Question

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

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

Answer 1

Answer: (B)

$(-\infty ,\infty )$

We have $f\left(x\right)=\frac{x}{1+\left|x\right|};x\in R$

$\Rightarrow f\left(x\right)=\{\frac{x}{1+x};x\ge 0\frac{x}{1-x};x\le 0\}$

To check for differentiability of the above function we start off by checking for differentiability of $f\left(x\right)$ at $x=0$ since the functional definition is altered at $x=0.$ So let's find the right hand derivative (RHD) and left hand derivative (LHD) at $x=0$

RHD :

${lim}_{x\to 0^{+}}\frac{f\left(x\right)-f\left(0\right)}{x-0}$

${lim}_{x\to 0^{+}}\left(\frac{\frac{x}{1+x}-\frac{0}{1+0}}{x}\right)={lim}_{x\to 0^{+}}\frac{x}{(1+x)x}={lim}_{x\to 0^{+}}\frac{1}{1+x}=1$

LHD :

${lim}_{x\to 0^{-}}\frac{f\left(x\right)-f\left(0\right)}{x-0}{lim}_{x\to 0^{-}}\left(\frac{\frac{x}{1-x}-\frac{0}{1-0}}{x}\right)={lim}_{x\to 0^{-}}\frac{x}{(1-x)x}={lim}_{x\to 0^{-}}\frac{1}{1-x}=1$

LHD $=$ RHD

The function is differentiable at $x=0$

Also, $f\left(x\right)=\frac{x}{1+x}$ is a well defined rational function of 'x' for all x greater than zero and hence will be continuous and differentiable through out its domain.

Similarly, $f\left(x\right)=\frac{x}{1+x}$ is a well defined rational function of 'x' for all 'x' less than zero and hence will be continuous ans differentiable through out its domain.

So, we can conclude from the above calculation and discussion that $f\left(x\right)=\frac{x}{1+\left|x\right|}$ is differentiable everywhere.

$\therefore f\left(x\right)=\frac{x}{1+\left|x\right|}$ is differentiable in $x\in (-\infty ,\infty )$

Hence, solved.