James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 4.3 How Derivatives Affect the Shape of a Graph 293
y
A
C
0 x
FIGURE 1
Let’s abbreviate the name of this test to
the I/D Test.
B
D
Many of the applications of calculus depend on our ability to deduce facts about a function
f from information concerning its derivatives. Because f 9sxd represents the slope of
the curve y − f sxd at the point sx, f sxdd, it tells us the direction in which the curve proceeds
at each point. So it is reasonable to expect that information about f 9sxd will provide
us with information about f sxd.
What Does f 9 Say About f ?
To see how the derivative of f can tell us where a function is increasing or decreasing,
look at Figure 1. (Increasing functions and decreasing functions were defined in Section
1.1.) Between A and B and between C and D, the tangent lines have positive slope and
so f 9sxd . 0. Between B and C, the tangent lines have negative slope and so f 9sxd , 0.
Thus it appears that f increases when f 9sxd is positive and decreases when f 9sxd is negative.
To prove that this is always the case, we use the Mean Value Theorem.
Increasing/Decreasing Test
(a) If f 9sxd . 0 on an interval, then f is increasing on that interval.
(b) If f 9sxd , 0 on an interval, then f is decreasing on that interval.
Proof
(a) Let x 1 and x 2 be any two numbers in the interval with x 1 , x 2 . According to the
definition of an increasing function (page 19), we have to show that f sx 1 d , f sx 2 d.
Because we are given that f 9sxd . 0, we know that f is differentiable on fx 1 , x 2 g. So,
by the Mean Value Theorem, there is a number c between x 1 and x 2 such that
1 f sx 2 d 2 f sx 1 d − f 9scdsx 2 2 x 1 d
Now f 9scd . 0 by assumption and x 2 2 x 1 . 0 because x 1 , x 2 . Thus the right side of
Equation 1 is positive, and so
f sx 2 d 2 f sx 1 d . 0 or f sx 1 d , f sx 2 d
This shows that f is increasing.
Part (b) is proved similarly.
n
ExamplE 1 Find where the function f sxd − 3x 4 2 4x 3 2 12x 2 1 5 is increasing
and where it is decreasing.
SOLUtion We start by differentiating f :
f 9sxd − 12x 3 2 12x 2 2 24x − 12xsx 2 2dsx 1 1d
_1 0 2 x
To use the IyD Test we have to know where f 9sxd . 0 and where f 9sxd , 0. To
solve these inequalities we first find where f 9sxd − 0, namely at x − 0, 2, and 21.
These are the critical numbers of f , and they divide the domain into four intervals
(see the number line at the left). Within each interval, f 9sxd must be always positive
or always negative. (See Examples 3 and 4 in Appendix A.) We can determine
which is the case for each interval from the signs of the three factors of f 9sxd, namely,
12x, x 2 2, and x 1 1, as shown in the following chart. A plus sign indicates that the
given expression is positive, and a minus sign indicates that it is negative. The last col-
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