Question

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

• A. $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$
• B. $\left(-\infty ,\infty \right)$
• C. $\left(0,\infty \right)$
• D. $\left(-\infty ,0\right)\cup \left(0,\infty \right)$

Answer 1

Answer: (B)

$\because f\left(x\right)=\frac{x}{1+|x|}$
Let $f\left(x\right)=\frac{g\left(x\right)}{h\left(x\right)}=\frac{x}{1+|x|}$
It is clear that $g(x)=x$ and $h(x)=1+|x|$ are differentiable on $\left(-\infty ,\infty \right)$ and $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ respectively.
Thus $f(x)$ is differentiable on $\left(-\infty ,0\right)\cup \left(0,\infty \right).$ For $x=0$
$\underset{x\to 0}{\lim }$$\frac{f\left(x\right)-f\left(0\right)}{x-0}=$ $\underset{x\to 0}{\lim }$$\frac{\frac{x}{1+|x|}}{x}$
$\underset{x\to 0}{\lim }$$\frac{1}{1+|x|}=1$
Thus $f(x)$ is differentiable on $\left(-\infty ,\infty \right).$