Graphs of Functions and Derivatives 127
(–4,6)
(3,2)
f(x)
y
10–1–2–3–4 2 3 x
Figure 7.4-5
Example 5
Iff(x)=
∣∣
ln(x+1)
∣∣
, find limx→ 0 − f′(x). (See Figure 7.4-6.)
[–2,5] by [–2,4]
Figure 7.4-6
The domain offis (−1,∞).
f(0)=
∣∣
ln(0+1)
∣∣
=
∣∣
ln(1)
∣∣
= 0
f(x)=
∣∣
ln(x+1)
∣∣
=
{
ln(x+1) ifx≥ 0
−ln(x+1) ifx< 0
Thus,f′(x)=
⎧
⎪⎪⎨
⎪⎪⎩
1
x+ 1
ifx≥ 0
−
1
x+ 1
ifx< 0
.
Therefore, limx→ 0 −f′(x)=xlim→ 0 −
(
−
1
x+ 1
)
=−1.