Question

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

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

Answer 1

Answer: (A)

$f\left(x\right)=\frac{x}{1-x}$ if $x\le 0$ and $f\left(x\right)=\frac{x}{1+x}$ if $x\ge 0$
At x = 0, left handed derivative
$={\lim }_{h\to 0}\frac{f\left(0-h\right)-f\left(0\right)}{-h}$
$={\lim }_{h\to 0}\frac{f(-h)-f(0)}{-h}={\lim }_{h\to 0}\frac{\frac{-h}{1+h}-0}{-h}$
$={\lim }_{h\to 0}\frac{1}{1+h}=1$
and right handed derivative
$={\lim }_{h\to 0}\frac{f\left(0+h\right)-f\left(0\right)}{h}$
$={\lim }_{h\to 0}\frac{f\left(h\right)-f\left(0\right)}{h}={\lim }_{h\to 0}\frac{\frac{h}{1+h}-0}{h}$
$={\lim }_{h\to 0}\frac{1}{1+h}=1$
$\therefore l.h.$derivative = r.h. derivative at x = 0
Hence f(x) is differentiable at x = 0
Further $f^{\prime }(x)=\frac{1}{(1-x{)}^2}$ , where $x\le 0$
and $f^{\prime }(x)=\frac{1}{(1+x{)}^2}$, where $x\ge 0$
Thus $f^{\prime }(x)$ exists for all values of x in the interval $(-\infty ,\infty )$