Limits Continuity And Differentiability Question 49
Question: What is the set of all points, where the function $ f(x)=\frac{x}{1+| x |} $ is differentiable?
Options:
A) $ (-\infty ,\infty ) $ only
B) $ (0,\infty ) $ only
C) $ (-\infty ,0)\cup (0,\infty ) $ only
D) $ (-\infty ,0) $ only
Show Answer
Answer:
Correct Answer: A
Solution:
Given $ f(x)=\frac{x}{1+| x |} $ $ = \begin{vmatrix} \frac{x}{1-x},x<0 \\ \frac{x}{1+x},x\ge 0 \\ \end{vmatrix} . $
$ ( \because | x |= \begin{vmatrix} x & if & x\ge 0 \\ -x & if & x<0 \\ \end{vmatrix} . ) $
$ \therefore LHD=f’({0^{-}})=\underset{h\to 0}{\mathop{\lim }}\frac{f(0-h)-f(0)}{-h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{f(-h)-f(0)}{-h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{\frac{-h}{1+| -h |}-0}{-h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{\frac{-h}{1+h}-0}{-h}=\underset{h\to 0}{\mathop{\lim }}\frac{1}{1+h}=1 $ and RHD $ =f’({0^{+}}) $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{f(0+h)-f(0)}{h} $
$ =\underset{h\to 0}{\mathop{\lim }}\frac{\frac{h}{1+h}-0}{h}=\underset{h\to 0}{\mathop{\lim }}\frac{1}{1+h}=1 $ Since, LHD = RHD
$ \therefore f(x) $ is differentiable at $ x=0 $ Hence, $ f(x) $ is differentiable in $ (-\infty ,\infty ) $ .
BETA
