Question

The set of all 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. $(-\infty ,0)$

Answer 1

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

$f\left(x\right)=\frac{x}{1+|x|}$
For $x\ge 0,f\left(x\right)=\frac{x}{1+x}$
For $x<0,f\left(x\right)=\frac{x}{1-x}$
Since the function changes its definition at $x=0$, we need to check for continuity and differentiablity only at $x=0$
Clearly,
$f\left(0^{+}\right)=f\left(0^{-}\right)=f\left(0\right)=0$
Hence $f\left(x\right)$ is continuous at $x=0$

Now,
$R.H.D.=f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{f(0+h)-f\left(0\right)}{h}\Rightarrow f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{\frac{h}{1+h}}{h}\Rightarrow f^{\prime }\left(0^{+}\right)=\frac{1}{1+h}=1L.H.D.=f^{\prime }\left(0^{-}\right)=\underset{h\to 0^{+}}{lim}\frac{f(0-h)-f\left(0\right)}{-h}$
$\Rightarrow f^{\prime }\left(0^{+}\right)=\underset{h\to 0^{+}}{lim}\frac{\frac{-h}{1-h}}{-h}=1\Rightarrow L.H.D.=R.H.D.$
So, $f\left(x\right)$ is differentiable at $x=0$
Hence, $f\left(x\right)$ for all $x\in R$