How to prove that $f$ is differentiable at every point beside $x=-1$?

Question

Consider $f(x) = |x+1|$.
I want to show that for every $x_0\neq-1$ , $\lim_{h \rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h}$ is exist.
So far i just wrote the definition, and looked on 2 options:When $x>-1$ and $x<-1$.
But, i got stuck when i go the definiton of differentiabily in those points, because what the meaning of $\lim_{h \rightarrow 0}\frac{|x_0+h+1|- |x_0+1|}{h}$?
tnx!

Answer 1

You need to break the problem into cases.

First suppose $x_0 +1 > 0$.

By definition the absolute value of a positive number is just the number itself so we have $\vert x_0+1\vert = x_0+1.$

The term $\vert x_0 +h+1\vert$ is a little more subtle. $h$ can be negative so its not impossible that $x_0+h+1<0$. We need to remember that ultimately $h\rightarrow 0$ which gives us the freedom to require that $\vert h \vert \leq \vert x_0+1\vert$. With this requirement we can be assured that $x_0+h+1>0$ and therefore $\vert x_0+h+1\vert = x_0+h+1$.

Now we can turn to the difference qoutient and see that things couldn't be simpler.

$$\frac{\vert x_0+h+1\vert - \vert x_0+1 \vert}{h} = \frac{(x_0+h+1) - (x_0+1)}{h} = \frac{h}{h} = 1$$

We could handle the case for $x_0+1 < 0$ in a similar manner.

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Now lets look at what happens at $x_0=-1$. In this case our difference quotient becomes,

$$\frac{\vert h \vert}{h}.$$

This quotient will be either $+1$ if $h>0$ or $-1$ if $h<0$, which means that the value of the limit depends on which side $h$ approaches $0$ from. This means that the limit in question doesn't exist and that the derivative is not defined for $x_0=-1$.

Answer 2

if $x \neq 1,$ the graph of $f$ is a line with slope $1$ if $x > 1$ and slope $-1.$

if you want to use the definition of the derivative, for $x > 1,$ you want to use $f(x) = x + 1$ without any of the absolute value symbol. the case $x < 1$ should be similar.