Let f(x) = x|x|. Then f(x) is:

The correct answer is Option (3) - Both Continuous and Differentiable at x = 0

$f(x)=\left\{\begin{matrix}x^2,&x≥0\\-x^2,&x<0\end{matrix}\right.$

$f(0)=0,\lim\limits_{x→0^-}(-x^2)=0, \lim\limits_{x→0^+}(x^2)=0$

f(x) is continuous

$f'(x)=\left\{\begin{matrix}2x,&x≥0\\-2x,&x<0\end{matrix}\right.$

$LHD=\lim\limits_{x→0^-}(-2x)=0=\lim\limits_{x→0^+}(2x)=RHD$

f(x) is differentiable