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d x2 ’ 1
(d)
dx x 2 + 1
[x · ln(sin x)]
(e)

d s(s+2)
(f) e
ds
80 CHAPTER 2 Foundations of Calculus


d sin(x 2 )
(g) e
dx
[ln(ex + x)]
(h)
5. Imitate the example in the text to do each of these falling body problems.
(a) A ball is dropped from a height of 100 feet. How long will it take
that ball to hit the ground?
(b) Suppose that the ball from part (a) is thrown straight down with an
initial velocity of 10 feet per second. Then how long will it take the
ball to hit the ground?
(c) Suppose that the ball from part (a) is thrown straight up with an
initial velocity of 10 feet per second. Then how long will it take the
ball to hit the ground?
6. Use the Chain Rule to perform each of these differentiations:
d
(a) sin(ln(cos x))
dx
d sin(cos x)
(b) e
dx
d
ln(esin x + x)
(c)
dx
d
arcsin(x 2 + tan x)
(d)
dx
d
arccos(ln x ’ ex /5)
(e)
dx
d
arctan(x 2 + ex )
(f)
dx
7. If a car has position p(t) = 6t 2 ’ 5t + 20 feet, where t is measured in
seconds, then what is the velocity of that car at time t = 4? What is the
average velocity of that car from t = 2 to t = 8? What is the greatest
velocity over the time interval [5, 10]?
8. In each of these problems, use the formula for the derivative of an inverse
function to ¬nd [f ’1 ] (1).
f (0) = 1, f (0) = 3
(a)
f (3) = 1, f (3) = 8
(b)
f (2) = 1, f (2) = π 2
(c)
f (1) = 1, f (1) = 40
(d)
CHAPTER 3



Applications of
the Derivative
3.1 Graphing of Functions
We know that the value of the derivative of a function f at a point x represents the
slope of the tangent line to the graph of f at the point (x, f (x)). If that slope is
positive, then the tangent line rises as x increases from left to right, hence so does
the curve (we say that the function is increasing). If instead the slope of the tangent
line is negative, then the tangent line falls as x increases from left to right, hence
so does the curve (we say that the function is decreasing). We summarize:
On an interval where f > 0 the graph of f goes uphill.
On an interval where f < 0 the graph of f goes downhill.
See Fig. 3.1.
With some additional thought, we can also get useful information from the second
derivative. If f = (f ) > 0 at a point, then f is increasing. Hence the slope of the
tangent line is getting ever greater (the graph is concave up). The picture must be as
in Fig. 3.2(a) or 3.2(b). If instead f = (f ) < 0 at a point then f is decreasing.
Hence the slope of the tangent line is getting ever less (the graph is concave down).
The picture must be as in Fig. 3.3(a) or 3.3(b).
Using information about the ¬rst and second derivatives, we can render rather
accurate graphs of functions. We now illustrate with some examples.
EXAMPLE 3.1
Sketch the graph of f (x) = x 2 .


81
Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.
82 CHAPTER 3 Applications of the Derivative




Fig. 3.1




Fig. 3.2




Fig. 3.3

SOLUTION
Of course this is a simple and familiar function, and you know that its graph
is a parabola. But it is satisfying to see calculus con¬rm the shape of the graph.
Let us see how this works.
83
CHAPTER 3 Applications of the Derivative

First observe that f (x) = 2x. We see that f < 0 when x < 0 and f > 0
when x > 0. So the graph is decreasing on the negative real axis and the graph
is increasing on the positive real axis.
Next observe that f (x) = 2. Thus f > 0 at all points. Thus the graph is
concave up everywhere.
Finally note that the graph passes through the origin.
We summarize this information in the graph shown in Fig. 3.4.




Fig. 3.4

EXAMPLE 3.2
Sketch the graph of f (x) = x 3 .
SOLUTION
First observe that f (x) = 3x 2 . Thus f ≥ 0 everywhere. The function is
always increasing.
Second observe that f (x) = 6x. Thus f (x) < 0 when x < 0 and
f (x) > 0 when x > 0. So the graph is concave down on the negative real
axis and concave up on the positive real axis.
Finally note that the graph passes through the origin.
We summarize our ¬ndings in the graph shown in Fig. 3.5.
You Try It: Use calculus to aid you in sketching the graph of f (x) = x 3 + x.
EXAMPLE 3.3
Sketch the graph of g(x) = x + sin x.
SOLUTION
We see that g (x) = 1 + cos x. Since ’1 ¤ cos x ¤ 1, it follows that
g (x) ≥ 0. Hence the graph of g is always increasing.
Now g (x) = ’ sin x. This function is positive sometimes and negative
sometimes. In fact
’ sin x is positive when kπ < x < (k + 1)π, k odd
84 CHAPTER 3 Applications of the Derivative




Fig. 3.5


and
’ sin x is negative when kπ < x < (k + 1)π, k even.
So the graph alternates being concave down and concave up. Of course it also
passes through the origin. We amalgamate all our information in the graph
shown in Fig. 3.6.



6

4

2

_F F 2F
_2




Fig. 3.6


EXAMPLE 3.4
Sketch the graph of h(x) = x/(x + 1 ).
85
CHAPTER 3 Applications of the Derivative

SOLUTION
First note that the function is unde¬ned at x = ’1.
We calculate that h (x) = 1/((x + 1)2 ). Thus the graph is everywhere
increasing (except at x = ’1).
We also calculate that h (x) = ’2/((x + 1)3 ). Hence h > 0 and the graph
is concave up when x < ’1. Likewise h < 0 and the graph is concave down
when x > ’1.
Finally, as x tends to ’1 from the left we notice that h tends to +∞ and as
x tends to ’1 from the right we see that h tends to ’∞.
Putting all this information together, we obtain the graph shown in Fig. 3.7.




Fig. 3.7


You Try It: Sketch the graph of the function k(x) = x · x + 1.
EXAMPLE 3.5
Sketch the graph of k(x) = x 3 + 3x 2 ’ 9x + 6.
SOLUTION
We notice that k (x) = 3x 2 + 6x ’ 9 = 3(x ’ 1)(x + 3). So the sign of k
changes at x = 1 and x = ’3. We conclude that
k is positive when x < ’3;
k is negative when ’3 < x < 1;
k is positive when x > 3.
Finally, k (x) = 6x + 6. Thus the graph is concave down when x < ’1 and
the graph is concave up when x > ’1.
Putting all this information together, and using the facts that k(x) ’ ’∞
when x ’ ’∞ and k(x) ’ +∞ when x ’ +∞, we obtain the graph shown
in Fig. 3.8.
86 CHAPTER 3 Applications of the Derivative




Fig. 3.8




3.2 Maximum/Minimum Problems
One of the great classical applications of the calculus is to determine the maxima
and minima of functions. Look at Fig. 3.9. It shows some (local) maxima and (local)
minima of the function f .




Fig. 3.9

Notice that a maximum has the characterizing property that it looks like a hump:
the function is increasing to the left of the hump and decreasing to the right of the
hump. The derivative at the hump is 0: the function neither increases nor decreases
87
CHAPTER 3 Applications of the Derivative

at a local maximum. This is sometimes called Fermat™s test. Also, we see that the
graph is concave down at a local maximum.
It is common to refer to the points where the derivative vanishes as critical points.
In some contexts, we will designate the endpoints of the domain of our function to
be critical points as well.
Now look at a local minimum. Notice that a minimum has the characterizing
property that it looks like a valley: the function is decreasing to the left of the valley
and increasing to the right of the valley. The derivative at the valley is 0: the function
neither increases nor decreases at a local minimum. This is another manifestation
of Fermat™s test. Also, we see that the graph is concave up at a local minimum.
Let us now apply these mathematical ideas to some concrete examples.
EXAMPLE 3.6
Find all local maxima and minima of the function k(x) = x 3 ’ 3x 2 ’ 24x + 5.
SOLUTION
We begin by calculating the ¬rst derivative:
k (x) = 3x 2 ’ 6x ’ 24 = 3(x + 2)(x ’ 4).
We notice that k vanishes only when x = ’2 or x = 4. These are the only
candidates for local maxima or minima. The second derivative is k (x) =
6x ’ 6. Now k (4) = 18 > 0, so x = 4 is the location of a local minimum.
Also k (’2) = ’18 < 0, so x = ’2 is the location of a local maximum.
A glance at the graph of this function, as depicted in Fig. 3.10, con¬rms our
calculations.
EXAMPLE 3.7
Find all local maxima and minima of the function g(x) = x + sin x.
SOLUTION
First we calculate that
g (x) = 1 + cos x.
Thus g vanishes at the points (2k + 1)π for k = . . . , ’2, ’1, 0, 1, 2, . . ..
Now g (x) = sin x. And g ((2k + 1)π ) = 0. Thus the second derivative test
is inconclusive. Let us instead look at the ¬rst derivative. We notice that it is
always ≥ 0. But, as we have already noticed, the ¬rst derivative changes sign at
a local maximum or minimum. We conclude that none of the points (2k + 1)π
is either a maximum nor a minimum. The graph in Fig. 3.11 con¬rms this
calculation.
You Try It: Find all local maxima and minima of the function g(x) = 2x 3 ’
15x 2 + 24x + 6.
88 CHAPTER 3 Applications of the Derivative




Y
FL
AM

Fig. 3.10
TE



6

4

2

_F F 2F
_2




Fig. 3.11


EXAMPLE 3.8
A box is to be made from a sheet of cardboard that measures 12 — 12 .
The construction will be achieved by cutting a square from each corner of
89
CHAPTER 3 Applications of the Derivative

the sheet and then folding up the sides (see Fig. 3.12). What is the box of
greatest volume that can be constructed in this fashion?




Fig. 3.12

SOLUTION
It is important in a problem of this kind to introduce a variable. Let x be the
side length of the squares that are to be cut from the sheet of cardboard. Then
the side length of the resulting box will be 12 ’ 2x (see Fig. 3.13). Also the
height of the box will be x. As a result, the volume of the box will be
V (x) = x · (12 ’ 2x) · (12 ’ 2x) = 144x ’ 48x 2 + 4x 3 .
Our job is to maximize this function V .



x



12 _ 2x

Fig. 3.13

Now V (x) = 144 ’ 96x + 12x 2 . We may solve the quadratic equation
144 ’ 96x + 12x 2 = 0
to ¬nd the critical points for this problem. Using the quadratic formula, we
¬nd that x = 2 and x = 6 are the critical points. Now V (x) = ’96 + 24x.
Since V (2) = ’48 < 0, we conclude that x = 2 is a local maximum for the
problem. In fact we can sketch a graph of V (x) using ideas from calculus and
see that x = 2 is a global maximum.
We conclude that if squares of side 2 are cut from the sheet of cardboard
then a box of maximum volume will result.
Observe in passing that if squares of side 6 are cut from the sheet then (there
will be no cardboard left!) the resulting box will have zero volume. This value
for x gives a minimum for the problem.
90 CHAPTER 3 Applications of the Derivative

EXAMPLE 3.9
A rectangular garden is to be constructed against the side of a garage. The
gardener has 100 feet of fencing, and will construct a three-sided fence;
the side of the garage will form the fourth side. What dimensions will give
the garden of greatest area?

SOLUTION
Look at Fig. 3.14. Let x denote the side of the garden that is perpendicular
to the side of the garage. Then the resulting garden has width x feet and length
100 ’ 2x feet. The area of the garden is
A(x) = x · (100 ’ 2x) = 100x ’ 2x 2 .



x

garage




100 _ 2x

x


Fig. 3.14

We calculate A (x) = 100 ’ 4x and ¬nd that the only critical point for the
problem is x = 25. Since A (x) = ’4 for all x, we determine that x = 25 is
a local maximum. By inspection, we see that the graph of A is a downward-
opening parabola. So x = 25 must also be the global maximum that we seek.
The optimal dimensions for the garden are
width = 25 ft. length = 50 ft.

You Try It: Find the right circular cylinder of greatest volume that can be
contained in a sphere of radius 1.
EXAMPLE 3.10
The sum of two positive numbers is 60. How can we choose them so as to
maximize their product?

SOLUTION
Let x be one of the two numbers. Then the other is 60 ’ x. Their product is

P (x) = x · (60 ’ x) = 60x ’ x 2 .
91
CHAPTER 3 Applications of the Derivative

Thus P is the quantity that we wish to maximize. Calculating the derivative,
we ¬nd that
P (x) = 60 ’ 2x.
Thus the only critical point for the problem is x = 30. Since P (x) ≡ ’2, we
¬nd that x = 30 is a local maximum. Since the graph of P is a downward-
opening parabola, we can in fact be sure that x = 30 is a global maximum.
We conclude that the two numbers that add to 60 and maximize the product
are 30 and 30.

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