Question

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

• A. $\left(-\infty ,\infty \right)$
• B. $[0,\infty )$
• C. $\left(-\infty ,0\right)\cup \left(0,\infty \right)$
• D. $\left(0,\infty \right)$

Answer 1

The correct option is A $\left(-\infty ,\infty \right)$

$f\left(x\right)=\frac{x}{1+\left|x\right|}$ is differentiable everywhere except probably at x = 0.
For x = 0,
$\left(\mathrm{LHD}\mathrm{at}x=0\right)=\underset{h\to 0}{\lim }\frac{f(0-h)-f\left(0\right)}{-h}=\underset{h\to 0}{\lim }\frac{\frac{-h}{1+h}-0}{-h}=1\left(\mathrm{RHD}\mathrm{at}x=0\right)=\underset{h\to 0}{\lim }\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{\lim }\frac{\frac{h}{1+h}-0}{h}=1\left(\mathrm{LHD}\mathrm{at}x=0\right)=\left(\mathrm{RHD}\mathrm{at}x=0\right)$
Therefore, f is differentiable at x = 0.
Hence, f is differentiable in $\left(-\infty ,\infty \right)$