Question

The set of all points, where the function $f\left(x\right)=\frac{x}{1+\left|x\right|}$ is differentiable, is

• A. $(-\infty ,\infty )$
• B. $(0,\infty )$
• C. $(-\infty ,0)\cup (0,\infty )$
• D. None of these

Answer 1

Answer: (A)

$(-\infty ,\infty )$

$f\left(x\right)=\frac{x}{1+\left|x\right|}=\{\begin{array}{c}\begin{array}{cc}\frac{x}{1-x},& x<0\end{array}\\ \begin{array}{cc}\frac{x}{1+x},& x\ge 0\end{array}\end{array}$

Let us check the differentiablility of $f\left(x\right)$ at $x=0$

$f\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{f(0-h)-f\left(0\right)}{-h}=\underset{h\to 0}{lim}\frac{\frac{0-h}{1-(0-h)}-0}{-h}$

$=\underset{h\to 0}{lim}\frac{-h}{(1+h)(-h)}=\underset{h\to 0}{lim}\frac{1}{1+h}=1$

$f\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{\frac{0+h}{1+(0+h)}-0}{h}$

$=\underset{h\to 0}{lim}\frac{1}{1+h}=1$

$\because f^{\prime }\left(0^{-}\right)=f^{\prime }\left(0^{+}\right)$

$\Rightarrow f\left(x\right)$ is differentiable at $x=0$

$\therefore f\left(x\right)$ is differentiable for all $x\in R$