y = log10 x ve units to the left toobtain the graph of y = log10 (x + 5). Note the vertical(b) Reect the graph of y = ln x about the x-axis to obtain the graph of y = ln x.asymptote of x = 5.y = log10 x 47. (a) 2 ln x = 1 ln x =y = log10 (x + 5) 1 2 x = e1/2y = ln xy = ln x = e(b) ex = 5 x = ln 5 x = ln 5 49. (a) 2x5 = 3Or: 2x5 log2 3 = x 5 x = 5 + log2 3. = 3 ln 2x5 = ln 3 (x 5) ln 2 = ln 3 x 5 =ln 3 ln 2 x=5+ln 3 ln 2 23. SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS31(b) ln x + ln(x 1) = ln(x(x 1)) = 1 x(x 1) = e1 x2 x e = 0. The quadratic formula (with a = 1, b = 1, and c = e) gives x = 1 1 1 + 4e , but we reject the negative root since the natural logarithm is not 2 dened for x < 0. So x = 51. (a) ex < 101 21+ 1 + 4e . ln ex < ln 10 x < ln 10 x (, ln 10)(b) ln x > 1 eln x > e1 53. (a) For f (x) = x > e1 x (1/e, ) 3 e2x , we must have 3 e2x 0 e2x 32x ln 3 x 1 2ln 3. Thus, the domainof f is (, 1 ln 3]. 2 (b) y = f (x) = x=1 2 3 e2x[note that y 0] y 2 = 3 e2xln(3 y 2 ). Interchange x and y: y =1 2 e2x = 3 y 2ln(3 x2 ). So f 1 (x) =1 2 2x = ln(3 y 2 ) ln(3 x2 ). For the domain of f 1 , we must have 3 x2 > 0 x2 < 3 |x| < 3 3 < x < 3 0 x < 3 since x 0. Note that the domain of f 1 , [0, 3 ), equals the range of f . 55. We see that the graph of y = f (x) = x3 + x2 + x + 1 is increasing, so f is 1-1.y 3 + y 2 + y + 1 and use your CAS to solve the equation for y.Enter x =Using Derive, we get two (irrelevant) solutions involving imaginary expressions, as well as one which can be simplied to the following: 3 4 6y = f 1 (x) = where D = 3 3 D 27x2 + 20 3 D + 27x2 20 + 3 2 3 27x4 40x2 + 16.Maple and Mathematica each give two complex expressions and one real expression, and the real expression is equivalent to that given by Derive. For example, Maples expression simplies to 48 120x2 + 81x4 80.M = 108x2 + 12 57. (a) n = 100 2t/3n = 2t/3 100this as t = f 1 (n) = 3 log2 3 2= 3n 100=since sin = 3 3 2and 3ln50,000 100ln 2(b) sin11 2= 4 4since tan = 1 and 4 since sin = 41 2 4andis in , . 2 2 4=3is in , . 2 2(b) cos1 (1) = since cos = 1 and is in [0, ]. 61. (a) arctan 1 =t 3 t = 3 log2n . Using formula (10), we can write 100ln(n/100) . This function tells us how long it will take to obtain n bacteria (given the number n). ln 2(b) n = 50,000 t = f 1 (50,000) = 3 59. (a) sin11 M 2/3 8 2M 1/3 , where 6 2M 1/3is in , . 2 2ln 500 ln 2 26.9 hours 24. 32CHAPTER 1 FUNCTIONS AND MODELS63. (a) In general, tan(arctan x) = x for any real number x. Thus, tan(arctan 10) = 10.(b) sin1 sin 7 = sin1 sin = sin1 3 3 [Recall that7 3= 3 3 2= 3since sin = 33 2and 3is in , 2 2.+ 2 and the sine function is periodic with period 2.]65. Let y = sin1 x. Then y 2 2 cos y 0, so cos(sin1 x) = cos y =1 sin2 y = 1 x2 .67. Let y = tan1 x. Then tan y = x, so from the triangle we see thatx . sin(tan1 x) = sin y = 1 + x2The graph of sin1 x is the reection of the graph of69.sin x about the line y = x.71. g(x) = sin1 (3x + 1).Domain (g) = {x | 1 3x + 1 1} = {x | 2 3x 0} = x | 2 x 0 = 2 , 0 . 3 3 Range (g) = y | y 2 2= , . 2 273. (a) If the point (x, y) is on the graph of y = f (x), then the point (x c, y) is that point shifted c units to the left. Since f is1-1, the point (y, x) is on the graph of y = f 1 (x) and the point corresponding to (x c, y) on the graph of f is (y, x c) on the graph of f 1 . Thus, the curves reection is shifted down the same number of units as the curve itself is shifted to the left. So an expression for the inverse function is g 1 (x) = f 1 (x) c. (b) If we compress (or stretch) a curve horizontally, the curves reection in the line y = x is compressed (or stretched) vertically by the same factor. Using this geometric principle, we see that the inverse of h(x) = f(cx) can be expressed as h1 (x) = (1/c) f 1 (x). 25. CHAPTER 1 REVIEW331 Review1. (a) A function f is a rule that assigns to each element x in a set A exactly one element, called f(x), in a set B. The set A iscalled the domain of the function. The range of f is the set of all possible values of f (x) as x varies throughout the domain. (b) If f is a function with domain A, then its graph is the set of ordered pairs {(x, f (x)) | x A}. (c) Use the Vertical Line Test on page 16. 2. The four ways to represent a function are: verbally, numerically, visually, and algebraically. An example of each is givenbelow. Verbally: An assignment of students to chairs in a classroom (a description in words) Numerically: A tax table that assigns an amount of tax to an income (a table of values) Visually: A graphical history of the Dow Jones average (a graph) Algebraically: A relationship between distance, rate, and time: d = rt (an explicit formula) 3. (a) An even function f satises f (x) = f (x) for every number x in its domain. It is symmetric with respect to the y-axis.(b) An odd function g satises g(x) = g(x) for every number x in its domain. It is symmetric with respect to the origin. 4. A function f is called increasing on an interval I if f (x1 ) < f (x2 ) whenever x1 < x2 in I. 5. A mathematical model is a mathematical description (often by means of a function or an equation) of a real-worldphenomenon. 6. (a) Linear function: f(x) = 2x + 1, f (x) = ax + b7.(b) Power function: f(x) = x2 , f(x) = xa (c) Exponential function: f (x) = 2x , f (x) = ax (d) Quadratic function: f (x) = x2 + x + 1, f (x) = ax2 + bx + c (e) Polynomial of degree 5: f(x) = x5 + 2 (f ) Rational function: f (x) =P (x) x , f (x) = where P (x) and x+2 Q(x)Q(x) are polynomials 8. (a)(b) 26. 34CHAPTER 1 FUNCTIONS AND MODELS(c)(d)(e)(f )(g)(h)9. (a) The domain of f + g is the intersection of the domain of f and the domain of g; that is, A B.(b) The domain of f g is also A B.(c) The domain of f /g must exclude values of x that make g equal to 0; that is, {x A B | g(x) 6= 0}.10. Given two functions f and g, the composite function f g is dened by (f g) (x) = f (g (x)). The domain of f g is theset of all x in the domain of g such that g(x) is in the domain of f . 11. (a) If the graph of f is shifted 2 units upward, its equation becomes y = f (x) + 2.(b) If the graph of f is shifted 2 units downward, its equation becomes y = f (x) 2. (c) If the graph of f is shifted 2 units to the right, its equation becomes y = f (x 2). (d) If the graph of f is shifted 2 units to the left, its equation becomes y = f (x + 2). (e) If the graph of f is reected about the x-axis, its equation becomes y = f (x). (f ) If the graph of f is reected about the y-axis, its equation becomes y = f (x). (g) If the graph of f is stretched vertically by a factor of 2, its equation becomes y = 2f (x). (h) If the graph of f is shrunk vertically by a factor of 2, its equation becomes y = 1 f (x). 2 (i) If the graph of f is stretched horizontally by a factor of 2, its equation becomes y = f1 x 2.(j) If the graph of f is shrunk horizontally by a factor of 2, its equation becomes y = f (2x). 12. (a) A function f is called a one-to-one function if it never takes on the same value twice; that is, if f (x1 ) 6= f (x2 ) wheneverx1 6= x2 . (Or, f is 1-1 if each output corresponds to only one input.)Use the Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more thanonce. (b) If f is a one-to-one function with domain A and range B, then its inverse function f 1 has domain B and range A and is dened by f 1 (y) = x f (x) = y for any y in B. The graph of f 1 is obtained by reecting the graph of f about the line y = x. 27. CHAPTER 1 REVIEW3513. (a) The inverse sine function f (x) = sin1 x is dened as follows:sin1 x = y Its domain is 1 x 1 and its range is sin y = xand y 2 2 y . 2 2(b) The inverse cosine function f (x) = cos1 x is dened as follows: cos1 x = ycos y = xand0yIts domain is 1 x 1 and its range is 0 y . (c) The inverse tangent function f (x) = tan1 x is dened as follows: tan1 x = y Its domain is R and its range is 1. False.tan y = xand 0 x > 6. D = (6, )Range:Range:1 31 , 3all reals except 0 (y = 0 is the horizontal asymptote for f .) R = (, 0) (0, )x + 6 > 0, so ln(x + 6) takes on all real numbers and, hence, the range is R.R = (, ) 9. (a) To obtain the graph of y = f (x) + 8, we shift the graph of y = f (x) up 8 units.(b) To obtain the graph of y = f (x + 8), we shift the graph of y = f (x) left 8 units. (c) To obtain the graph of y = 1 + 2f (x), we stretch the graph of y = f (x) vertically by a factor of 2, and then shift the resulting graph 1 unit upward. (d) To obtain the graph of y = f (x 2) 2, we shift the graph of y = f (x) right 2 units (for the 2 inside the parentheses), and then shift the resulting graph 2 units downward. (e) To obtain the graph of y = f (x), we reect the graph of y = f (x) about the x-axis. (f ) To obtain the graph of y = f 1 (x), we reect the graph of y = f(x) about the line y = x (assuming f is oneto-one). 11. y = sin 2x: Start with the graph of y = sin x, compress horizontally by a factor of 2, and reect about the x-axis.13. y =1 (1 2+ ex ):Start with the graph of y = ex , shift 1 unit upward, and compress vertically by a factor of 2.15. f (x) =1 : x+2Start with the graph of f (x) = 1/x and shift 2 units to the left. 29. CHAPTER 1 REVIEW3717. (a) The terms of f are a mixture of odd and even powers of x, so f is neither even nor odd.(b) The terms of f are all odd powers of x, so f is odd. 22(c) f (x) = e(x) = ex = f (x), so f is even. (d) f (x) = 1 + sin(x) = 1 sin x. Now f (x) 6= f (x) and f(x) 6= f(x), so f is neither even nor odd. 19. f (x) = ln x,D = (0, ); g(x) = x2 9, D = R.(a) (f g)(x) = f (g(x)) = f(x2 9) = ln(x2 9). Domain: x2 9 > 0 x2 > 9 |x| > 3 x (, 3) (3, ) (b) (g f)(x) = g(f (x)) = g(ln x) = (ln x)2 9. Domain: x > 0, or (0, ) (c) (f f )(x) = f(f (x)) = f (ln x) = ln(ln x). Domain: ln x > 0 x > e0 = 1, or (1, ) (d) (g g)(x) = g(g(x)) = g(x2 9) = (x2 9)2 9. Domain: x R, or (, ) Many models appear to be plausible. Your choice depends on whether you21.think medical advances will keep increasing life expectancy, or if there is bound to be a natural leveling-off of life expectancy. A linear model, y = 0.2493x 423.4818, gives us an estimate of 77.6 years for theyear 2010.23. We need to know the value of x such that f (x) = 2x + ln x = 2. Since x = 1 gives us y = 2, f 1 (2) = 1. 25. (a) e2 ln 3 = eln 32= 32 = 9(b) log10 25 + log10 4 = log10 (25 4) = log10 100 = log10 102 = 2 (c) tan arcsin 1 = tan = 2 61 3(d) Let = cos1 4 , so cos = 4 . Then sin cos1 5 5 27. (a)4 5= sin = 1 cos2 =14 2 59 25== 3. 5The population would reach 900 in about 4.4 years.(b) P =100,000 100P + 900P et = 100,000 900P et = 100,000 100P 100 + 900etet =1000 P 100,000 100P t = ln 900P 9P t = ln1000 P 9Prequired for the population to reach a given number P . (c) P = 900 t = ln9 900 1000 900= ln 81 4.4 years, as in part (a)., or ln 9P 1000 P; this is the time 30. PRINCIPLES OF PROBLEM SOLVING By using the area formula for a triangle,(base) (height), in two ways, we see that 4y 1 . Since 42 + y 2 = h2 , y = h2 16, and (4) (y) = 1 (h) (a), so a = 2 2 h 4 h2 16 . a= h1.3. |2x 1| =2x 1 if x 1 2x if x 3, then f (x) = x2 4x + 3. This enables us to sketch the graph for x 0. Then we use the fact that f is an evenfunction to reect this part of the graph about the y-axis to obtain the entire graph. Or, we could consider also the cases x < 3, 3 x < 1, and 1 x < 0.7. Remember that |a| = a if a 0 and that |a| = a if a < 0. Thus,x + |x| =2x if x 0 0if x < 0andy + |y| =2y 0if y 0 if y < 0We will consider the equation x + |x| = y + |y| in four cases. (1) x 0, y 0 2x = 2y(2) x 0, y < 0 2x = 0x=yx=0(3) x < 0, y 0 0 = 2y(4) x < 0, y < 0 0=00=yCase 1 gives us the line y = x with nonnegative x and y. Case 2 gives us the portion of the y-axis with y negative. Case 3 gives us the portion of the x-axis with x negative. Case 4 gives us the entire third quadrant.39 31. 40CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING9. |x| + |y| 1. The boundary of the region has equation |x| + |y| = 1. In quadrantsI, II, III, and IV, this becomes the lines x + y = 1, x + y = 1, x y = 1, and x y = 1 respectively.11. (log2 3)(log3 4)(log4 5) (log31 32) = 13. ln x2 2x 2 0ln 3 ln 2ln 4 ln 3ln 5 ln 4ln 32 ln 31=ln 25 5 ln 2 ln 32 = = =5 ln 2 ln 2 ln 2 x2 2x 2 e0 = 1 x2 2x 3 0 (x 3)(x + 1) 0 x [1, 3].Since the argument must be positive, x2 2x 2 > 0 x 1 3x 1+ 3>0 x , 1 3 1 + 3, . The intersection of these intervals is 1, 1 3 1 + 3, 3 .15. Let d be the distance traveled on each half of the trip. Let t1 and t2 be the times taken for the rst and second halves of the trip.For the rst half of the trip we have t1 = d/30 and for the second half we have t2 = d/60. Thus, the average speed for the entire trip is2d 120d 120d 60 2d total distance = = = = 40. The average speed for the entire trip = d d 60 total time t1 + t2 2d + d 3d + 30 60is 40 mi/h. 17. Let Sn be the statement that 7n 1 is divisible by 6. S1 is true because 71 1 = 6 is divisible by 6. Assume Sk is true, that is, 7k 1 is divisible by 6. In other words, 7k 1 = 6m for some positive integer m. Then 7k+1 1 = 7k 7 1 = (6m + 1) 7 1 = 42m + 6 = 6(7m + 1), which is divisible by 6, so Sk+1 is true. Therefore, by mathematical induction, 7n 1 is divisible by 6 for every positive integer n. 19. f0 (x) = x2 and fn+1 (x) = f0 (fn (x)) for n = 0, 1, 2, . . ..f1 (x) = f0 (f0 (x)) = f0 x2 = x22= x4 , f2 (x) = f0 (f1 (x)) = f0 (x4 ) = (x4 )2 = x8 , n+1f3 (x) = f0 (f2 (x)) = f0 (x8 ) = (x8 )2 = x16 , . . .. Thus, a general formula is fn (x) = x2. 32. 2LIMITS AND DERIVATIVES2.1 The Tangent and Velocity Problems 1. (a) Using P (15, 250), we construct the following table:slope = mP QtQ5(5, 694)694250 515(10, 444)444250 1015= 194 = 38.8 520(20, 111)111250 2015= 139 = 27.8 525(25, 28)28250 251530(30, 0)0250 3015closest to P (t = 10 and t = 20), we have= 444 = 44.4 1010(b) Using the values of t that correspond to the points38.8 + (27.8) = 33.3 2= 222 = 22.2 10 = 250 = 16.6 15(c) From the graph, we can estimate the slope of the tangent line at P to be300 9= 33.3.3. (a)xQmP Q(i)0.5(0.5, 0.333333)0.333333(ii)0.9(0.9, 0.473684)0.99(0.99, 0.497487)0.999(0.999, 0.499750)1 2= 1 (x 1) or y = 1 x + 1 . 4 4 40.251256(iv)(c) y 0.263158(iii)(b) The slope appears to be 1 . 40.250125(v)1.5(1.5, 0.6)0.2(vi)1.1(1.1, 0.523810)0.238095(vii)1.01(1.01, 0.502488)0.248756(viii)1.001(1.001, 0.500250)0.2498755. (a) y = y(t) = 40t 16t2 . At t = 2, y = 40(2) 16(2)2 = 16. The average velocity between times 2 and 2 + h is40(2 + h) 16(2 + h)2 16 24h 16h2 y(2 + h) y(2) = = = 24 16h, if h 6= 0. (2 + h) 2 h h (ii) [2, 2.1]: h = 0.1, vave = 25.6 ft/s (i) [2, 2.5]: h = 0.5, vave = 32 ft/svave =(iii) [2, 2.05]: h = 0.05, vave = 24.8 ft/s(iv) [2, 2.01]: h = 0.01, vave = 24.16 ft/s(b) The instantaneous velocity when t = 2 (h approaches 0) is 24 ft/s. 41 33. 42CHAPTER 2LIMITS AND DERIVATIVES7. (a) (i) On the interval [1, 3], vave =10.7 1.4 9.3 s(3) s(1) = = = 4.65 m/s. 31 2 2(ii) On the interval [2, 3], vave =s(3) s(2) 10.7 5.1 = = 5.6 m/s. 32 1(iii) On the interval [3, 5], vave =25.8 10.7 15.1 s(5) s(3) = = = 7.55 m/s. 53 2 2(iv) On the interval [3, 4], vave =17.7 10.7 s(4) s(3) = = 7 m/s. 43 1(b)Using the points (2, 4) and (5, 23) from the approximate tangent line, the instantaneous velocity at t = 3 is about23 4 6.3 m/s. 529. (a) For the curve y = sin(10/x) and the point P (1, 0):xQmP QxQ2(2, 0)00.5(0.5, 0)1.5(1.5, 0.8660)1.73210.6(0.6, 0.8660)1.4(1.4, 0.4339)1.08470.7(0.7, 0.7818)0.8(0.8, 1)4.33010.9(0.9, 0.3420)1.3 1.2 1.1(1.3, 0.8230) (1.2, 0.8660)(1.1, 0.2817)2.7433 2.8173mP Q 0 2.1651 2.6061 53.4202As x approaches 1, the slopes do not appear to be approaching any particular value. We see that problems with estimation are caused by the frequent(b)oscillations of the graph. The tangent is so steep at P that we need to take x-values much closer to 1 in order to get accurate estimates of its slope.(c) If we choose x = 1.001, then the point Q is (1.001, 0.0314) and mP Q 31.3794. If x = 0.999, then Q is (0.999, 0.0314) and mP Q = 31.4422. The average of these slopes is 31.4108. So we estimate that the slope of the tangent line at P is about 31.4. 34. SECTION 2.2 THE LIMIT OF A FUNCTION432.2 The Limit of a Function 1. As x approaches 2, f (x) approaches 5. [Or, the values of f (x) can be made as close to 5 as we like by taking x sufcientlyclose to 2 (but x 6= 2).] Yes, the graph could have a hole at (2, 5) and be dened such that f (2) = 3. 3. (a) lim f (x) = means that the values of f (x) can be made arbitrarily large (as large as we please) by taking x x3sufciently close to 3 (but not equal to 3). (b) lim f(x) = means that the values of f (x) can be made arbitrarily large negative by taking x sufciently close to 4 x4+through values larger than 4. 5. (a) f (x) approaches 2 as x approaches 1 from the left, so lim f (x) = 2. x1(b) f (x) approaches 3 as x approaches 1 from the right, so lim f (x) = 3. x1+(c) lim f (x) does not exist because the limits in part (a) and part (b) are not equal. x1(d) f (x) approaches 4 as x approaches 5 from the left and from the right, so lim f (x) = 4. x5(e) f(5) is not dened, so it doesnt exist. 7. (a) lim g(t) = 1(b) lim g(t) = 2t0t0+(c) lim g(t) does not exist because the limits in part (a) and part (b) are not equal. t0(d) lim g(t) = 2(e) lim g(t) = 0t2t2+(f ) lim g(t) does not exist because the limits in part (d) and part (e) are not equal. t2(g) g(2) = 1(h) lim g(t) = 3 t49. (a) lim f (x) = (b) lim f(x) = (d) lim f (x) = (e) lim f (x) = x7x6(c) lim f (x) = x3x0x6+(f ) The equations of the vertical asymptotes are x = 7, x = 3, x = 0, and x = 6. 11. (a) lim f (x) = 1 x0(b) lim f(x) = 0 x0+(c) lim f (x) does not exist because the limits x0in part (a) and part (b) are not equal. 13. lim f (x) = 2, x1lim f (x) = 2, f (1) = 2x1+15. lim f (x) = 4, x3+lim f (x) = 2, lim f(x) = 2,x3f (3) = 3, f (2) = 1x2 35. 44CHAPTER 2LIMITS AND DERIVATIVESx2 2x : x217. For f (x) =19. For f (x) =x2xf (x)2.5x0.7142861.9ex 1 x : x2xf (x)f (x)0.65517210.718282 0.5948852.10.6774191.950.6610170.52.050.6721311.990.6655520.10.5170922.010.6677741.9950.6661100.050.5084392.005 2.0010.667221 0.6667781.9990.6665560.010.501671It appears that lim2It appears that limx221. For f (x) =xx 2x = 0. = 2 . 6 3 x2 x 2 x+42 : x0.2360680.50.2426410.10.2484570.050.2492240.010.249844It appears that limx0limx3+27. limx123. For f (x) =f (x)125.x0xf (x)xf (x)10.3678790.10.4837420.010.4983370.50.4261230.050.491770ex 1 x = 0.5 = 1 . 2 x2x6 1 : x10 1xf (x)xf (x)10.2679490.50.9853371.50.1833690.2583430.90.7193971.10.4841190.10.2515820.950.6601861.050.5407830.2507860.990.6120181.010.5880220.2501560.9990.6012001.0010.5988000.5 0.05 0.01 x+42 = 0.25 = 1 . 4 xIt appears that limx1x6 1 = 0.6 = 3 . 5 x10 1x+2 = since the numerator is negative and the denominator approaches 0 from the positive side as x 3+ . x+32x = since the numerator is positive and the denominator approaches 0 through positive values as x 1. (x 1)229. Let t = x2 9. Then as x 3+ , t 0+ , and lim ln(x2 9) = lim ln t = by (3). x3+31.lim x csc x = limx2 x2 t0+x = since the numerator is positive and the denominator approaches 0 through negative sin xvalues as x 2 . 33. (a) f (x) =1 . x3 1x 0.5From these calculations, it seems that lim f (x) = and lim f (x) = .x1x1+0.9 0.99 0.999 0.9999 0.99999f (x)xf (x)1.141.50.423.691.13.0233.71.0133.0333.71.001333.03333.71.00013333.033,333.71.0000133,333.3 36. SECTION 2.2 THE LIMIT OF A FUNCTION45(b) If x is slightly smaller than 1, then x3 1 will be a negative number close to 0, and the reciprocal of x3 1, that is, f (x), will be a negative number with large absolute value. So lim f(x) = . x1If x is slightly larger than 1, then x3 1 will be a small positive number, and its reciprocal, f (x), will be a large positive number. So lim f (x) = . x1+(c) It appears from the graph of f that lim f (x) = and lim f (x) = .x1x1+35. (a) Let h(x) = (1 + x)1/x .x 0.001 0.0001 0.00001 0.000001 0.000001 0.00001 0.0001 0.001(b)h(x) 2.71964 2.71842 2.71830 2.71828 2.71828 2.71827 2.71815 2.71692It appears that lim (1 + x)1/x 2.71828, which is approximately e. x0In Section 3.6 we will see that the value of the limit is exactly e. 37. For f (x) = x2 (2x/1000):(a)(b)xf (x)xf (x)1 0.8 0.6 0.4 0.2 0.10.998000 0.638259 0.358484 0.158680 0.038851 0.0089280.04 0.02 0.01 0.005 0.0030.000572 0.000614 0.000907 0.000978 0.0009930.050.001465It appears that lim f (x) = 0. x00.0010.001000It appears that lim f (x) = 0.001. x0 37. 46CHAPTER 2LIMITS AND DERIVATIVES39. No matter how many times we zoom in toward the origin, the graphs of f (x) = sin(/x) appear to consist of almost-verticallines. This indicates more and more frequent oscillations as x 0.There appear to be vertical asymptotes of the curve y = tan(2 sin x) at x 0.9041.and x 2.24. To nd the exact equations of these asymptotes, we note that the graph of the tangent function has vertical asymptotes at x = must have 2 sin x = 2+ n, or equivalently, sin x = 4 2+ n. Thus, we+ n. Since 21 sin x 1, we must have sin x = and so x = sin1 4 4(correspondingto x 0.90). Just as 150 is the reference angle for 30 , sin1 reference angle for sin1 . So x = sin1 4 vertical asymptotes (corresponding to x 2.24).2.3 Calculating Limits Using the Limit Laws 1. (a) lim [f (x) + 5g(x)] = lim f (x) + lim [5g(x)] x2x2x2= lim f (x) + 5 lim g(x) x2x2= 4 + 5(2) = 6 (b) lim [g(x)]3 = x23lim g(x)x2[Limit Law 6]= ( 2)3 = 8 (c) limx2f (x) = =lim f (x)x2 4=2[Limit Law 11][Limit Law 1] [Limit Law 3] 4 4is theare also equations of 38. SECTION 2.3(d) limx2lim [3f (x)] 3f (x) x2 = g(x) lim g(x)CALCULATING LIMITS USING THE LIMIT LAWS47[Limit Law 5]x23 lim f (x) =x2[Limit Law 3]lim g(x)x2=3(4) = 6 2(e) Because the limit of the denominator is 0, we cant use Limit Law 5. The given limit, limx2g(x) , does not exist because the h(x)denominator approaches 0 while the numerator approaches a nonzero number. (f ) limx2lim [g(x) h(x)] g(x) h(x) x2 = f (x) lim f (x)[Limit Law 5]x2=lim g(x) lim h(x)x2x2[Limit Law 4]lim f (x)x2=2 0 =0 4 [Limit Laws 1 and 2]3. lim (3x4 + 2x2 x + 1) = lim 3x4 + lim 2x2 lim x + lim 1 x2x2x2x2x2= 3 lim x4 + 2 lim x2 lim x + lim 1[3]= 3(2)4 + 2(2)2 (2) + (1)[9, 8, and 7]x2x2x2x2= 48 + 8 + 2 + 1 = 59 5. lim (1 + x8 3 x ) (2 6x2 + x3 ) = lim (1 + 3 x ) lim (2 6x2 + x3 ) x8[Limit Law 4]x8 lim 1 + lim 3 x lim 2 6 lim x2 + lim x3 x8 x8 x8 x8 = 1 + 3 8 2 6 82 + 83=x8= (3)(130) = 390 7. limx11 + 3x 1 + 4x2 + 3x43=31 + 3x x1 1 + 4x2 + 3x4[6]lim3lim (1 + 3x)=x1[5]lim (1 + 4x2 + 3x4 )x13lim 1 + 3 lim x=x1lim 1 + 4 limx1= 9. limx4 16 x2 = = =x1x1 x2 + 31 + 3(1) 1 + 4(1)2 + 3(1)4[2, 1, and 3]lim x4x13=4 8lim (16 x2 )=[11]lim 16 lim x23[2]x4 x4x416 (4)2 = 0[7 and 9]1 23=1 8[7, 8, and 9][1, 2, and 3] [7, 10, 9] 39. 48CHAPTER 2LIMITS AND DERIVATIVES11. limx2 + x 6 (x + 3)(x 2) = lim = lim (x + 3) = 2 + 3 = 5 x2 x2 x2 x213. limx2 x + 6 does not exist since x 2 0 but x2 x + 6 8 as x 2. x215. limt2 9 (t + 3)(t 3) t3 3 3 6 6 = lim = lim = = = 2t2 + 7t + 3 t3 (2t + 1)(t + 3) t3 2t + 1 2(3) + 1 5 5x2x2t3(4 + h)2 16 (16 + 8h + h2 ) 16 h(8 + h) 8h + h2 = lim = lim = lim = lim (8 + h) = 8 + 0 = 8 h0 h0 h0 h0 h h h h17. limh019. By the formula for the sum of cubes, we havelimx221. limt9x+2 x+2 1 1 1 = lim = lim = = . x3 + 8 x2 (x + 2)(x2 2x + 4) x2 x2 2x + 4 4+4+4 12 3+ t 3 t 9t = lim 3 t t9 3 t= lim 3 + t9 t =3+ 9=6 x+23 x+23 x+2+3 (x + 2) 9 = lim = lim 23. lim x7 x7 x7 x7 x + 2 + 3 x7 (x 7) x + 2 + 3 = limx725.x7 1 1 1 = lim = = x7 6 (x 7) x + 2 + 3 x+2+3 9+31 x+4 1 + 1 1 x+4 1 x = lim 4x = lim = lim = = lim 4 x4 4 + x x4 4 + x x4 4x(4 + x) x4 4x 4(4) 16 (4 x )(4 + x ) 4 x 16 x 27. lim = lim = lim x16 16x x2 x16 (16x x2 )(4 + x ) x16 x(16 x)(4 + x ) = limx1629. limt01 1 t t 1+t1 1 = x(4 + x ) 16 4 + 16= limt0=1 1 = 16(8) 128 1 1+t 1+ 1+t 1 1+t = lim t0 t 1+t t t+1 1+ 1+t1 = lim t0 1+t 1+ 1+t1 = 1+0 1+ 1+0t = lim t0 t 1+t 1+ 1+t =1 2(b)31. (a)xx 2 lim 3 1 + 3x 1x0f (x)0.001 0.0001 0.00001 0.000001 0.000001 0.00001 0.0001 0.0010.6661663 0.6666167 0.6666617 0.6666662 0.6666672 0.6666717 0.6667167 0.6671663The limit appears to be2 . 3 40. SECTION 2.3(c) limx0CALCULATING LIMITS USING THE LIMIT LAWS49 x 1 + 3x + 1 x 1 + 3x + 1 = lim = lim x0 x0 (1 + 3x) 1 3x 1 + 3x + 1 x 1 + 3x 1 1 + 3x + 1= 1 lim 1 + 3x + 1 3 x0=1 3=1 3[Limit Law 3]lim (1 + 3x) + lim 1[1 and 11]lim 1 + 3 lim x + 1[1, 3, and 7]x0x0x0x01 1+30+1 3 2 1 = (1 + 1) = 3 3[7 and 8]=33. Let f(x) = x2 , g(x) = x2 cos 20x and h(x) = x2 . Then1 cos 20x 1 x2 x2 cos 20x x2 f (x) g(x) h(x).So since lim f (x) = lim h(x) = 0, by the Squeeze Theorem we have x0x0lim g(x) = 0.x035. We have lim (4x 9) = 4(4) 9 = 7 and lim x2 4x + 7 = 42 4(4) + 7 = 7. Since 4x 9 f (x) x2 4x + 7 x4x4for x 0, lim f (x) = 7 by the Squeeze Theorem. x437. 1 cos(2/x) 1 x4 x4 cos(2/x) x4 . Since lim x4 = 0 and lim x4 = 0, we have x0x0lim x cos(2/x) = 0 by the Squeeze Theorem. 4x039. |x 3| =if x 3 0x3if x 3 < 0(x 3)=x3 3xif x 3if x < 3Thus, lim (2x + |x 3|) = lim (2x + x 3) = lim (3x 3) = 3(3) 3 = 6 and x3+x3+x3+lim (2x + |x 3|) = lim (2x + 3 x) = lim (x + 3) = 3 + 3 = 6. Since the left and right limits are equal,x3x3x3lim (2x + |x 3|) = 6.x341. 2x3 x2 = x2 (2x 1) = x2 |2x 1| = x2 |2x 1||2x 1| =2x 1 (2x 1)if 2x 1 0if 2x 1 < 0=2x 1 (2x 1)if x 0.5if x < 0.5So 2x3 x2 = x2 [(2x 1)] for x < 0.5. Thus,limx0.52x 1 2x 1 1 1 1 = lim = lim = 4. = = |2x3 x2 | x0.5 x2 [(2x 1)] x0.5 x2 (0.5)2 0.25 41. 50CHAPTER 2LIMITS AND DERIVATIVES1 1 x |x|43. Since |x| = x for x < 0, we have limx0= limx0denominator approaches 0 and the numerator does not. 45. (a)1 1 x x= limx02 , which does not exist since the x(b) (i) Since sgn x = 1 for x > 0, lim sgn x = lim 1 = 1. x0+x0+(ii) Since sgn x = 1 for x < 0, lim sgn x = lim 1 = 1. x0x0(iii) Since lim sgn x 6= lim sgn x, lim sgn x does not exist. x0x0x0+(iv) Since |sgn x| = 1 for x 6= 0, lim |sgn x| = lim 1 = 1. x047. (a) (i) lim F (x) = lim x1+x1+(ii) lim F (x) = lim x1x1x2 1 x2 1 = lim = lim (x + 1) = 2 |x 1| x1+ x 1 x1+x0(c)x2 1 x2 1 = lim = lim (x + 1) = 2 |x 1| x1 (x 1) x1(b) No, lim F (x) does not exist since lim F (x) 6= lim F (x). x1x1x1+49. (a) (i) [[x]] = 2 for 2 x < 1, sox2+(ii) [[x]] = 3 for 3 x < 2, sox2lim [[x]] = lim [[x]] =lim (2) = 2x2+lim (3) = 3.x2The right and left limits are different, so lim [[x]] does not exist. x2(iii) [[x]] = 3 for 3 x < 2, solim [[x]] =x2.4lim (3) = 3.x2.4(b) (i) [[x]] = n 1 for n 1 x < n, so lim [[x]] = lim (n 1) = n 1. xnxn(ii) [[x]] = n for n x < n + 1, so lim [[x]] = lim n = n. xn+xn+(c) lim [[x]] exists a is not an integer. xa51. The graph of f (x) = [[x]] + [[x]] is the same as the graph of g(x) = 1 with holes at each integer, since f (a) = 0 for anyinteger a. Thus, lim f (x) = 1 and lim f (x) = 1, so lim f (x) = 1. However, x2x2+x2f (2) = [[2]] + [[2]] = 2 + (2) = 0, so lim f (x) 6= f (2). x253. Since p(x) is a polynomial, p(x) = a0 + a1 x + a2 x2 + + an xn . Thus, by the Limit Laws,lim p(x) = lim a0 + a1 x + a2 x2 + + an xn = a0 + a1 lim x + a2 lim x2 + + an lim xnxaxaxa= a0 + a1 a + a2 a2 + + an an = p(a) Thus, for any polynomial p, lim p(x) = p(a). xaxaxa 42. SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT55. lim [f(x) 8] = lim x1x151f (x) 8 f(x) 8 (x 1) = lim lim (x 1) = 10 0 = 0. x1 x1 x1 x1Thus, lim f (x) = lim {[f (x) 8] + 8} = lim [f (x) 8] + lim 8 = 0 + 8 = 8. x1x1Note: The value of limx1x1x1f (x) 8 f (x) 8 does not affect the answer since its multiplied by 0. Whats important is that lim x1 x 1 x1exists. 57. Observe that 0 f (x) x2 for all x, and lim 0 = 0 = lim x2 . So, by the Squeeze Theorem, lim f (x) = 0. x0x0x059. Let f(x) = H(x) and g(x) = 1 H(x), where H is the Heaviside function dened in Exercise 1.3.57.Thus, either f or g is 0 for any value of x. Then lim f (x) and lim g(x) do not exist, but lim [f (x)g(x)] = lim 0 = 0. x0x0x0x061. Since the denominator approaches 0 as x 2, the limit will exist only if the numerator also approaches0 as x 2. In order for this to happen, we need limx23x2 + ax + a + 3 = 0 3(2)2 + a(2) + a + 3 = 0 12 2a + a + 3 = 0 a = 15. With a = 15, the limit becomes limx23(2 + 3) 3(x + 2)(x + 3) 3(x + 3) 3 3x2 + 15x + 18 = lim = lim = = = 1. x2 (x 1)(x + 2) x2 x 1 x2 + x 2 2 1 32.4 The Precise Definition of a Limit 1. On the left side of x = 2, we need |x 2| 0) 17.8412 cm.(b) |A 1000| 5 5 r2 1000 5 1000 5 r2 1000 + 5 995 r1005 17.7966 r 17.8858.1000 995 0.04466 and1005 1000 0.04455. Soif the machinist gets the radius within 0.0445 cm of 17.8412, the area will be within 5 cm2 of 1000. (c) x is the radius, f (x) is the area, a is the target radius given in part (a), L is the target area (1000), is the tolerance in the area (5), and is the tolerance in the radius given in part (b). 13. (a) |4x 8| = 4 |x 2| < 0.1 |x 2| 0 such that if 0 < |x 1| < , then|(2x + 3) 5| < . But |(2x + 3) 5| < |2x 2| < 2 |x 1| < |x 1| < /2. So if we choose = /2, then 0 < |x 1| < |(2x + 3) 5| < . Thus, lim (2x + 3) = 5 by the denition of a limit. x117. Given > 0, we need > 0 such that if 0 < |x (3)| < , then|(1 4x) 13| < . But |(1 4x) 13| < |4x 12| < |4| |x + 3| < |x (3)| < /4. So if we choose = /4, then 0 < |x (3)| < |(1 4x) 13| < . Thus, lim (1 4x) = 13 by the denition of x3xa limit.19. Given > 0, we need > 0 such that if 0 < |x 3| < , thenSo choose = 5. Then 0 < |x 3| < of a limit, limx3x 3 < 5 5 |x 3| < 5 1 5|x 3| < |x 3| < 5.|x 3| < 5x 3 < . By the denition 5 5x 3 = . 5 521. Given > 0, we need > 0 such that if 0 < |x 2| < , then(x + 3)(x 2) 5 < x2 Then 0 < |x 2| < x2 + x 6 5 < x2|x + 3 5| < [x 6= 2] |x 2| < . So choose = . |x 2| < |x + 3 5| < (x + 3)(x 2) 5 < [x 6= 2] x2x2 + x 6 x2 + x 6 5 < . By the denition of a limit, lim = 5. x2 x2 x2 23. Given > 0, we need > 0 such that if 0 < |x a| < , then |x a| < . So = will work. 25. Given > 0, we need > 0 such that if 0 < |x 0| < , then x2 0 < Then 0 < |x 0| < x2 < |x| < . Take = .x2 0 < . Thus, lim x2 = 0 by the denition of a limit. x027. Given > 0, we need > 0 such that if 0 < |x 0| < , then |x| 0 < . But |x| = |x|. So this is true if we pick = .Thus, lim |x| = 0 by the denition of a limit. x0 45. 54CHAPTER 2 LIMITS AND DERIVATIVES29. Given > 0, we need > 0 such that if 0 < |x 2| < , then(x 2)2 < . So take = . Then 0 < |x 2| < x2 4x + 5 1 < |x 2| < x2 4x + 4 < (x 2)2 < . Thus,lim x2 4x + 5 = 1 by the denition of a limit.x231. Given > 0, we need > 0 such that if 0 < |x (2)| < , thenx2 1 3 < or upon simplifying we needx2 4 < whenever 0 < |x + 2| < . Notice that if |x + 2| < 1, then 1 < x + 2 < 1 5 < x 2 < 3 |x 2| < 5. So take = min {/5, 1}. Then 0 < |x + 2| < |x 2| < 5 and |x + 2| < /5, sox2 1 3 = |(x + 2)(x 2)| = |x + 2| |x 2| < (/5)(5) = . Thus, by the denition of a limit, lim (x2 1) = 3. x233. Given > 0, we let = min 2, 8. If 0 < |x 3| < , then |x 3| < 2 2 < x 3 < 2 4 < x + 3 < 8 |x + 3| < 8. Also |x 3| < 8 , so x2 9 = |x + 3| |x 3| < 8 8= . Thus, lim x2 = 9. x335. (a) The points of intersection in the graph are (x1 , 2.6) and (x2 , 3.4)with x1 0.891 and x2 1.093. Thus, we can take to be the smaller of 1 x1 and x2 1. So = x2 1 0.093.(b) Solving x3 + x + 1 = 3 + gives us two nonreal complex roots and one real root, which is 2/3 216 + 108 + 12 336 + 324 + 812 12 x() = . Thus, = x() 1. 1/3 2 6 216 + 108 + 12 336 + 324 + 81 (c) If = 0.4, then x() 1.093 272 342 and = x() 1 0.093, which agrees with our answer in part (a). 37. 1. Guessing a value for Given > 0, we must nd > 0 such that | x a| < whenever 0 < |x a| < . But |x a| < (from the hint). Now if we can nd a positive constant C such that x + a > C then | x a| = x+ a |x a| |x a| < < , and we take |x a| < C. We can nd this number by restricting x to lie in some interval C x+ a centered at a. If |x a| < 1 a, then 1 a < x a < 1 a 2 2 2 C=1 2a = min+1 2a< x < 3a 2 a is a suitable choice for the constant. So |x a| 1 2a x+ a > |x a| < | x a| = x+ a1 a 2+1 a, 21 a 2+ a . If 0 < |x a| < , then a (as in part 1). Also |x a| 0, we let = min+ 46. SECTION 2.5 CONTINUITY39. Suppose that lim f (x) = L. Given = x01 2,there exists > 0 such that 0 < |x| < 55 |f(x) L| < 1 . Take any rational 2number r with 0 < |r| < . Then f (r) = 0, so |0 L| < 1 , so L |L| < 1 . Now take any irrational number s with 2 2 0 < |s| < . Then f (s) = 1, so |1 L| < 1 . Hence, 1 L < 1 , so L > 1 . This contradicts L < 1 , so lim f(x) does not 2 2 2 2 x0exist. 41.1 1 > 10,000 (x + 3)4 < (x + 3)4 10,0001 |x + 3| < 4 10,000|x (3)| 0 so that ln x < M whenever 0 < x < ; that is, x = eln x < eM whenever 0 < x < . Thissuggests that we take = eM . If 0 < x < eM , then ln x < ln eM = M . By the denition of a limit, lim ln x = . x0+2.5 Continuity 1. From Denition 1, lim f (x) = f (4). x43. (a) The following are the numbers at which f is discontinuous and the type of discontinuity at that number: 4 (removable),2 ( jump), 2 ( jump), 4 (innite).(b) f is continuous from the left at 2 sincelim f (x) = f (2). f is continuous from the right at 2 and 4 sincex2lim f(x) = f (2) and lim f (x) = f (4). It is continuous from neither side at 4 since f (4) is undened.x2+x4+5. The graph of y = f (x) must have a discontinuity at x = 3 and must show that lim f(x) = f (3). x37. (a)(b) There are discontinuities at times t = 1, 2, 3, and 4. A person parking in the lot would want to keep in mind that the charge will jump at the beginning of each hour.9. Since f and g are continuous functions,lim [2f (x) g(x)] = 2 lim f (x) lim g(x)x3x3x3= 2f(3) g(3)[by Limit Laws 2 and 3] [by continuity of f and g at x = 3]= 2 5 g(3) = 10 g(3) Since it is given that lim [2f (x) g(x)] = 4, we have 10 g(3) = 4, so g(3) = 6. x3 47. 56CHAPTER 2LIMITS AND DERIVATIVES11. lim f (x) = lim x1x1x + 2x344=lim x + 2 lim x3x1= 1 + 2(1)3x14= (3)4 = 81 = f(1).By the denition of continuity, f is continuous at a = 1. 13. For a > 2, we havelim (2x + 3) 2x + 3 xa = xa x 2 lim (x 2)lim f (x) = limxa[Limit Law 5]xa2 lim x + lim 3 =xaxalim x lim 2xa=[1, 2, and 3]xa2a + 3 a2[7 and 8]= f (a)Thus, f is continuous at x = a for every a in (2, ); that is, f is continuous on (2, ). 15. f (x) = ln |x 2| is discontinuous at 2 since f(2) = ln 0 is not dened.17. f (x) =exif x < 02if x 0xThe left-hand limit of f at a = 0 is lim f (x) = lim ex = 1. The x0x0right-hand limit of f at a = 0 is lim f (x) = lim x2 = 0. Since these x0+x0+limits are not equal, lim f (x) does not exist and f is discontinuous at 0. x0 cos x 19. f (x) = 0 1 x2if x < 0 if x = 0 if x > 0lim f (x) = 1, but f (0) = 0 6= 1, so f is discontinuous at 0.x021. F (x) =x is a rational function. So by Theorem 5 (or Theorem 7), F is continuous at every number in its domain, x2 + 5x + 6x | x2 + 5x + 6 6= 0 = {x | (x + 3)(x + 2) 6= 0} = {x | x 6= 3, 2} or (, 3) (3, 2) (2, ). 23. By Theorem 5, the polynomials x2 and 2x 1 are continuous on (, ). By Theorem 7, the root function continuous on [0, ). By Theorem 9, the composite function 2x 1 is continuous on its domain, [ 1 , ). 2 2 1 By part 1 of Theorem 4, the sum R(x) = x + 2x 1 is continuous on [ 2 , ). x is 48. SECTION 2.5 CONTINUITY5725. By Theorem 7, the exponential function e5t and the trigonometric function cos 2t are continuous on (, ).By part 4 of Theorem 4, L(t) = e5t cos 2t is continuous on (, ). 27. By Theorem 5, the polynomial t4 1 is continuous on (, ). By Theorem 7, ln x is continuous on its domain, (0, ).By Theorem 9, ln t4 1 is continuous on its domain, which is t | t4 1 > 0 = t | t4 > 1 = {t | |t| > 1} = (, 1) (1, ) 29. The function y =1 is discontinuous at x = 0 because the 1 + e1/xleft- and right-hand limits at x = 0 are different.31. Because we are dealing with root functions, 5 + x is continuous on [0, ), x + 5 is continuous on [5, ), so the 5+ x is continuous on [0, ). Since f is continuous at x = 4, lim f (x) = f (4) = 7 . quotient f (x) = 3 x4 5+x 33. Because x2 x is continuous on R, the composite function f (x) = ex2 xis continuous on R, solim f (x) = f (1) = e1 1 = e0 = 1.x135. f (x) =x2 xif x < 1 if x 1By Theorem 5, since f (x) equals the polynomial x2 on (, 1), f is continuous on (, 1). By Theorem 7, since f (x) equals the root function x on (1, ), f is continuous on (1, ). At x = 1, lim f(x) = lim x2 = 1 and x1x1 lim f (x) = lim x = 1. Thus, lim f(x) exists and equals 1. Also, f (1) = 1 = 1. Thus, f is continuous at x = 1.x1+x1x1+We conclude that f is continuous on (, ). 1 + x2 37. f (x) = 2 x (x 2)2if x 0if 0 < x 2 if x > 2f is continuous on (, 0), (0, 2), and (2, ) since it is a polynomial on each of these intervals. Now lim f (x) = lim (1 + x2 ) = 1 and lim f (x) = lim (2 x) = 2, so f is x0x0x0+x0+discontinuous at 0. Since f (0) = 1, f is continuous from the left at 0. Also, lim f (x) = lim (2 x) = 0, x2x2lim f (x) = lim (x 2)2 = 0, and f (2) = 0, so f is continuous at 2. The only number at which f is discontinuous is 0.x2+x2+ 49. 58CHAPTER 2LIMITS AND DERIVATIVES x + 2 39. f (x) = ex 2xif x < 0 if 0 x 1 if x > 1f is continuous on (, 0) and (1, ) since on each of these intervalsit is a polynomial; it is continuous on (0, 1) since it is an exponential. Now lim f (x) = lim (x + 2) = 2 and lim f (x) = lim ex = 1, so f is discontinuous at 0. Since f (0) = 1, f is x0x0x0+x0+continuous from the right at 0. Also lim f (x) = lim ex = e and lim f (x) = lim (2 x) = 1, so f is discontinuous x1x1x1+x1+at 1. Since f (1) = e, f is continuous from the left at 1. 41. f (x) =cx2 + 2x 3x cxif x < 2 if x 2f is continuous on (, 2) and (2, ). Now lim f (x) = lim x2lim f (x) = limx2+x2+x2cx2 + 2x = 4c + 4 andx3 cx = 8 2c. So f is continuous 4c + 4 = 8 2c 6c = 4 c = 2 . Thus, for f 3to be continuous on (, ), c = 2 . 3 43. (a) f (x) =(x2 + 1)(x2 1) (x2 + 1)(x + 1)(x 1) x4 1 = = = (x2 + 1)(x + 1) [or x3 + x2 + x + 1] x1 x1 x1for x 6= 1. The discontinuity is removable and g(x) = x3 + x2 + x + 1 agrees with f for x 6= 1 and is continuous on R. (b) f (x) =x(x2 x 2) x(x 2)(x + 1) x3 x2 2x = = = x(x + 1) [or x2 + x] for x 6= 2. The discontinuity x2 x2 x2is removable and g(x) = x2 + x agrees with f for x 6= 2 and is continuous on R. (c) lim f (x) = lim [[sin x]] = lim 0 = 0 and lim f (x) = lim [[sin x]] = lim (1) = 1, so lim f (x) does not x x x x +x +x +xexist. The discontinuity at x = is a jump discontinuity. 45. f (x) = x2 + 10 sin x is continuous on the interval [31, 32], f (31) 957, and f(32) 1030. Since 957 < 1000 < 1030,there is a number c in (31, 32) such that f (c) = 1000 by the Intermediate Value Theorem. Note: There is also a number c in (32, 31) such that f (c) = 1000. 47. f (x) = x4 + x 3 is continuous on the interval [1, 2], f (1) = 1, and f(2) = 15. Since 1 < 0 < 15, there is a number cin (1, 2) such that f (c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation x4 + x 3 = 0 in the interval (1, 2). 49. f (x) = cos x x is continuous on the interval [0, 1], f (0) = 1, and f(1) = cos 1 1 0.46. Since 0.46 < 0 < 1, thereis a number c in (0, 1) such that f (c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos x x = 0, or cos x = x, in the interval (0, 1). 50. SECTION 2.5 CONTINUITY5951. (a) f (x) = cos x x3 is continuous on the interval [0, 1], f (0) = 1 > 0, and f (1) = cos 1 1 0.46 < 0. Since1 > 0 > 0.46, there is a number c in (0, 1) such that f (c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos x x3 = 0, or cos x = x3 , in the interval (0, 1). (b) f (0.86) 0.016 > 0 and f (0.87) 0.014 < 0, so there is a root between 0.86 and 0.87, that is, in the interval (0.86, 0.87). 53. (a) Let f (x) = 100ex/100 0.01x2 . Then f (0) = 100 > 0 andf (100) = 100e1 100 63.2 < 0. So by the Intermediate Value Theorem, there is a number c in (0, 100) such that f (c) = 0. This implies that 100ec/100 = 0.01c2 . (b) Using the intersect feature of the graphing device, we nd that the root of the equation is x = 70.347, correct to three decimal places. 55. () If f is continuous at a, then by Theorem 8 with g(h) = a + h, we havelim f (a + h) = f lim (a + h) = f (a).h0h0() Let > 0. Since lim f(a + h) = f (a), there exists > 0 such that 0 < |h| < h0|f (a + h) f (a)| < . So if 0 < |x a| < , then |f (x) f (a)| = |f (a + (x a)) f (a)| < . Thus, lim f (x) = f (a) and so f is continuous at a. xa57. As in the previous exercise, we must show that lim cos(a + h) = cos a to prove that the cosine function is continuous. h0lim cos(a + h) = lim (cos a cos h sin a sin h) = lim (cos a cos h) lim (sin a sin h)h0h0=59. f (x) =lim cos ah00 if x is rational 1 if x is irrationalh0lim cos h lim sin ah0h0h0lim sin h = (cos a)(1) (sin a)(0) = cos ah0is continuous nowhere. For, given any number a and any > 0, the interval (a , a + )contains both innitely many rational and innitely many irrational numbers. Since f (a) = 0 or 1, there are innitely many numbers x with 0 < |x a| < and |f (x) f (a)| = 1. Thus, lim f(x) 6= f (a). [In fact, lim f (x) does not even exist.] xa61. If there is such a number, it satises the equation x3 + 1 = xxa x3 x + 1 = 0. Let the left-hand side of this equation becalled f (x). Now f (2) = 5 < 0, and f (1) = 1 > 0. Note also that f (x) is a polynomial, and thus continuous. So by the Intermediate Value Theorem, there is a number c between 2 and 1 such that f(c) = 0, so that c = c3 + 1. 63. f (x) = x4 sin(1/x) is continuous on (, 0) (0, ) since it is the product of a polynomial and a composite of atrigonometric function and a rational function. Now since 1 sin(1/x) 1, we have x4 x4 sin(1/x) x4 . Because 51. 60CHAPTER 2 LIMITS AND DERIVATIVESlim (x4 ) = 0 and lim x4 = 0, the Squeeze Theorem gives us lim (x4 sin(1/x)) = 0, which equals f(0). Thus, f isx0x0x0continuous at 0 and, hence, on (, ). 65. Dene u(t) to be the monks distance from the monastery, as a function of time, on the rst day, and dene d(t) to be hisdistance from the monastery, as a function of time, on the second day. Let D be the distance from the monastery to the top of the mountain. From the given information we know that u(0) = 0, u(12) = D, d(0) = D and d(12) = 0. Now consider the function u d, which is clearly continuous. We calculate that (u d)(0) = D and (u d)(12) = D. So by the Intermediate Value Theorem, there must be some time t0 between 0 and 12 such that (u d)(t0 ) = 0 u(t0 ) = d(t0 ). So at time t0 after 7:00 AM, the monk will be at the same place on both days.2.6 Limits at Infinity; Horizontal Asymptotes 1. (a) As x becomes large, the values of f (x) approach 5.(b) As x becomes large negative, the values of f (x) approach 3. 3. (a) lim f (x) = x2(d) lim f (x) = 1 x5. f (0) = 0,(b)lim f (x) = x1(e) lim f (x) = 2f(1) = 1,lim f (x) = 0,xx7. lim f (x) = , x2lim f (x) = 0,xf is oddlim f (x) = x0(c)lim f (x) = x1+(f ) Vertical: x = 1, x = 2; Horizontal: y = 1, y = 2 lim f(x) = ,xlim f (x) = ,x0+9. f (0) = 3,lim f (x) = 4,x0lim f (x) = 2,x0+lim f (x) = ,xlim f (x) = ,x4+lim f (x) = ,x4lim f (x) = 3x11. If f (x) = x2/2x , then a calculator gives f(0) = 0, f (1) = 0.5, f(2) = 1, f (3) = 1.125, f (4) = 1, f (5) = 0.78125,f (6) = 0.5625, f (7) = 0.3828125, f (8) = 0.25, f (9) = 0.158203125, f(10) = 0.09765625, f (20) 0.00038147, f (50) 2.2204 1012 , f (100) 7.8886 1027 . It appears that lim x2/2x = 0. x 52. SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES13. limx3x2 x + 4 (3x2 x + 4)/x2 = lim 2 + 5x 8 x (2x2 + 5x 8)/x2 2x=[divide both the numerator and denominator by x2 (the highest power of x thatappears in the denominator)]lim (3 1/x + 4/x2 )x[Limit Law 5]lim (2 + 5/x 8/x2 )x=lim 3 lim (1/x) + lim (4/x2 )xx=xxx3 lim (1/x) + 4 lim (1/x2 ) xxx=3 0 + 4(0) 2 + 5(0) 8(0)=x3 2[Theorem 5 of Section 2.5]lim (1/x) lim (1/x) 1 1/x 0 0 x x = lim = = = = =0 2x + 3 x (2x + 3)/x lim (2 + 3/x) lim 2 + 3 lim (1/x) 2 + 3(0) 2 x17.lim[Limit Laws 7 and 3]2 + 5 lim (1/x) 8 lim (1/x2 ) x15. lim[Limit Laws 1 and 2]lim 2 + lim (5/x) lim (8/x2 )xxx(1 x x2 )/x2 1 x x2 = lim = 2 7 x (2x2 7)/x2 2x =xlim (1/x2 1/x 1) lim (2 7/x2 )xxlim 2 7 lim (1/x2 )xxxxlim (1/x2 ) lim (1/x) lim 1x61=001 1 = 2 7(0) 219. Divide both the numerator and denominator by x3 (the highest power of x that occurs in the denominator).5 x3 + 5x 5 lim 1 + 2 1+ 2 x 3 x x3 + 5x x x lim = lim = lim = 4 1 x 2x3 x2 + 4 x 2x3 x2 + 4 x 4 1 2 + 3 lim 2 + 3 x x x x x x3 1 lim 1 + 5 lim 2 1 1 + 5(0) x x x = = = 1 1 2 0 + 4(0) 2 + 4 lim 3 lim 2 lim x x x x x 21. First, multiply the factors in the denominator. Then divide both the numerator and denominator by u4 .5 4u4 + 5 4+ 4 4u4 + 5 4u4 + 5 u4 u = lim = lim lim = lim 5 2 u (u2 2)(2u2 1) u 2u4 5u2 + 2 u 2u4 5u2 + 2 u 2 2 + 4 u u u4 5 1 lim 4 + 4 lim 4 + 5 lim 4 u u 4 4 + 5(0) u u u = =2 = = = 1 1 2 5(0) + 2(0) 2 2 5 lim 2 5 lim 2 + 2 lim 4 lim 2 2 + 4 u u u u u u u u 53. 62CHAPTER 2 LIMITS AND DERIVATIVES (9x6 x)/x6 lim 9x6 x 9x6 x /x3 x = lim = x x (x3 + 1)/x3 x3 + 1 lim (1 + 1/x3 )23. lim[since x3 = x6 for x > 0]x9 1/x5limx=lim 1 + lim (1/x3 ) x x = 90 =3 25. lim 9x2 + x 3x = lim xxlim 9 lim (1/x5 )x=x1+0 2 9x2 + x 3x 9x2 + x + 3x 9x2 + x (3x)2 = lim x 9x2 + x + 3x 9x2 + x + 3x9x2 + x 9x2 1/x x = lim = lim x x 9x2 + x + 3x 9x2 + x + 3x 1/x x/x 1 1 1 1 = lim = = = lim = x 3+3 6 9+3 9x2 /x2 + x/x2 + 3x/x x 9 + 1/x + 3 x2 + ax x2 + bx = lim27. limxx x2 + ax x2 + bx x2 + ax + x2 + bx x2 + ax + x2 + bx(x2 + ax) (x2 + bx) [(a b)x]/x = lim = lim 2 + ax + 2 + bx 2 + ax + x x x x x x2 + bx / x2 = limx29. limxab ab ab = = 2 1+0+ 1+0 1 + a/x + 1 + b/xx + x3 + x5 (x + x3 + x5 )/x4 = lim x (1 x2 + x4 )/x4 1 x2 + x4 = limx[divide by the highest power of x in the denominator]1/x3 + 1/x + x = 1/x4 1/x2 + 1because (1/x3 + 1/x + x) and (1/x4 1/x2 + 1) 1 as x . 31.1 lim (x4 + x5 ) = lim x5 ( x + 1) [factor out the largest power of x] = because x5 and 1/x + 1 1xxas x . Or:limx33. limxx4 + x5 = lim x4 (1 + x) = . x1 ex (1 ex )/ex 1/ex 1 01 1 = = = lim = lim x (1 + 2ex )/ex x 1/ex + 2 1 + 2ex 0+2 235. Since 1 cos x 1 and e2x > 0, we have e2x e2x cos x e2x . We know that lim (e2x ) = 0 and x2xlim ex2x= 0, so by the Squeeze Theorem, lim (e xcos x) = 0. (b)37. (a)xf (x)10,0000.49996251,000,0000.4999996100,000From the graph of f (x) = x2 + x + 1 + x, we estimatethe value of lim f (x) to be 0.5. x0.4999962From the table, we estimate the limit to be 0.5. 54. SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES(c) limx x2 + x + 1 x2 x2 + x + 1 x x2 + x + 1 + x = lim x2 + x + 1 + x = lim x x x2 + x + 1 x x2 + x + 1 x (x + 1)(1/x) 1 + (1/x) = lim = lim x x2 + x + 1 x (1/x) x 1 + (1/x) + (1/x2 ) 1 1+0 1 = = 2 1+0+01 Note that for x < 0, we have x2 = |x| = x, so when we divide the radical by x, with x < 0, we get 1 2 1 2 x + x + 1 = x + x + 1 = 1 + (1/x) + (1/x2 ). x x2 2x + 1 2+ 2x + 1 x = lim = lim 39. lim x x 2 x x 2 x 1 x =1 1 lim 2 + x x = x 2 2 lim 1 x x x1 x = 2 lim 1 lim x x x lim 2 + limxx2+0 = 2, so y = 2 is a horizontal asymptote. 10The denominator x 2 is zero when x = 2 and the numerator is not zero, so we investigate y = f (x) =2x + 1 as x approaches 2. x2lim f (x) = because asx2x 2 the numerator is positive and the denominator approaches 0 throughnegative values. Similarly, lim f (x) = . Thus, x = 2 is a vertical asymptote. x2+The graph conrms our work. 1 1 2x2 + x 1 1 1 lim 2 + 2 2+ 2 x 2 x x 2x2 + x 1 x x x = = lim 41. lim = lim 2 1 x x2 + x 2 x x2 + x 2 x 2 1 1+ 2 lim 1 + 2 x x x x x x2 1 1 lim 2 lim 2 + lim 2+00 x x x x x = 2, so y = 2 is a horizontal asymptote. = = 1 1 1 + 0 2(0) 2 lim 2 lim 1 + lim x x x x x y = f (x) =(2x 1)(x + 1) 2x2 + x 1 = , so lim f(x) = , x2 + x 2 (x + 2)(x 1) x2lim f (x) = , lim f (x) = , and lim f (x) = . Thus, x = 2 x1x2+x1+and x = 1 are vertical asymptotes. The graph conrms our work.43. y = f (x) =x2x(x2 1) x(x + 1)(x 1) x(x + 1) x3 x = = = = g(x) for x 6= 1. 6x + 5 (x 1)(x 5) (x 1)(x 5) x5The graph of g is the same as the graph of f with the exception of a hole in the graph of f at x = 1. By long division, g(x) =x2 + x 30 =x+6+ . x5 x5As x , g(x) , so there is no horizontal asymptote. The denominator of g is zero when x = 5. lim g(x) = and lim g(x) = , so x = 5 is a x5x5+vertical asymptote. The graph conrms our work.63 55. 64CHAPTER 2 LIMITS AND DERIVATIVES45. From the graph, it appears y = 1 is a horizontal asymptote.3x3 + 500x2 3x3 + 500x2 3 + (500/x) x3 = lim 3 = lim lim x x3 + 500x2 + 100x + 2000 x x + 500x2 + 100x + 2000 x 1 + (500/x) + (100/x2 ) + (2000/x3 ) x3 3+0 = 3, so y = 3 is a horizontal asymptote. = 1+0+0+0 The discrepancy can be explained by the choice of the viewing window. Try [100,000, 100,000] by [1, 4] to get a graph that lends credibility to our calculation that y = 3 is a horizontal asymptote.47. Lets look for a rational function.(1)lim f(x) = 0 degree of numerator < degree of denominatorx(2) lim f (x) = there is a factor of x2 in the denominator (not just x, since that would produce a sign x0change at x = 0), and the function is negative near x = 0. (3) lim f (x) = and lim f(x) = vertical asymptote at x = 3; there is a factor of (x 3) in the x3x3+denominator. (4) f (2) = 0 2 is an x-intercept; there is at least one factor of (x 2) in the numerator. Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us f (x) =2x as one possibility. x2 (x 3)49. y = f (x) = x4 x6 = x4 (1 x2 ) = x4 (1 + x)(1 x). The y-intercept isf (0) = 0. The x-intercepts are 0, 1, and 1 [found by solving f (x) = 0 for x]. Since x4 > 0 for x 6= 0, f doesnt change sign at x = 0. The function does change sign at x = 1 and x = 1. As x , f (x) = x4 (1 x2 ) approaches because x4 and (1 x2 ) . 51. y = f (x) = (3 x)(1 + x)2 (1 x)4 . The y-intercept is f (0) = 3(1)2 (1)4 = 3.The x-intercepts are 3, 1, and 1. There is a sign change at 3, but not at 1 and 1. When x is large positive, 3 x is negative and the other factors are positive, so lim f (x) = . When x is large negative, 3 x is positive, soxlim f(x) = .xsin x 1 1 for x > 0. As x , 1/x 0 and 1/x 0, so by the Squeeze x x x sin x = 0. Theorem, (sin x)/x 0. Thus, lim x x53. (a) Since 1 sin x 1 for all x, 56. SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES(b) From part (a), the horizontal asymptote is y = 0. The function y = (sin x)/x crosses the horizontal asymptote whenever sin x = 0; that is, at x = n for every integer n. Thus, the graph crosses the asymptote an innite number of times.55. Divide the numerator and the denominator by the highest power of x in Q(x).(a) If deg P < deg Q, then the numerator 0 but the denominator doesnt. So lim [P (x)/Q(x)] = 0. x(b) If deg P > deg Q, then the numerator but the denominator doesnt, so lim [P (x)/Q(x)] = x(depending on the ratio of the leading coefcients of P and Q). 5 x 1/ x = lim 57. lim x x x 1 1/ x limx5 5 = = 5 and 10 1 (1/x) 10ex 21 1/ex 10 (21/ex ) 10 0 10ex 21 5 x = = 5. Since , = lim < f (x) < x 2ex 1/ex 2 2 2ex x1we have lim f(x) = 5 by the Squeeze Theorem. x59. (a) lim v(t) = lim v 1 egt/v tt= v (1 0) = v (b) We graph v(t) = 1 e9.8t and v(t) = 0.99v , or in this case, v(t) = 0.99. Using an intersect feature or zooming in on the point of intersection, we nd that t 0.47 s.61. Let g(x) =3x2 + 1 and f (x) = |g(x) 1.5|. Note that 2x2 + x + 1lim g(x) =x3 2and lim f(x) = 0. We are interested in nding the xx-value at which f (x) < 0.05. From the graph, we nd that x 14.804, so we choose N = 15 (or any larger number). 63. For = 0.5, we need to nd N such that2.5 < 4x2 + 1 (2) < 0.5 x+1 4x2 + 1 < 1.5 whenever x N. We graph the three parts of this x+1inequality on the same screen, and see that the inequality holds for x 6. So we choose N = 6 (or any smaller number). 4x2 + 1 For = 0.1, we need 2.1 < < 1.9 whenever x N. From the x+1 graph, it seems that this inequality holds for x 22. So we choose N = 22 (or any smaller number).65 57. 66