The function $f(x) =|x^3|$ is

differentiable everywhere

The range of the function x³ is (– ∞, ∞), and the range of f(x) is [0, ∞), f is clearly differentiable except possibly at the point x = 0.

Now, clearly by definition Rf'(0) = Lf'(0) = 0

so that, f is differentiable at x = 0 and hence every where.

Hence (A) is the correct answer.

Alternative solution:

Here \(f(x)=\left\{\begin{array}-x^3, & x<0 \\ x^3, & x > 0\end{array}\right.\)

so that \(f'(x)=\left\{\begin{array}-3x^2, & x<0 \\ 3x^2, & x > 0\end{array}\right.\)

⇒ the function is differentiable everywhere including x = 0.