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

156 Chapter 2 Limits and Derivatives
ExamplE 5 Where is the function f sxd − | x | differentiable?
SOLUTION If x . 0, then | x |
and hence | x 1 h | − x 1 h. Therefore, for x . 0, we have
f 9sxd − lim
h l 0
− x and we can choose h small enough that x 1 h . 0
| x 1 h | 2 | x |
h
h
− lim
h l 0 h − lim 1 − 1
h l 0
sx 1 hd 2 x
− lim
h l 0 h
and so f is differentiable for any x . 0.
Similarly, for x , 0 we have | x | − 2x and h can be chosen small enough that
− 2sx 1 hd. Therefore, for x , 0,
x 1 h , 0 and so | x 1 h |
f 9sxd − lim
| x 1 h | 2 | x | 2sx 1 hd 2 s2xd
− lim
h l 0 h
h l 0 h
2h
− lim
h l 0 h
− lim s21d − 21
h l 0
and so f is differentiable for any x , 0.
For x − 0 we have to investigate
y
f 9s0d − lim
h l 0
f s0 1 hd 2 f s0d
h
− lim
| 0 1 h | 2 | 0 |
− lim
| h |
h l 0 h
hl0 h
(if it exists)
0
(a) y=ƒ=|x|
y
1
x
Let’s compute the left and right limits separately:
and
lim
| h |
h l0 1 h
− lim h
h l0 1 h − lim 1 − 1
h l0 1
lim
| h |
h l0 2 h
− lim 2h
h l0 2 h
− lim s21d − 21
h l02 Since these limits are different, f 9s0d does not exist. Thus f is differentiable at all x
except 0.
A formula for f 9 is given by
0
_1
x
f 9sxd −H 1 21
if x . 0
if x , 0
FIGURE 5
(b) y=fª(x)
and its graph is shown in Figure 5(b). The fact that f 9s0d does not exist is reflected
geometrically in the fact that the curve y − | x | does not have a tangent line at s0, 0d.
[See Figure 5(a).]
n
Both continuity and differentiability are desirable properties for a function to have.
The following theorem shows how these properties are related.
4 Theorem If f is differentiable at a, then f is continuous at a.
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