Cul. 6

Chapter 4 Overview
In the past, when virtually all graphing was done by hand—often laboriously—derivatives
were the key tool used to sketch the graph of a function. Now we can graph a function
quickly, and usually correctly, using a grapher. However, confirmation of much of what we
see and conclude true from a grapher view must still come from calculus.
This chapter shows how to draw conclusions from derivatives about the extreme values
of a function and about the general shape of a function’s graph. We will also see
how a tangent line captures the shape of a curve near the point of tangency, how to deduce
rates of change we cannot measure from rates of change we already know, and
how to find a function when we know only its first derivative and its value at a single
point. The key to recovering functions from derivatives is the Mean Value Theorem, a
theorem whose corollaries provide the gateway to integral calculus, which we begin in
Chapter 5.
Extreme Values of Functions
Absolute (Global) Extreme Values
One of the most useful things we can learn from a function’s derivative is whether the
function assumes any maximum or minimum values on a given interval and where
these values are located if it does. Once we know how to find a function’s extreme values,
we will be able to answer such questions as “What is the most effective size for a
dose of medicine?” and “What is the least expensive way to pipe oil from an offshore
well to a refinery down the coast?” We will see how to answer questions like these in
Section 4.4.
4.1
What you’ll learn about
• Absolute (Global) Extreme Values
• Local (Relative) Extreme Values
• Finding Extreme Values
. . . and why
Finding maximum and minimum
values of functions, called optimization,
is an important issue in
real-world problems.
DEFINITION Absolute Extreme Values
Let f be a function with domain D. Then f c is the
(a) absolute maximum value on D if and only if f x  f c for all x in D.
(b) absolute minimum value on D if and only if f x  f c for all x in D.
Absolute (or global) maximum and minimum values are also called absolute extrema
(plural of the Latin extremum). We often omit the term “absolute” or “global” and just say
maximum and minimum.
Example 1 shows that extreme values can occur at interior points or endpoints of
intervals.
EXAMPLE 1 Exploring Extreme Values
On p2, p2, f x  cos x takes on a maximum value of 1 (once) and a minimum
value of 0 (twice). The function gx  sin x takes on a maximum value of 1 and a
minimum value of 1 ( Figure 4.1). Now try Exercise 1.
Functions with the same defining rule can have different extrema, depending on the
Figure 4.1 (Example 1) domain.
x
y
0
1
–1
y  sin x
––
2
–
y  cos x
––
2
188 Chapter 4 Applications of Derivatives
EXAMPLE 2 Exploring Absolute Extrema
The absolute extrema of the following functions on their domains can be seen in Figure 4.2.
Function Rule Domain D Absolute Extrema on D
(a) y  x2 ,  No absolute maximum.
Absolute minimum of 0 at x  0.
(b) y  x2 0, 2 Absolute maximum of 4 at x  2.
Absolute minimum of 0 at x  0.
(c) y  x2 0, 2 Absolute maximum of 4 at x  2.
No absolute minimum.
(d) y  x2 0, 2 No absolute extrema.
Now try Exercise 3.
Example 2 shows that a function may fail to have a maximum or minimum value. This
cannot happen with a continuous function on a finite closed interval.
THEOREM 1 The Extreme Value Theorem
If f is continuous on a closed interval a, b, then f has both a maximum value and a
minimum value on the interval. (Figure 4.3)
Figure 4.2 (Example 2)
x
y
2
(a) abs min only
y  x2
D  (–, )
x
y
2
(b) abs max and min
y  x2
D  [0, 2]
x
y
2
(c) abs max only
y  x2
D  (0, 2]
x
y
2
(d) no abs max or min
y  x2
D  (0, 2)
Figure 4.3 Some possibilities for a continuous function’s maximum (M) and
minimum (m) on a closed interval [a, b].
x
a
y  f (x)
(x2, M)
Maximum and minimum
at interior points
x2 b
M
x1
(x1, m)
m
x
a b
y  f (x)
M
m
Maximum and minimum
at endpoints
x
a
y  f (x)
Maximum at interior point,
minimum at endpoint
x2
M
b
m
x
a
y  f(x)
Minimum at interior point,
maximum at endpoint
x1
M
b
m
Section 4.1 Extreme Values of Functions 189
Local (Relative) Extreme Values
Figure 4.4 shows a graph with five points where a function has extreme values on its domain
a, b. The function’s absolute minimum occurs at a even though at e the function’s value is
smaller than at any other point nearby. The curve rises to the left and falls to the right around
c, making f c a maximum locally. The function attains its absolute maximum at d.
THEOREM 2 Local Extreme Values
If a function f has a local maximum value or a local minimum value at an interior
point c of its domain, and if f exists at c, then
f c  0.
Local extrema are also called relative extrema.
An absolute extremum is also a local extremum, because being an extreme value
overall makes it an extreme value in its immediate neighborhood. Hence, a list of local extrema
will automatically include absolute extrema if there are any.
Finding Extreme Values
The interior domain points where the function in Figure 4.4 has local extreme values are
points where either f is zero or f does not exist. This is generally the case, as we see from
the following theorem.
Figure 4.4 Classifying extreme values.
x
a b
y  f (x)
c e d
Local maximum.
No greater value of
f nearby.
Absolute maximum.
No greater value of f anywhere.
Also a local maximum.
Local minimum.
No smaller value of
f nearby.
Local minimum.
No smaller
value of f nearby.
Absolute minimum.
No smaller value
of f anywhere. Also a
local minimum.
DEFINITION Local Extreme Values
Let c be an interior point of the domain of the function f. Then f c is a
(a) local maximum value at c if and only if f x  f c for all x in some open
interval containing c.
(b) local minimum value at c if and only if f x  f c for all x in some open
interval containing c.
A function f has a local maximum or local minimum at an endpoint c if the appropriate
inequality holds for all x in some half-open domain interval containing c.
190 Chapter 4 Applications of Derivatives
EXAMPLE 3 Finding Absolute Extrema
Find the absolute maximum and minimum values of f x  x2 3 on the interval
2, 3.
SOLUTION
Solve Graphically Figure 4.5 suggests that f has an absolute maximum value of
about 2 at x  3 and an absolute minimum value of 0 at x  0.
Confirm Analytically We evaluate the function at the critical points and endpoints
and take the largest and smallest of the resulting values.
The first derivative
f x 

2
3

x13 

3
2
3 x


has no zeros but is undefined at x  0. The values of f at this one critical point and at
the endpoints are
Critical point value: f 0  0;
Endpoint values: f 2  223 3 4;
f 3  32 3 3 9.
We can see from this list that the function’s absolute maximum value is 3 9  2.08,
and occurs at the right endpoint x  3. The absolute minimum value is 0, and occurs
at the interior point x  0. Now try Exercise 11.
In Example 4, we investigate the reciprocal of the function whose graph was drawn in
Example 3 of Section 1.2 to illustrate “grapher failure.”
EXAMPLE 4 Finding Extreme Values
Find the extreme values of f x 

4
1
x 2

.
SOLUTION
Solve Graphically Figure 4.6 suggests that f has an absolute minimum of about 0.5 at
x  0. There also appear to be local maxima at x2 and x  2. However, f is not defined
at these points and there do not appear to be maxima anywhere else.
continued
Figure 4.5 (Example 3)
[–2, 3] by [–1, 2.5]
y  x2/3
Figure 4.6 The graph of
f x 

4
1
x 2

.
(Example 4)
[–4, 4] by [–2, 4]
Because of Theorem 2, we usually need to look at only a few points to find a function’s
extrema. These consist of the interior domain points where f  0 or f does not exist (the
domain points covered by the theorem) and the domain endpoints (the domain points not
covered by the theorem). At all other domain points, f  0 or f  0.
The following definition helps us summarize these findings.
Thus, in summary, extreme values occur only at critical points and endpoints.
DEFINITION Critical Point
A point in the interior of the domain of a function f at which f  0 or f does not
exist is a critical point of f.
Section 4.1 Extreme Values of Functions 191
Confirm Analytically The function f is defined only for 4  x2  0, so its domain
is the open interval 2, 2. The domain has no endpoints, so all the extreme values
must occur at critical points. We rewrite the formula for f to find f :
f x 

4
1
x 2

 4  x212.
Thus,
f x

1
2

4  x23/22x 
4 
x
x232
.
The only critical point in the domain 2, 2 is x  0. The value
f 0 

4
1
0 2



1
2


is therefore the sole candidate for an extreme value.
To determine whether 12 is an extreme value of f, we examine the formula
f x 

4
1
x 2

.
As x moves away from 0 on either side, the denominator gets smaller, the values of f
increase, and the graph rises. We have a minimum value at x  0, and the minimum is
absolute.
The function has no maxima, either local or absolute. This does not violate Theorem 1
(The Extreme Value Theorem) because here f is defined on an open interval. To invoke
Theorem 1’s guarantee of extreme points, the interval must be closed.
Now try Exercise 25.
While a function’s extrema can occur only at critical points and endpoints, not every
critical point or endpoint signals the presence of an extreme value. Figure 4.7 illustrates
this for interior points. Exercise 55 describes a function that fails to assume an extreme
value at an endpoint of its domain.
Figure 4.7 Critical points without extreme values. (a) y  3x2 is 0 at x  0, but
y  x3 has no extremum there. (b) y  13x23 is undefined at x  0, but y  x13
has no extremum there.
–1
x
y
–1 1
1
y  x3
0
(a)
–1
x
y
–1 1
1
y  x1/3
(b)
EXAMPLE 5 Finding Extreme Values
Find the extreme values of
5  2x2, x  1
f x  {x  2, x  1.
continued
192 Chapter 4 Applications of Derivatives
SOLUTION
Solve Graphically The graph in Figure 4.8 suggests that f 0  0 and that f 1
does not exist. There appears to be a local maximum value of 5 at x  0 and a local
minimum value of 3 at x  1.
Confirm Analytically For x
1, the derivative is
f x  {

d
d
x

5  2x24x, x  1


d
d
x

x  2  1, x  1.
The only point where f  0 is x  0. What happens at x  1?
At x  1, the right- and left-hand derivatives are respectively
lim
h?0

f
1  h
h
  f 1

 lim
h?0


1

h

h
 2 3

 lim
h?0


h
h

1,
lim
h?0

f
1  h
h
  f 1

 lim
h?0

5
 21 
h
h2 3


 lim
h?0


2h
h
2  h

4.
Since these one-sided derivatives differ, f has no derivative at x  1, and 1 is a second
critical point of f.
The domain ,  has no endpoints, so the only values of f that might be local extrema
are those at the critical points:
f 0  5 and f 1  3.
From the formula for f, we see that the values of f immediately to either side of x  0
are less than 5, so 5 is a local maximum. Similarly, the values of f immediately to either
side of x  1 are greater than 3, so 3 is a local minimum. Now try Exercise 41.
Most graphing calculators have built-in methods to find the coordinates of points where
extreme values occur. We must, of course, be sure that we use correct graphs to find these
values. The calculus that you learn in this chapter should make you feel more confident
about working with graphs.
EXAMPLE 6 Using Graphical Methods
Find the extreme values of f x  ln

1 
x
x2
.
SOLUTION
Solve Graphically The domain of f is the set of all nonzero real numbers. Figure 4.9
suggests that f is an even function with a maximum value at two points. The coordinates
found in this window suggest an extreme value of about 0.69 at approximately
x  1. Because f is even, there is another extreme of the same value at approximately
x1. The figure also suggests a minimum value at x  0, but f is not defined
there.
Confirm Analytically The derivative
f x 

x 
1
1


x
x
2
2

is defined at every point of the function’s domain. The critical points where f x  0 are
x  1 and x1. The corresponding values of f are both ln 12ln 2  0.69.
Now try Exercise 37.
Figure 4.8 The function in Example 5.
[–5, 5] by [–5, 10]
Figure 4.9 The function in Example 6.
[–4.5, 4.5] by [–4, 2]
Maximum
X = .9999988 Y = –.6931472
Section 4.1 Extreme Values of Functions 193
Section 4.1 Exercises
Finding Extreme Values
Let f x 

x 2 
x
1

, 2  x  2.
1. Determine graphically the extreme values of f and where they occur. Find f at
these values of x.
2. Graph f and f or NDER  f x, x, x in the same viewing window. Comment
on the relationship between the graphs.
3. Find a formula for f x.
EXPLORATION 1
In Exercises 1–4, find the first derivative of the function.
1. f x 4x
2. f x 

9
2
x2



(9 
2
x
x

2)3/2
3. gx  cos ln x 

sin (
x
ln x)
4. hx  e2x 2e2x
In Exercises 5–8, match the table with a graph of f (x).
5. 6.
7. 8.
In Exercises 9 and 10, find the limit for
f x 

9
2
x 2

.
9. lim
x?3
f x  10. lim
x?3
f x 
In Exercises 11 and 12, let
x3  2x, x  2
f x  {x  2, x  2.
11. Find (a) f 1, 1 (b) f 3, 1 (c) f 2. Undefined
12. (a) Find the domain of f . x
2
(b) Write a formula for f x. f (x) 
3x2  2, x  2
1, x  2
a b c
(d)
a b c
(c)
a b c
(b)
a b c
(a)
x f x
a does not exist
b does not exist
c 1.7
x f x
a does not exist
b 0
c 2
x f x
a 0
b 0
c 5
x f x
a 0
b 0
c 5
Quick Review 4.1 (For help, go to Sections 1.2, 2.1, 3.5, and 3.6.)
In Exercises 1–4, find the extreme values and where they occur.
1. 2.
x
y
–1 1
1
–1
x
y
2
2
–2 0
3. 4.
2
(1, 2)
–1
–3 2
x
y
x
y
0 2
5
1



24 x
(c) (b)
(d) (a)
1. Minima at (2, 0) and (2, 0), maximum at (0, 2)
2. Local minimum at (1, 0), local maximum at (1, 0)
3. Maximum at (0, 5)
4. Local maximum at (3, 0), local
minimum at (2, 0), maximum at
(1, 2), minimum at (0, 1)
194 Chapter 4 Applications of Derivatives
In Exercises 5–10, identify each x-value at which any absolute extreme
value occurs. Explain how your answer is consistent with the
Extreme Value Theorem. See page 195.
5. 6.
7. 8.
9. 10.
In Exercises 11–18, use analytic methods to find the extreme values
of the function on the interval and where they occur. See page 195.
11. f x 

1
x

 ln x, 0.5  x  4
12. gx  ex, 1  x  1
13. hx  ln x  1, 0 x  3
14. kx  ex2,   x  
15. f x  sin (x 

p
4

), 0 x 

7
4
p


16. gx  sec x, 

p
2

 x 

3
2
p


17. f x  x25, 3  x  1
18. f x  x35, 2  x  3
In Exercises 19–30, find the extreme values of the function and where
they occur.
19. y  2x 2  8x  9 20. y  x3  2x  4
21. y  x3  x2  8x  5 22. y  x3  3x2  3x  2
23. y x2 1  24. y 

x 2
1
1


25. y 

1
1
x 2

26. y 

3 1
1
x 2


27. y 32xx2
28. y 

3
2

x4  4x3  9x2  10
29. y 

x 2 
x
1

30. y 

x 2 
x 
2x
1
2


x
y
0 a c b
y  g(x)
x
y
0 a c b
y  g(x)
x
y
0 a b
y  h(x)
c
x
y
0 a b
y  f (x)
c
x
y
0 a c b
y  f (x)
x
y
0 a c1 b
y  h(x)
c2
Group Activity In Exercises 31–34, find the extreme values of the
function on the interval and where they occur.
31. f x  x  2  x  3 , 5  x  5
32. gx  x  1  x  5 , 2  x  7
33. hx  x  2  x  3 ,   x  
34. kx  x  1  x  3 ,   x  
In Exercises 35–42, identify the critical point and determine the local
extreme values.
35. y  x 23x  2 36. y  x 23x2  4
37. y  x4x2 38. y  x23x
39.
4  2x, x  1
y  {x  1, x  1
40.
3  x, x  0
y  {3  2x  x2, x  0
41. x2  2x  4, x  1
y  {x2  6x  4, x  1


1
4

x2 

1
2

x 

1
4
5

, x  1
42. y  {x3  6x2  8x, x  1
43. Writing to Learn The function
Vx  x10  2x16  2x, 0 x  5,
models the volume of a box.
(a) Find the extreme values of V.
(b) Interpret any values found in (a) in terms of volume of
the box.
44. Writing to Learn The function
Px  2x 

20
x
0

, 0 x  ,
models the perimeter of a rectangle of dimensions x by 100x.
(a) Find any extreme values of P.
(b) Give an interpretation in terms of perimeter of the rectangle
for any values found in (a).
Standardized Test Questions
You should solve the following problems without using a
graphing calculator.
45. True or False If f (c) is a local maximum of a continuous
function f on an open interval (a, b), then f (c)  0. Justify your
answer.
46. True or False If m is a local minimum and M is a local maximum
of a continuous function f on (a, b), then m  M. Justify
your answer.
47. Multiple Choice Which of the following values is the absolute
maximum of the function f (x)  4x  x2  6 on the interval
[0, 4]? E
(A) 0 (B) 2 (C) 4 (D) 6 (E) 10
45. False. For example, the maximum could occur at a corner, where f (c)
would not exist.
Min value 1 at
x  2
Min value
1 at x  0
Max value 2 at x  1;
min value 0 at x1, 3
Min value 0
at x1, 1
None
Local max at (0, 1)
Local min
at (0, 1)
21. Local max at (2, 17); local min at 

4
3

, 

4
2
1
7


Max value is 144 at x  2.
Min value is 40 at x  10.
The largest volume of the box is 144 cubic units and
it occurs when x  2.
The smallest perimeter is 40 units
and it occurs when x  10, which makes it a 10 by 10 square.
Section 4.1 Extreme Values of Functions 195
48. Multiple Choice If f is a continuous, decreasing function on
[0, 10] with a critical point at (4, 2), which of the following statements
must be false? E
(A) f (10) is an absolute minimum of f on [0, 10].
(B) f (4) is neither a relative maximum nor a relative minimum.
(C) f (4) does not exist.
(D) f (4)  0
(E) f (4)  0
49. Multiple Choice Which of the following functions has exactly
two local extrema on its domain? B
(A) f (x)  ?x  2?
(B) f (x)  x3  6x  5
(C) f (x)  x3  6x  5
(D) f (x)  tan x
(E) f (x)  x  ln x
50. Multiple Choice If an even function f with domain all real
numbers has a local maximum at x  a, then f (a) B
(A) is a local minimum.
(B) is a local maximum.
(C) is both a local minimum and a local maximum.
(D) could be either a local minimum or a local maximum.
(E) is neither a local minimum nor a local maximum.
Explorations
In Exercises 51 and 52, give reasons for your answers.
51. Writing to Learn Let f x  x  223.
(a) Does f 2 exist? No
(b) Show that the only local extreme value of f occurs at x  2.
(c) Does the result in (b) contradict the Extreme Value Theorem?
(d) Repeat parts (a) and (b) for f x  x  a23, replacing 2 by a.
52. Writing to Learn Let f x  x3  9x .
(a) Does f 0 exist? No (b) Does f 3 exist? No
(c) Does f 3 exist? No (d) Determine all extrema of f.
Extending the Ideas
53. Cubic Functions Consider the cubic function
f x  ax3  bx2  cx  d.
(a) Show that f can have 0, 1, or 2 critical points. Give examples
and graphs to support your argument.
(b) How many local extreme values can f have? Two or none
54. Proving Theorem 2 Assume that the function f has a local
maximum value at the interior point c of its domain and that f (c)
exists.
(a) Show that there is an open interval containing c such that
f x  f c  0 for all x in the open interval.
(b) Writing to Learn Now explain why we may say
lim
x?c

f
x
x



c
f c

 0.
(c) Writing to Learn Now explain why we may say
lim
x?c

f
x
x



c
f c

 0.
(d) Writing to Learn Explain how parts (b) and (c) allow us
to conclude f c  0.
(e) Writing to Learn Give a similar argument if f has a local
minimum value at an interior point.
55. Functions with No Extreme Values at Endpoints
(a) Graph the function
sin

1
x

, x  0
f x {0, x  0.
Explain why f 0  0 is not a local extreme value of f.
(b) Group Activity Construct a function of your own that fails
to have an extreme value at a domain endpoint.
Answers:
5. Maximum at x  b, minimum at x  c2;
Extreme Value Theorem applies, so both the max
and min exist.
6. Maximum at x  c, minimum at x  b;
Extreme Value Theorem applies, so both the max
and min exist.
7. Maximum at x  c, no minimum;
Extreme Value Theorem doesn’t apply, since the
function isn’t defined on a closed interval.
8. No maximum, no minimum;
Extreme Value Theorem doesn’t apply, since the
function isn’t continuous or defined on a closed
interval.
9. Maximum at x  c, minimum at x  a;
Extreme Value Theorem doesn’t apply, since the
function isn’t continuous.
10. Maximum at x  a, minimum at x  c;
Extreme Value Theorem doesn’t apply, since the
function isn’t continuous.
11. Maximum value is

1
4

 ln 4 at x  4; minimum value is 1 at x  1;
local maximum at 

1
2

, 2  ln 2
12. Maximum value is e at x1; minimum value is

1
e

at x  1.
13. Maximum value is ln 4 at x  3; minimum value is 0 at x  0.
14. Maximum value is 1 at x  0
15. Maximum value is 1 at x 


4

; minimum value is 1 at x 

5
4


;
local minimum at 0, ; local maximum at 

7
4


,
0

16. Local minimum at (0, 1); local maximum at (, 1)
17. Maximum value is 32/5 at x3; minimum value is 0 at x  0
18. Maximum value is 33/5 at x  3
1

2
51. (b) The derivative is defined and nonzero for x
2. Also, f (2)  0, and
f (x)  0 for all x
2.
(c) No, because (, ) is not a closed interval.
(d) The answers are the same as (a) and (b) with 2 replaced by a.
Minimum value is 0 at x3,
x  0, and x  3; local maxima
at (3, 63) and (3, 63)
196 Chapter 4 Applications of Derivatives
Mean Value Theorem
Mean Value Theorem
The Mean Value Theorem connects the average rate of change of a function over an interval
with the instantaneous rate of change of the function at a point within the interval. Its powerful
corollaries lie at the heart of some of the most important applications of the calculus.
The theorem says that somewhere between points A and B on a differentiable curve,
there is at least one tangent line parallel to chord AB (Figure 4.10).
The hypotheses of Theorem 3 cannot be relaxed. If they fail at even one point, the
graph may fail to have a tangent parallel to the chord. For instance, the function f x  x
is continuous on 1, 1 and differentiable at every point of the interior 1, 1 except
x  0. The graph has no tangent parallel to chord AB (Figure 4.11a). The function
gx  int x is differentiable at every point of 1, 2 and continuous at every point of
1, 2 except x  2. Again, the graph has no tangent parallel to chord AB (Figure 4.11b).
The Mean Value Theorem is an existence theorem. It tells us the number c exists
without telling how to find it. We can sometimes satisfy our curiosity about the value of
c but the real importance of the theorem lies in the surprising conclusions we can draw
from it.
4.2
What you’ll learn about
• Mean Value Theorem
• Physical Interpretation
• Increasing and Decreasing
Functions
• Other Consequences
. . . and why
The Mean Value Theorem is an
important theoretical tool to
connect the average and instantaneous
rates of change.
Figure 4.10 Figure for the Mean Value
Theorem.
x
y
0
Slope f (c)
a
Tangent parallel to chord
c b
y  f (x)
Slope f(b)  f(a)
————–
b  a
B
A
THEOREM 3 Mean Value Theorem for Derivatives
If y  f x is continuous at every point of the closed interval a, b and differentiable
at every point of its interior a, b, then there is at least one point c in a, b at
which
f c 

f b
b
 

f
a
a

.
Rolle’s Theorem
The first version of the Mean Value Theorem
was proved by French mathematician
Michel Rolle (1652–1719). His version
had f a  f b  0 and was proved
only for polynomials, using algebra and
geometry.
Rolle distrusted calculus and spent
most of his life denouncing it. It is
ironic that he is known today only for
an unintended contribution to a field
he tried to suppress.
y
x
0
f (c) = 0
a
y = f (x)
c b
Figure 4.11 No tangent parallel to chord AB.
x
y
B (1, 1)
a  –1 b  1
(a)
A (–1, 1)
y  |x|, –1 ? x ? 1
y
x
y  int x, 1? x ? 2
A (1, 1)
B (2, 2)
a  1 b  2
(b)
2
1
0
EXAMPLE 1 Exploring the Mean Value Theorem
Show that the function f x  x2 satisfies the hypotheses of the Mean Value Theorem
on the interval 0, 2. Then find a solution c to the equation
f c 

f b
b
 

f
a
a


on this interval.
continued
Section 4.2 Mean Value Theorem 197
SOLUTION
The function f x  x2 is continuous on 0, 2 and differentiable on 0, 2. Since
f 0  0 and f 2  4, the Mean Value Theorem guarantees a point c in the interval
0, 2 for which
f c 

f b
b
 

f
a
a


2c 

f 2
2



0
f 0

 2 f x  2x
c  1.
Interpret The tangent line to f x  x2 at x  1 has slope 2 and is parallel to the
chord joining A0, 0 and B2, 4 ( Figure 4.12).
Now try Exercise 1.
EXAMPLE 2 Exploring the Mean Value Theorem
Explain why each of the following functions fails to satisfy the conditions of the Mean
Value Theorem on the interval [–1, 1].
(a) f (x)  x2  1 (b)
SOLUTION
(a) Note that x2  1  |x|  1, so this is just a vertical shift of the absolute value
function, which has a nondifferentiable “corner” at x  0. (See Section 3.2.) The
function f is not differentiable on (–1, 1).
(b) Since limx?1– f (x)  limx?1– x3  3  4 and limx?1+ f (x)  limx?1+ x2  1  2, the
function has a discontinuity at x  1. The function f is not continuous on [–1, 1].
If the two functions given had satisfied the necessary conditions, the conclusion of the
Mean Value Theorem would have guaranteed the existence of a number c in (– 1, 1)
such that f (c) 

f (1
1
)


(
f

(
1)
1)

 0. Such a number c does not exist for the function in
part (a), but one happens to exist for the function in part (b) (Figure 4.13).
Figure 4.12 (Example 1)
y = x2
x
y
1
(1, 1)
B(2, 4)
A(0, 0) 2
Figure 4.13 For both functions in Example 2,

f (1
1
)


(
f

(
1)
1)

 0 but neither
function satisfies the conditions of the Mean Value Theorem on the interval
[– 1, 1]. For the function in Example 2(a), there is no number c such that
f (c)  0. It happens that f (0)  0 in Example 2(b).
y
x
1
2
3
–2 –1 0 1 2
(a)
y
(b)
– 4 x
4
4 0
x3  3 for x  1
f x  { x2  1 for x  1
Now try Exercise 3.
198 Chapter 4 Applications of Derivatives
EXAMPLE 3 Applying the Mean Value Theorem
Let f x 1x2 , A  1, f 1, and B  1, f 1. Find a
tangent to f in the interval 1, 1 that is parallel to the secant AB.
SOLUTION
The function f (Figure 4.14) is continuous on the interval [–1, 1] and
f x 

1


x
x 2


is defined on the interval 1, 1. The function is not differentiable at x1 and x  1,
but it does not need to be for the theorem to apply. Since f 1  f 1 0, the tangent
we are looking for is horizontal. We find that f  0 at x  0, where the graph has the
horizontal tangent y  1. Now try Exercise 9.
Physical Interpretation
If we think of the difference quotient  f b  f ab  a as the average change in f
over a, b and f c as an instantaneous change, then the Mean Value Theorem says that
the instantaneous change at some interior point must equal the average change over the entire
interval.
EXAMPLE 4 Interpreting the Mean Value Theorem
If a car accelerating from zero takes 8 sec to go 352 ft, its average velocity for the 8-sec
interval is 352 8  44 ft sec, or 30 mph. At some point during the acceleration, the theorem
says, the speedometer must read exactly 30 mph (Figure 4.15).
Now try Exercise 11.
Increasing and Decreasing Functions
Our first use of the Mean Value Theorem will be its application to increasing and decreasing
functions.
Figure 4.15 (Example 4)
Figure 4.14 (Example 3)
y 
x
y
–1 0 1
1 ???1? ?? x2 , –1 ? x ? 1
t
s
0
5
80
160 At this point,
the car’s speed
was 30 mph.
Time (sec)
(8, 352)
240
320
400
Distance (ft)
s  f(t)
Monotonic Functions
A function that is always increasing on
an interval or always decreasing on an
interval is said to be monotonic there.
The Mean Value Theorem allows us to identify exactly where graphs rise and fall.
Functions with positive derivatives are increasing functions; functions with negative derivatives
are decreasing functions.
COROLLARY 1 Increasing and Decreasing Functions
Let f be continuous on a, b and differentiable on a, b.
1. If f  0 at each point of a, b, then f increases on a, b.
2. If f  0 at each point of a, b, then f decreases on a, b.
DEFINITIONS Increasing Function, Decreasing Function
Let f be a function defined on an interval I and let x1 and x2 be any two points in I.
1. f increases on I if x1 x2 ? f x1   f x2 .
2. f decreases on I if x1 x2 ? f x1   f x2 .
Section 4.2 Mean Value Theorem 199
Proof Let x1 and x2 be any two points in a, b with x1 x2. The Mean Value Theorem
applied to f on x1, x2  gives
f x2   f x1   f cx2  x1
for some c between x1 and x2. The sign of the right-hand side of this equation is the same
as the sign of f c because x2  x1 is positive. Therefore,
(a) f x1   f x2  if f  0 on a, b ( f is increasing), or
(b) f x1   f x2  if f  0 on a, b ( f is decreasing). ?
EXAMPLE 5 Determining Where Graphs Rise or Fall
The function y  x2 ( Figure 4.16) is
(a) decreasing on , 0 because y  2x  0 on , 0.
(b) increasing on 0,  because y  2x  0 on 0, . Now try Exercise 15.
EXAMPLE 6 Determining Where Graphs Rise or Fall
Where is the function f x  x3  4x increasing and where is it decreasing?
SOLUTION
Solve Graphically The graph of f in Figure 4.17 suggests that f is increasing from
 to the x-coordinate of the local maximum, decreasing between the two local extrema,
and increasing again from the x-coordinate of the local minimum to . This information
is supported by the superimposed graph of f x  3x2  4.
Confirm Analytically The function is increasing where f x  0.
3x2  4  0
x2 

4
3


x

4
3

or x 

4
3


The function is decreasing where f x  0.
3x2  4  0
x2 

4
3




4
3

 x 

4
3


In interval notation, f is increasing on , 43], decreasing on 43, 43,
and increasing on 43, .