Graphs of Functions and Derivatives 127(–4,6)(3,2)f(x)y10–1–2–3–4 2 3 xFigure 7.4-5Example 5
Iff(x)=
∣∣
ln(x+1)∣∣
, find limx→ 0 − f′(x). (See Figure 7.4-6.)[–2,5] by [–2,4]
Figure 7.4-6The domain offis (−1,∞).
f(0)=∣∣
ln(0+1)∣∣
=∣∣
ln(1)∣∣
= 0f(x)=∣∣
ln(x+1)∣∣
={
ln(x+1) ifx≥ 0
−ln(x+1) ifx< 0Thus,f′(x)=
⎧
⎪⎪⎨⎪⎪⎩1
x+ 1
ifx≥ 0−1
x+ 1
ifx< 0.
Therefore, limx→ 0 −f′(x)=xlim→ 0 −
(
−1
x+ 1)
=−1.