Functions Question 597
Question: The set of all those points, where the function $ f(x)=\frac{x}{1+|x|} $ is differentiable, is
Options:
A) $ (-\infty ,\infty ) $
B) $ [0,\infty ] $
C) $ (-\infty ,,0)\cup (0,\infty ) $
D) $ (0,\infty ) $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ h(x)=x,x\in (-\infty ,\infty ) $ ; $ g(x)=1+|x|,x\in (-\infty ,\infty ) $ Here h is differentiable in $ (-\infty ,\infty ) $ but $ |x| $ is not differentiable at $ x=0 $ . Therefore g is differentiable in $ (-\infty ,0)\cup (0,\infty ) $ and $ g(x)\ne 0,x\in $ $ (-\infty ,\infty ) $ , therefore $ f(x)=\frac{h(x)}{g(x)}=\frac{x}{1+|x|} $ It is differentiable in $ (-\infty ,0)\cup (0,\infty ) $ for $ x=0 $ $ \underset{h\to 0}{\mathop{\lim }},\frac{f(h)-f(0)}{h-0}=\underset{h\to 0}{\mathop{\lim }},\frac{\frac{h}{1+|h|}-0}{h}=\underset{h\to 0}{\mathop{\lim }},\frac{1}{1+|h|}=1 $ Therefore f is differentiable at $ x=0 $ , so f is differentiable in $ (-\infty ,\infty ) $ .
BETA
