Differentiability of an absolute function.

Question

Check the differentiability of $f(x)=x|x|$, $x$ is in $\mathbb{R}$. I know that it is differentiable when $x>0$ and $x<0$. I am not sure about the case when $x=0$. I found that as $$\lim \limits_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ exists and equal to $0$. But still something seems wrong. Is function differentiable at $0$? Thank you.

Answer 1

You are not wrong. The reason is that multiplication by $x$ helps your function become `smoother' at 0. Here are some graphs:

the green line is your function: you can see that it matches exactly the black $x^2$ graph for $x>0$ and is a reflection across the $x$-axis for $x<0$. It has a similar shape to $x^3$ (drawn in blue), but the derivative of $f$ is the black angular graph, $$f'(x) = \begin{cases} 2x & x≥0 \\ -2x & x<0\end{cases}$$ So the function multiplied by $x$ is differentiable, but its derivative is not. In contrast, $x^n$ is infinitely often differentiable, so the $|x|$ term does remove some 'smoothness'.