# The set of points where the function f (x) = x |x| is differentiable is

Question:

The set of points where the function f (x) = x |x| is differentiable is

(a) $(-\infty, \infty)$

(b) $(-\infty, 0) \cup(0, \infty)$

(c) $(0, \infty)$

(d) $[0, \infty]$

Solution:

(a) $(-\infty, \infty)$

We have,

$f(x)=x|x|$

$\Rightarrow f(x)=\left\{\begin{array}{cc}-x^{2}, & x<0 \\ 0, & x=0 \\ x^{2}, & x>0\end{array}\right.$

When, $x<0$, we have

$f(x)=-x^{2}$ which being a polynomial function is continuous and differentiable in $(-\infty, 0)$

When, $x>0$, we have

$f(x)=x^{2}$ which being a polynomial function is continuous and differentiable in $(0, \infty)$

Thus possible point of non-differentiability of $f(x)$ is $x=0$

Now, $\operatorname{LHD}($ at $x=0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{x \rightarrow 0^{-}} \frac{-x^{2}-0}{x}$

$=\lim _{h \rightarrow 0} \frac{-(-h)^{2}}{-h}$

$=\lim _{h \rightarrow 0} h$

$=0$

And RHD $($ at $x=0)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{x \rightarrow 0^{+}} \frac{x^{2}-0}{x}$

$=\lim _{h \rightarrow 0} \frac{h^{2}}{h}$

$=\lim _{h \rightarrow 0} h$

$=0$

$\therefore \operatorname{LHD}($ at $x=0)=\mathrm{RHD}($ at $x=0)$

So, $f(x)$ is also differentiable at $x=0$

i. e. $f(x)$ is differentiable in $(-\infty, \infty)$