Question

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

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

Answer 1

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

Given$f\left(x\right)=2x|x|$

we know that $|x|=x\forall x>0,|x|=-x\forall x<0$

then

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

$f^1\left(x\right)=\begin{array}{cc}-4x& x<0\\ 4x& x>0\end{array}$
$f\left(x\right)$ is differentiable everywhere except $x=0$

So,

$f^1\left(0^{-}\right)=0$
$f^1\left(0^{+}\right)=0$
$\because f^1\left(0^{-}\right)=f^1\left(0^{+}\right)$ so $f\left(x\right)$ is differentiable at $x=0$
Whice mean f(x) is differentiable in $(-\infty ,\infty )$