Let $f:\mathbb R \to \mathbb R$ defined by $f(x) = x|x|$, Is the function continous at all points? If it is, then is it differentiable at all points?

Yes, the function is continuous everywhere but there is a slight confusion in differentiability. What I tried is,

$$\large{f(x) = \begin{cases} x^2 & x\geq 0 \\ -x^2 & x \leq 0 \end{cases}}$$

Then I applied left hand derivative limit and right hand derivative limit at $0$, then LDL comes to be $-x$ and RDL comes to be $x$, then the function is not differentiable at $0$. Am i right? and is the function not differentiable at some other points too.

  • $\begingroup$ the function is once, but, not twice differentiable at zero. $\endgroup$ – James S. Cook May 19 '15 at 6:37

As you wrote, it is continuous everywhere. But it's also differentiable everywhere.

You might think that it's not differentiable because $|x|$ is not differentiable at $x=0$, where the graph makes a sharp turn.

But remember that you're dealing with $x|x|$, so as $f(x)$ approaches $0$, $x|x|$ approaches from the negative left (since $x$ is negative, $x|x|$ is too), and $x|x|$ approaches from the positive right. So there's no sharp turn that would cause it to be non-differentiable. Really, a picture paints a thousand words. Just plot $f(x)$ to see that it is clearly differentiable:

enter image description here

  • $\begingroup$ No problem. I just tried to supply the intuitition. You can make your proof rigorous by using @learnmore's both-sided limits. $\endgroup$ – Newb May 19 '15 at 6:57

RHD:$\lim _{h\to 0}\dfrac{f(h)-f(0)}{h-0}=\lim_{h\to 0}\dfrac{h^2}{h}=0$

LHD:$\lim _{h\to 0}\dfrac{f(0)-f(-h)}{h}=\lim_{h\to 0}\dfrac{-h^2}{h}=0$

Both are equal and hence differentiable

  • $\begingroup$ Yep...I forgot to apply the limit.... $\endgroup$ – Sam Christopher May 19 '15 at 6:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.