Let $x \in \mathbb{R}$. Show by applying the definition, that the absolute value function$ f(x):= |x|$ is
not differentiable in $x_{0} = 0$, although it is everywhere continuous. Is f differentiable for $x_{0} \neq 0 $?

Question

Let $x \in \mathbb{R}$. Show by applying the definition, that the absolute value function$ f(x):= |x|$ is
not differentiable in $x_{0} = 0$, although it is everywhere continuous. Is f differentiable for $x_{0} \neq 0 $?

anonymous · Answer

You have to distinguish two cases, x>0 and x <0. if x> 0, the derivative is 1 if x<0, the derivative is -1.

anonymous · Answer

Yeah F is differentiable for $$_{?}\neq0 with _{=0}$$ because f(x)=$$\left| x \right|$$ means f(x)=x when x is positive and -x when x is negative

anonymous · Answer

$$
\lim_{x\to 0} \frac { f(x) -f(0)}{x-0}=\lim_{x\to 0} \frac {|x|}{x}=\pm 1
$$
depending if x >- or x <0
The limit does not exist and f is not differentiable at zero.

anonymous · Answer

But f is differentiable at any other x different from 0.

anonymous · Answer

ok Mr Elias it looks very good, this is our end solution right?

anonymous · Answer

yes

anonymous · Answer

Thank you Mr @eliassaab very much and thank you too @pape

anonymous · Answer

yw

anonymous · Answer

you welcome