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

yess it is

anonymous · Answer

can you prove it?

anonymous · Answer

$$for  x \neq 0$$
if x>0: f(x)=x : differetiable in R+
if x<0 : f(x)=-x : differentiable in R-

anonymous · Answer

ok thanks for consideration, but i need more detailed solution so that my mentor accepts it, if you ask me about this question, i have no idea how to do it...

zzr0ck3r · Answer

sec let me think of how to show

anonymous · Answer

ok..

zzr0ck3r · Answer

do you know about the difference quotent?

anonymous · Answer

no unfortunately

zzr0ck3r · Answer

if you take the left and right side limits of the quotent rule you will get two different numbers, nad thus no tanget line can be drawn at 0. so it is not differentiable at 0

anonymous · Answer

ok...

zzr0ck3r · Answer

the quotent rule is the standard way of obtaining a derviative
its lim H->0 (f(x+h) - f(x))/h this would take to long to explaine why, but think about derivatives as slopes and to find a slope we do (y_1-y_o)/(x_1-x_0) its sort of like that:)

zzr0ck3r · Answer

who expects you to explane all this calc stuff if you have never seen where it comes from?

anonymous · Answer

our universtiy is like that, they show less and expects much... unfortunately.. and i am not participating in lectures this year, because i had same lecture in same professour 2 or 3 years before