fourteen

Standardized Test Questions
You may use a graphing calculator to solve the following
problems.
51. True or False If f is differentiable and increasing on (a, b),
then f (c)  0 for every c in (a, b). Justify your answer.
52. True or False If f is differentiable and f (c) > 0 for every c in
(a, b), then f is increasing on (a, b). Justify your answer.
x3  x2  x  C
ex  C ln (x  1)  C
about 14.142 m/sec
51. False. For example, the function x3 is increasing on (1, 1), but f (0)  0.
52. True. In fact, f is increasing on [a, b] by Corollary 1 to the Mean Value
Theorem.
23. (a) Local max at  (2.67, 3.08); local min at (4, 0)
(b) On (, 8/3] (c) On [8/3, 4]
(a) None (b) On (, ) (c) None
24. (a) Local min at  (2, 7.56)
(b) On [2, )
(c) On (, 2]
204 Chapter 4 Applications of Derivatives
53. Multiple Choice If f (x)  cos x, then the Mean Value
Theorem guarantees that somewhere between 0 and p/3, f (x) 
(A) 

2
3
p

(B) 


2
3

(C) 

1
2

(D) 0 (E)

1
2


54. Multiple Choice On what interval is the function g(x) 
ex36x28 decreasing? B
(A) (, 2] (B) [0, 4] (C) [2, 4] (D) (4, ) (E) no interval
55. Multiple Choice Which of the following functions is an
antiderivative of


1
x

? E
(A) 


1
2x3

(B) 


2
x

(C)


2
x

(D) x  5 (E) 2x  10
56. Multiple Choice All of the following functions satisfy the
conditions of the Mean Value Theorem on the interval [– 1, 1]
except D
(A) sin x (B) sin1 x (C) x5/3 (D) x3/5 (E)

x 
x
2

Explorations
57. Analyzing Derivative Data Assume that f is continuous on
2, 2 and differentiable on 2, 2. The table gives some
values of f (x.
(a) Estimate where f is increasing, decreasing, and has local
extrema.
(b) Find a quadratic regression equation for the data in the table
and superimpose its graph on a scatter plot of the data.
(c) Use the model in part (b) for f and find a formula for f that
satisfies f 0  0.
58. Analyzing Motion Data Priya’s distance D in meters from a
motion detector is given by the data in Table 4.1.
(a) Estimate when Priya is moving toward the motion detector;
away from the motion detector.
(b) Writing to Learn Give an interpretation of any local
extreme values in terms of this problem situation.
(c) Find a cubic regression equation D  f t for the data in
Table 4.1 and superimpose its graph on a scatter plot of the data.
(d) Use the model in (c) for f to find a formula for f . Use this
formula to estimate the answers to (a).
Extending the Ideas
59. Geometric Mean The geometric mean of two positive
numbers a and b is ab. Show that for f x  1x on any
interval a, b of positive numbers, the value of c in the
conclusion of the Mean Value Theorem is c  ab.
60. Arithmetic Mean The arithmetic mean of two numbers
a and b is a  b2. Show that for f x  x2 on any interval
a, b, the value of c in the conclusion of the Mean Value
Theorem is c  a  b2.
61. Upper Bounds Show that for any numbers a and b,
sin b  sin a  b  a .
62. Sign of f Assume that f is differentiable on a  x  b and
that f b  f a. Show that f is negative at some point
between a and b.
63. Monotonic Functions Show that monotonic increasing and
decreasing functions are one-to-one.
Table 4.1 Motion Detector Data
t (sec) D (m) t (sec) D (m)
0.0 3.36 4.5 3.59
0.5 2.61 5.0 4.15
1.0 1.86 5.5 3.99
1.5 1.27 6.0 3.37
2.0 0.91 6.5 2.58
2.5 1.14 7.0 1.93
3.0 1.69 7.5 1.25
3.5 2.37 8.0 0.67
4.0 3.01
x f (x x f (x
2 7 0.25 4.81
1.75 4.19 0.5 4.25
1.5 1.75 0.75 3.31
1.25 0.31 1 2
1 2 1.25 0.31
0.75 3.31 1.5 1.75
0.5 4.25 1.75 4.19
0.25 4.81 2 7
0 5
10. (a) y  x  , or y  0.707x  0.707
(b) y  x  , or y  0.707x  0.354
1
2
2
1

2
1

2
1

2
15. (a) Local maximum at 

5
2

,

2
4
5

 (b) On ,

5
2


15. (c) On 

5
2

, 
16. (a) Local minimum at 

1
2

, 

4
4
9

 (b) On 

1
2

, 
16. (c) On ,

1
2


17. (a) None (b) None (c) On (, 0) and (0, )
18. (a) None (b) On (, 0) (c) On (0, )
19. (a) None (b) On (, ) (c) None
20. (a) None (b) None (c) On (, )
21. (a) Local maximum at (2, 4) (b) None (c) On [2, )
22. (a) Local maximum at (0, 9); local minima at (5, 16)
and (5, 16) (b) On [5, 0] and [5, )
(c) On (, 5] and [0, 5]
A
Answers:
Section 4.3 Connecting f? and f? with the Graph of f 205
Connecting f and f  with the Graph of f
First Derivative Test for Local Extrema
As we see once again in Figure 4.18, a function f may have local extrema at some critical
points while failing to have local extrema at others. The key is the sign of f in a critical
point’s immediate vicinity. As x moves from left to right, the values of f increase where
f  0 and decrease where f  0.
At the points where f has a minimum value, we see that f  0 on the interval immediately
to the left and f  0 on the interval immediately to the right. (If the point is an endpoint,
there is only the interval on the appropriate side to consider.) This means that the
curve is falling (values decreasing) on the left of the minimum value and rising (values increasing)
on its right. Similarly, at the points where f has a maximum value, f  0 on the
interval immediately to the left and f  0 on the interval immediately to the right. This
means that the curve is rising (values increasing) on the left of the maximum value and
falling (values decreasing) on its right.
4.3
What you’ll learn about
• First Derivative Test for Local
Extrema
• Concavity
• Points of Inflection
• Second Derivative Test for Local
Extrema
• Learning about Functions from
Derivatives
. . . and why
Differential calculus is a powerful
problem-solving tool precisely
because of its usefulness for analyzing
functions.
Figure 4.18 A function’s first derivative tells how the graph rises and falls.
x
y  f(x)
a c b 1 c2 c5 c4 c3
Absolute min
Absolute max
f undefined
Local min
f  0
Local max
f  0 No extreme
f  0
No extreme
f  0
f  0
f  0 f  0
f  0
Local min
f  0
f  0
THEOREM 4 First Derivative Test for Local Extrema
The following test applies to a continuous function f x.
At a critical point c:
1. If f changes sign from positive to negative at c  f  0 for x  c and f  0 for
x  c, then f has a local maximum value at c.
continued
f  0
c
local max
(a) f (c)  0
f  0 f  0
c
local max
(b) f (c) undefined
f  0
206 Chapter 4 Applications of Derivatives
Here is how we apply the First Derivative Test to find the local extrema of a function. The
critical points of a function f partition the x-axis into intervals on which f is either positive
or negative. We determine the sign of f in each interval by evaluating f for one value of x in
the interval. Then we apply Theorem 4 as shown in Examples 1 and 2.
EXAMPLE 1 Using the First Derivative Test
For each of the following functions, use the First Derivative Test to find the local extreme
values. Identify any absolute extrema.
(a) f (x)  x3  12x  5 (b) g(x)  (x2  3)ex
continued
2. If f changes sign from negative to positive at c  f  0 for x  c and f  0 for
x  c, then f has a local minimum value at c.
3. If f does not change sign at c  f has the same sign on both sides of c, then f
has no local extreme value at c.
At a left endpoint a:
If f  0 ( f  0) for x  a, then f has a local maximum (minimum) value at a.
At a right endpoint b:
If f  0 ( f  0) for x  b, then f has a local minimum (maximum) value at b.
f  0 f  0 f  0
c
local
min
(a) f (c)  0
f  0
c
local min
(b) f (c) undefined
f  0
c
no extreme
(a) f (c)  0
f  0
c
(b) f (c) undefined
f  0
no extreme
f  0
f  0
a
local max
f  0
a
local min
f  0
b
local max
f  0
b
local min
Section 4.3 Connecting f? and f? with the Graph of f 207
SOLUTION
(a) Since f is differentiable for all real numbers, the only possible critical points are the
zeros of f . Solving f (x)  3x2  12  0, we find the zeros to be x  2 and x2. The
zeros partition the x-axis into three intervals, as shown below:
Figure 4.19 The graph of
f x  x3  12x  5.
Figure 4.20 The graph of
gx  x2  3ex.
[–5, 5] by [–25, 25]
[–5, 5] by [–8, 5]
Using the First Derivative Test, we can see from the sign of f on each interval that there is
a local maximum at x2 and a local minimum at x  2. The local maximum value is
f (2)  11, and the local minimum value is f(2) 21. There are no absolute extrema,
as the function has range (, ) (Figure 4.19).
(b) Since g is differentiable for all real numbers, the only possible critical points are the
zeros of g . Since g (x)  (x2  3) • ex  (2x) • ex  (x2  2x  3) • ex, we find the zeros
of g to be x  1 and x3. The zeros partition the x-axis into three intervals, as shown
below:
Figure 4.21 The graph of y  x3 is
concave down on , 0 and concave up
on 0, .
y decreases
y increases
x
y
0
y  x3
CONCAVE UP
CONCAVE DOWN
x
Sign of f + – +
–2 2

x
Sign of g + – +
–3 1

If a function y  f x has a second derivative, then we can conclude that y increases if
y  0 and y decreases if y  0.
DEFINITION Concavity
The graph of a differentiable function y  f (x) is
(a) concave up on an open interval I if y is increasing on I.
(b) concave down on an open interval I if y is decreasing on I.
Using the First Derivative Test, we can see from the sign of f on each interval that there is
a local maximum at x3 and a local minimum at x  1. The local maximum value is
g(3)  6e3  0.299, and the local minimum value is g(1)2e  5.437. Although
this function has the same increasing–decreasing–increasing pattern as f, its left end
behavior is quite different. We see that limx? g(x)  0, so the graph approaches the
y-axis asymptotically and is therefore bounded below. This makes g(1) an absolute
minimum. Since limx? g(x)  , there is no absolute maximum (Figure 4.20).
Now try Exercise 3.
Concavity
As you can see in Figure 4.21, the function y  x3 rises as x increases, but the portions defined
on the intervals , 0 and 0,  turn in different ways. Looking at tangents as we
scan from left to right, we see that the slope y of the curve decreases on the interval ,
0 and then increases on the interval 0, . The curve y  x3 is concave down on , 0
and concave up on 0, . The curve lies below the tangents where it is concave down, and
above the tangents where it is concave up.
208 Chapter 4 Applications of Derivatives
EXAMPLE