Question

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

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

Answer 1

Answer: (A), (B), (C), (D)

$\left(0,\infty \right)$

$f\left(x\right)=\frac{x}{1+\left|x\right|}=\left(\begin{array}{c}\frac{x}{1+x},\mathrm{if}x\ge 0\\ \frac{x}{1-x},\mathrm{if}x<0\end{array}\mathrm{Looking}\mathrm{at}f\right(x),\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}f(x)\mathrm{is}\mathrm{differentiable}\mathrm{at}\mathrm{every}\mathrm{where}\mathrm{except}0,\mathrm{where}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{check}\mathrm{its}\mathrm{differentiability}.(\mathrm{LHD}\mathrm{at}x=0)=\underset{h\to 0^{-}}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{\frac{-h}{1+h}-0}{-h}=\underset{h\to 0^{-}}{\lim }\frac{1}{1+h}=1(\mathrm{RHD}\mathrm{at}x=0)=\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{\frac{h}{1+h}-0}{h}=\underset{h\to 0^{+}}{\lim }\frac{1}{1+h}=1\because (\mathrm{LHD}\mathrm{at}x=0)=(\mathrm{RHD}\mathrm{at}x=0)\therefore f\mathrm{is}\mathrm{differentiable}\mathrm{at}x=0.\mathrm{Hence},f\mathrm{is}\mathrm{differentiable}\mathrm{in}(-\infty ,\infty )$