Question

The set of all point where the function $f\left(x\right)=2x|x|$ is differentiable is -

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

Answer 1

Answer: (A)

$(-\infty ,\infty )$

$f\left(x\right)=2x\left|x\right|=\{\begin{array}{c}{2x}^{2}x\ge 0\\ {-2x}^{2}x<0\end{array}$
Since ${2x}^2$ and $-{2x}^2$ are differentiable function
$f\left(x\right)$ is differentiable, except possibly at $x=0$
Now $f^{\prime }(0+)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}$ $[\because f\left(0\right)=0]$
$=\underset{h\to 0}{lim}\frac{{2h}^2}{h}=\underset{h\to 0+}{lim}2h=0$
and $f(0-)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0-}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$
$=\underset{h\to 0-}{lim}\frac{-{2h}^2}{h}=\underset{h\to 0}{lim}-2h=0$
Hence $f$ is differentiable everywhere.