Question

The set of all those points, where the function

$f\left(x\right)=\frac{x}{1+|x|}$

is differentiable, is

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

Answer 1

The correct option is A

$(-\infty ,\infty )$

Let $h\left(x\right)=x,xϵ(-\infty ,\infty );g\left(x\right)=1+|x|,xϵ(-\infty ,\infty )$
Here h is differentiable in $(-\infty ,\infty )$ but $|x|$ is not differentiable at $x=0$.
Therefore g is differentiable in $(-\infty ,0)\cup (0,\infty )\text{and}g\left(x\right)\ne 0,\forall x\in (-\infty ,\infty ).$

$\therefore f\left(x\right)=\frac{h\left(x\right)}{g\left(x\right)}=\frac{x}{1+|x|}$
It is differentiable in $(-\infty ,0)\cup (0,\infty )$ for x = 0
$\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h-0}=\underset{h\to 0}{lim}\frac{\frac{h}{1+|h|}}{h}=\underset{h\to 0}{lim}\frac{1}{1+|h|}=1$
Therefore f is differentiable at x = 0, so f is differentiable in $(-\infty ,\infty ).$