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Worksheet
We can usually recognize when a function is increasing or decreasing by looking at the graph of its
derivative. We can obtain information about the behavior of a given function f by studying the behavior
of the different derivatives of that function. If f is a given function, then its derivative f (also called
the first derivative of the function) is also a function.
The first derivative f provides information about the steepness, or the slope of a tangent line to the
curve at any instant. If f is positive, then the function f must be increasing. This means that the
function f is changing by having some quantity added to it so that f is actually growing. Similarly, if f
is negative, then the function f must be decreasing. This means that f is changing by having some
quantity subtracted from it so that f is actually shrinking.
Increasing function
If we wish to find the range of values of x for which a function is increasing, just find the range of
values of x where the derivative is positive.
That is, find the range of values of x for which
)( x f > 0.
Decreasing function
If we wish to find the range of values of x for which a function is decreasing, just find the range of
values of x where the derivative is negative.
That is, find the range of values of x for which
)( x f < 0.
Maximum and minimum points of a graph
A positive derivative indicates that the function is increasing.
A negative derivative indicates that the function is decreasing.
In this section, we will learn about how to use of the derivative of a function to detect whether a
function is increasing or decreasing. In addition, we learn how to find the absolute maximum,absolute minimum, local maximum and local minimum of a function.
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o A function f ( x) is said to be an increasing function on the interval (a, b) if for x1 < x2,f( x1) < f( x2) for any two numbers x1 and x2. At a point where a curve is rising, the slope of the
tangent line is positive.
o A function f (x) is said to be a decreasing function on the interval (a, b) if for x1 < x2, f(x1) >
f(x2) for any two numbers x1 and x2. At a point where a curve is falling, the slope of the tangentline is negative.
HORIZONTAL TANGENT LINE
If )( x f = 0, then the function is said to be stationary or constant at that point. At this point, the
slope of the curve is zero.
A tangent line to the curve at this point is parallel to the x-axis and it is called a horizontal tangent line. The curve is neither increasing nor decreasing at this point.
The slope of a curve at a point is the same as the slope of the tangent line to the curve at thatpoint. Similarly, the slope of a curve at a point is the same as the instantaneous rate at which the
height of the curve is changing at that point.
In general, if f is a function that is continuous on the closed interval [a, b], and differentiable on the
open interval (a, b), then we can conclude that:
1. If )( x f > 0 for all x in (a, b), then f is increasing on the interval [a, b].
2. If )( x f < 0 for all x in (a, b), then f is decreasing on the interval [a, b].
3. If )( x f = 0 for all x in (a, b), then f is constant on the interval [a, b]
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THE SLOPE OF A CURVE
In general, the slope of the tangent line is the same as the slope of the curve at the point of
tangency. Therefore to find the slope of a curve at a point, just find the instantaneous rate of
change of the function that the curve represents and substitute the value of the independent
variable into the derivative.
Assume y = f(x) is the equation of the curve, then the slope m at the point (x 1 , y1), where (x1 , y1)
is a point on the curve, is given by
m = )(1
x f or m =1 x x
dx
dy
EXAMPLE 1
Find the slope of the curve )( x f = x4
+ 2 x3 – x + 3 at the point x = -1.
Determine where the curve is rising or falling as it passes through this point.
Solution
We must first obtain the derivative of the function using the following format:
Therefore,
o The slope of the curve at the point (-1, 3) is 9.
o The curve is rising as it passes through that point.
x )( x f )( x f
x x4
+ 2 x3 – x + 3 4 x
3+ 6 x
2 – 1
-1 f(-1) = (-1)4 + 2(-1)3 – (-1)+ 3
= 3)1( f = 4(-1)
3+ 6(-1)
2 – 1
= 9
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Many of our applications of derivatives can be reduced to finding minimum and maximum
values of a function.
We will begin by differentiating between two types of minimum or maximum values, namely:o absolute (or global) and
o relative (or local).We call the minimum and maximum points of a function the extrema of the function.
The Absolute Extrema
An absolute maximum or absolute minimum point is a point that is higher or lower
than all the possible values of the function.
Absolute Maximum
Suppose y is a continuous function of x given by y = f( x), defined over some given
interval [a, b], then there will be some point c in the interval such that
f(c ) > f( x) for all x on the interval [a, b].
The value of f(c) is called the absolute maximum on [a, b]. We say the function has an
absolute maximum at x = c.
Absolute Minimum
Suppose y is a continuous function of x given by y = f( x), defined over some given
interval [a, b], then there will be some point c in the interval such that
f(c ) < f( x) for all x on the interval [a, b].
The value of f(c) is called the absolute minimum on [a, b]. We say the function has anabsolute minimum at x = c.
The maximum and minimum points of a graph
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The Relative Extrema
On the other hand, when there are other points in the domain of the function we are working on,
other than the absolute maximum or minimum, which are higher or lower than the points
surrounding them, we call the higher or lower value relative (or local) maximum or minimum.
Relative Maximum
The relative (or local) maximum point of a function is the point of the function that ishigher than the neighboring points just to the right or to the left of the graph.
We say that f(x) has a relative (or local) maximum at x=c
if f(c ) > f( x) for every x in some open interval around x=c.
Relative Minimum
The relative (or local) minimum point of a function is the point of the function that islower than the neighboring points just to the right or to the left of the graph. We say that f(x) has a relative (or local) maximum at x=c
if f(c ) < f( x) for every x in some open interval around x=c.
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ANOTHER GRAPHICAL LOOK AT EXTREMES
SITUATION WHERE THERE IS A RELATIVE MAXIMUM OR MINIMUM
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ANOTHER SITUATION WHERE THERE IS A RELATIVE MAXIMUM OR MIMMUM
It is also possible that )( x f does
not exist but there may be a
maximum or a minimum.
This is when there is no jumps inthe graph but the graph changes
suddenly from steep downhill to
a steep uphill or vice versawithout being flat for a moment.
An example of such a graph is
given here.
SITUATION WHERE THERE IS NO RELATIVE MAXIMUM OR MINIMUM
It is also possible that a graph may have no maximum or minimum points. If the slope of thecurve jumps from positive to negative or vice versa without the curve being flat for a moment,
then the slope of the function at that point would be undefined . From the diagram below, as x
passes through the value of zero, )( x f changes from negative to positive. However, both
)( x f and )( x f can never take on x = 0 so that it is no minimum point. Also the curve is never
flat.
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To find the maximum or minimum points of a curve, first find all the critical points of the
function
To find the critical points of a function:
First, find the first derivative of the function. Then,
a) Find all values c such that )(c f = 0. If the derivative is a rational function, set numerator
to zero and solve for x.
b) Find all values c for which the derivative does not exist but f(c) is defined. To do this, set
the denominator of the derivative to zero and solve for x.
Secondly, find the maximum or minimum.
Let c-
(any number smaller than c) be a test point to the left of c and let c+
(any number largerthan c) be a test point to the right of c. Then,
a) f(c) is a maximum if )(
c f > 0 (increasing), )(
c f < 0 (decreasing), and )(c f is defined at
x = c.
In general, if the sign of )( x f changes from negative to positive at c, and
)(c f is defined at c, then f(x) has a relative (or local ) minimum at x = c.
b) f(c) is a minimum if )(
c f < 0 (decreasing), )(
c f > 0 (increasing), and )(c f is defined at
x = c.If the sign of )( x f changes from positive to negative at c, and )(c f is
defined at c, then f(x) has a relative (or local ) maximum at x = c.
If the critical value cannot be found in the domain of the original function, then that particular
critical value cannot be classified as a maximum or a minimum. Maximum and minimum valuesmust always exist in the domain of the original function.
The critical points of a function f are the points in the domain of the f where either
a) the tangent line to the curve is horizontal, that is, )( x f = 0, or
b) the points at which )( x f does not exist.
CRITICAL POINTS OF A FUNCTION
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EXAMPLE 2
Given the function f(x) = 27 x - x3, - 7 < x < 7.
Find (a) the critical values and
(b) determine the intervals for which the function is increasing and decreasing.Find (c) the relative maximum and minimum values.
Also, find (d) the absolute maximum and minimum values.
Solution We must first obtain the derivative of the function using the following format:
Determine the signs of f
At this point, we have several x-values: the endpoints (x = -7 and x = 7), the critical values(x = -3 and x = 3). Arrange these values in order from smallest to largest .
(If there are no endpoints, then arrange starting from the lowest endpoint (-infinity) to the highest
endpoint (+infinity)).
x )( x f )( x f
x f(x) = 27 x - x3 )( x f = 27 - 3x2
)( x f = 3(9 - x2)
)( x f = 3(3 - x )(3 + x)
Critical values:
a) Set )( x f = 0 and solve for x.
3(3 - x )(3 + x) = 0
x = - 3 or x = 3
b) There are no points at which
)( x f does not exist.
Endpoint
- 7
f(-7) = 27(-7) – (-7)3
= 154
Endpoint
7
f(7) = 27(7) – (7)3
= -154
Critical Value
3
f(3) = 27(3) – (3)3
= 54
Critical Value
-3
f(-3) = 27(-3) – (-3)3
= -54
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Arrange values in order from smallest to largest:
x = -7 (endpt) -3(crit) 3(crit) 7(endpt)
Use the arranged values above and test points to determine the signs of f and by creating a
table such as the example below.
Thus, the function is:
o increasing on (-3, 3) and decreasing on [-7, -3) (3, 7].
Also, the function f has:
o local minimum at the point (-3, -54), local maximum at (3, 54),
o absolute maximum at (-7, 154) and absolute minimum at (7, -154).
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In this section, we will learn about how to use of the second derivative of a function to detectintervals for which the function is concave upward or concave downward. In addition, we
learn how to find the absolute maximum, absolute minimum, local maximum and local
minimum of a function using the second derivative.
So far we know that the first derivative f provides information about the steepness, or the slope
of a tangent line to the curve at any instant. What information does the second derivative
f provides?
Suppose f is a differentiable function, then, when we take the derivative of the first derivative, we
obtain a new function, f , called the second derivative of f . The second derivative function f ,
can help us to gain information about the behavior of the first derivative function, f .
The second derivative can tell us more about whether the first derivative f is increasing (i.e.,
f > 0), whether the first derivative f is decreasing (i.e., f < 0), or whether the first derivative
f is constant (i.e., f = 0). Thus, f is the rate of growth of f .
If f is positive, then the function f must be increasing. This means that the function f is
changing by having some quantity added to it so that f is actually growing. A differentiable
function is concave upward on an interval if its derivative f is increasing on that interval (that
is, if f > 0).
Similarly, if f is negative, then the function f must be decreasing. This means that f is
changing by having some quantity subtracted from it so that f is actually shrinking. A
differentiable function is concave downward on an interval if its derivative f is decreasing on
that interval (that is, if f < 0).
Concave upward
If we wish to find the range of values of x for which a function is concave upward, just find the
range of values of x where the second derivative f is positive.
That is, find the range of values of x for which
)('' x f > 0.
Using the second derivative to determine concavity
A positive second derivative indicates that the first derivative function is increasing.
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Concave downward
If we wish to find the range of values of x for which a function is concave downward, just find the
range of values of x where the second derivative f is negative.
That is, find the range of values of x for which
)('' x f < 0.
Graphical illustration
Lets now explore the meaning of the phrase concave upward and concave downward. To do this
lets consider the following diagrams.
In both Figure 1 and Figure 2, f is decreasing for every x in the given interval (a, b) since the
first derivative, which is the same as the slope of the tangent line, decreases as we move from left
to right. In figure 1, for instance, the slope of the tangent line to the curve decreases from higher
positive values towards zero, and in figure 2, the slope decreases from zero towards highernegative values.
A negative second derivative indicates that the first derivative function is decreasing.
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In both Figure 3 and Figure 4, f is increasing for every x in the given interval (a, b) since the
first derivative, which is the same as the slope of the tangent line, increases as we move from leftto right. In figure 3, for instance, the slope of the tangent line to the curve increases from highernegative values towards zero, and in figure 4, the slope increases from zero towards higher
positive values.
An inflection point
A point P on a curve is called an inflection point if f is continuous there and the curve changes
from concave upward to concave downward or from concave downward to concave upward.
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TEST FOR CONCAVITY
Clearly, we can lean about the concavity of a given function by studying the behavior of its first
derivative function, f . We now know that graph is concave down if f is decreasing (i.e.,
f < 0), on the interval and a graph is concave up if f is increasing (i.e., f > 0), on the
interval. Thus, in order to identify the intervals for which a graph is concave down or concave up,it is easier to use the second derivative as the test for concavity.
SECOND DERIVATIVE TEST FOR MAXIMUM AND MINIMUM VALUES
For a given function, at the point where the tangent line is horizontal, a relative maximum occurs
where the curve is concave down, and a relative minimum occurs where the curve is concave up.We can therefore use the sign of the second derivative to determine relative maximum orminimum at the point where the tangent line is horizontal.
o If )(c f = 0 and )(c f > 0, then f has a local or relative minimum at c.
o If )(c f = 0 and )(c f < 0, then f has a local or relative maximum at c.
Suppose f is a function whose second derivative f exists on an open interval I .
a) If f ( x) < 0 for every x in the open interval I , then the graph of f is concave
downward on that interval.
b) If f ( x) > 0 for every x in the open interval I , then the graph of f is concave
upward on that interval.c) If f ( x) = 0 for every x in the open interval I , then the graph of f is linear and
there is no concavity defined on that interval.
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EXAMPLE 3
Given the function f(x) = 2 x3
+ 3 x2 – 36 x + 4. Find the (a) the critical values, and (b) the relative
maximum and minimum values, if any (c) the intervals for which the function is increasing and
for which the function is decreasing; (d) the point of inflection; and (e) the intervals for which the
function is concave up or concave down.Solution
x )( x f )(' x f )('' x f
x )( x f =
2 x3 + 3 x2 – 36 x + 4
Domain:(- , )
)(' x f = 6 x2
+ 6 x – 36
= 6( x2 + x - 6)
= 6( x - 2)( x + 3)
Critical values:
a) Set )( x f = 0 and solve
for x.
6( x - 2)( x + 3) = 0
x = - 3 or x = 2
b) There are no points at
which )( x f does not exist.
)('' x f = 12 x + 6
Points of inflection
a) Set f ( x) = 0 and solve
for x.
6(2 x + 1) = 0 x = - ½
b) There are no points at
which f ( x) does not exist .
Critical
Value -3 )3( f = 85 )3( f = 0 f (-3) = - 30 < 0
Maximum at (-3, 85)
Critical
Value 2 )2( f = - 40 )2( f = 0 f (2) = 30 > 0
Minimum at (2, - 40)
Inflection
point - ½ )(21 f = - 13
a) The critical points occurredat x = -3 and x = 2.
b) There is a maximum at (-3,85) and a minimum at (2, -40)
c) The curve is increasing on
(- , -3) (2, )and decreasing on (-3, 2).
d) The point of inflection
occurred at (- ½, -13).
e) The curve is concave down
on the interval ( - , - ½ )and concave up on (- ½, )
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Question #1 Find the (a) the critical values, and (b) the relative maximum and minimum values, if
any (c) the intervals for which the function is increasing and for which the function is
decreasing; (d) the point of inflection; and (e) the intervals for which the function is
concave up or concave down, and (f) sketch the graph of f.
)( x f = x3
–
9 x2
–
24 x –
19
Questions
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Question #2 Find the (a) the critical values, and (b) the relative maximum and minimum values, if
any (c) the intervals for which the function is increasing and for which the function is
decreasing; (d) the point of inflection; and (e) the intervals for which the function is
concave up or concave down, and (f) sketch the graph of f.
)( x f = 2 x3
–
9 x2
–
12 x
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Question #3 Find the (a) the critical values, and (b) the relative maximum and minimum values, if
any (c) the intervals for which the function is increasing and for which the function is
decreasing; (d) the point of inflection; and (e) the intervals for which the function is
concave up or concave down, and (f) sketch the graph of f.
)( x f = –
2 + 12 x –
x3
.
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Question #4 Find the (a) the critical values, and (b) the relative maximum and minimum values, if
any (c) the intervals for which the function is increasing and for which the function is
decreasing; (d) the point of inflection; and (e) the intervals for which the function is
concave up or concave down, and (f) sketch the graph of f.
)( x f = 4 x3
–
3 x4
.
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Question #5 Find the (a) the critical values, and (b) the relative maximum and minimum values, if
any (c) the intervals for which the function is increasing and for which the function is
decreasing; (d) the point of inflection; and (e) the intervals for which the function is
concave up or concave down, and (f) sketch the graph of f.
)( x f = 1 + 6 x2
–
2 x3
.
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Question 1
Question 2
Question 3
Multiple choice Questions
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Question 4
Question 5
Question 6
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Question 7
Question 8
Question 9
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Question 10
Question 11
Question 12
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Question 13
Question 14
Question 15
Question 16
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Question 17
Question 18
Question 19
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Question 20
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Question 22
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Question 23
Question 24
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Question 25