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Problem: Let $f: \mathbb R -> \mathbb R$ be given by $f(x):= x.|x|$. Show that $f$ is continuous and differentiable on $\mathbb R$.

My solution:

By using the Differentialquotient we prove that $f$ is differentiable $ lim_{x\to x_0} = \left(\frac{ x.|x| - (x_0 .|x_0|)} {x-x_0}\right) =\left(\frac{(x-x_0).(|x|-|x_0|)} {x-x_0}\right)=|x|-|x_0|=|x|$

And as the function is differentiable, we know that then the function is continuous $ lim_{x\to x_0}= (x.|x|) - (x_0 .|x_0|)= ( |x|- 0) - (x -0)=|x|-x =0$

So is my solution right? And if you want to prove that a function is differentiable is it enough to do it this way? Or for my case do you have to prove that the function is differentiable at $x<0$, $x>0$ and $x=0$ in order to prove that the function is at whole differentiable on $\mathbb R$ ?

So if I prove if the function is differentiable at $x<0$, $x>0$ and $x=0$.

For $x>0$ I get $ lim_{x\to x_0}=\left(\frac{-x^2+x_0^2} {x-x_0}\right)=-(x+x_0)=-x$

For $x>0$ I get $ lim_{x\to x_0}=\left(\frac{x^2-x_0^2} {x-x_0}\right)=(x+x_0)=x$

For $x=0$ I get $ lim_{x\to x_0}=\left(\frac{0-0} {x-x_0}\right)=0$

Kai
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    How is: $$ \left(\frac{ x.|x| - (x_0 .|x_0|)} {x-x_0}\right) =\left(\frac{(x-x_0).(|x|.|x_0|)} {x-x_0}\right) $$ ? – Clement Yung Jun 27 '20 at 10:14
  • Ops, you are right.I will correct it. – Kai Jun 27 '20 at 10:15
  • To prove that a function is differentiable on $\mathbb{R}$ you must, by definition, prove that it is differentiable at any $x\in\mathbb{R}$. Here, you can distinguish the case where $x\neq 0$ and the case where $x=0$. – charlus Jun 27 '20 at 10:21
  • @charlus So, if it exist a limit for x<0,x>0 and x=0 then f is differentiable , however the limits at these points doesn't have to be equal? And my prove is then not enogh ? – Kai Jun 27 '20 at 10:26
  • $f$ is differentible on $A$ if $f$ is differntible at every point in $A$. So the answer is yess. – VN_nmd Jun 27 '20 at 10:26
  • I think that you are confusing two things: 1) differentiability at $0$ 2) continuity of the differential at $0$. Can you see the difference between the two? – charlus Jun 27 '20 at 10:27
  • not really.. I think for differentiablity of function you alway prove it at 0, and if a function us then at every point differentiable then schould be continual – Kai Jun 27 '20 at 10:42

2 Answers2

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As your function $f(x)=x|x|$ is odd $f(-x)=-f(x)$, then it is enough to consider the case $x \geqslant 0$. For $x>0$ it is $f(x)=x^2$ well known differentiable function, so we need to consider definition only of $x=0$ $$\lim _{x \to 0}\frac{x|x|}{x} = \lim _{x \to 0}|x| = 0$$.

zkutch
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For $x<0$ (resp. $x>0$) the function is $-x^2$ (resp. $x^2$) which are differentiable*. The only doubt is about $x=0$. But

$$\lim_{h\to0}\frac{|h|h}{h}=\lim_{h\to0}|h|=0.$$

*You can always find neighborhoods of $x$ that keep a constant sign.