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

280 Chapter 4 Applications of Differentiation
y
0
y=˛
x
Example 5 If f sxd − x 3 , then f 9sxd − 3x 2 , so f 9s0d − 0. But f has no maximum
or minimum at 0, as you can see from its graph in Figure 12. (Or observe that x 3 . 0
for x . 0 but x 3 , 0 for x , 0.) The fact that f 9s0d − 0 simply means that the curve
y − x 3 has a horizontal tangent at s0, 0d. Instead of having a maximum or minimum at
s0, 0d, the curve crosses its horizontal tangent there. n
Figure 12
If f sxd − x 3 , then f 9s0d − 0,
but f has no maximum
or minimum.
y
0
y=|x|
Figure 13
If f sxd − | x | , then f s0d − 0 is
a minimum value, but f 9s0d does
not exist.
x
Example 6 The function f sxd − | x | has its (local and absolute) minimum value at
0, but that value can’t be found by setting f 9sxd − 0 because, as was shown in Example
2.8.5, f 9s0d does not exist. (See Figure 13.) n
warning Examples 5 and 6 show that we must be careful when using Fermat’s Theorem.
Example 5 demonstrates that even when f 9scd − 0 there need not be a maximum
or minimum at c. (In other words, the converse of Fermat’s Theorem is false in general.)
Fur thermore, there may be an extreme value even when f 9scd does not exist (as in
Example 6).
Fermat’s Theorem does suggest that we should at least start looking for extreme values
of f at the numbers c where f 9scd − 0 or where f 9scd does not exist. Such numbers
are given a special name.
6 Definition A critical number of a function f is a number c in the domain of
f such that either f 9scd − 0 or f 9scd does not exist.
Figure 14 shows a graph of the function
f in Example 7. It supports our answer
because there is a horizontal tangent
when x − 1.5 fwhere f 9sxd − 0g and
a vertical tangent when x − 0 fwhere
f 9sxd is undefinedg.
3.5
Example 7 Find the critical numbers of f sxd − x 3y5 s4 2 xd.
SOLUtion The Product Rule gives
f 9sxd − x 3y5 s21d 1 s4 2 xd( 3 5 x22y5 ) − 2x 3y5 1
−
25x 1 3s4 2 xd
5x 2y5
−
12 2 8x
5x 2y5
3s4 2 xd
5x 2 y5
_0.5 5
_2
Figure 14
[The same result could be obtained by first writing f sxd − 4x 3y5 2 x 8y5 .] Therefore
f 9sxd − 0 if 12 2 8x − 0, that is, x − 3 2 , and f 9sxd does not exist when x − 0. Thus the
critical numbers are 3 2 and 0.
n
In terms of critical numbers, Fermat’s Theorem can be rephrased as follows (compare
Definition 6 with Theorem 4):
7 If f has a local maximum or minimum at c, then c is a critical number of f.
To find an absolute maximum or minimum of a continuous function on a closed
interval, we note that either it is local [in which case it occurs at a critical number by (7)]
or it occurs at an endpoint of the interval, as we see from the examples in Figure 8. Thus
the following three-step procedure always works.
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