Let $f(x) = (1/2)|x|$
a). What is the value of $f′(x)$ on $(−∞,0)$?
b). What is the value of $f′(x)$ on $(0,∞)$?
c). What is the value of $|f′(x)|$ on $(−∞, 0) ∪ (0, ∞)$
d). By part c). we can conclude that $|f(x)−f(y)|/|x-y| ≤ 1/2$ for all $x,y ∈ R$. So can you conclude $f$ is a contraction?
e). Is there a unique fixed point of $f$? If so, what is it?
Partial answers and hint:
You have $f(x)-f(y)=\frac{1}{2}\left( |x|-|y| \right)$ then $$|f(x)-f(y)|=\frac{1}{2}||x|-|y||\le \frac{1}{2}|x-y|,$$ so $f$ is a contractive map.
Also observe that $f(0)=0$.
For $x>0$ you have $f(x)=\frac{x}{2}$ and $f'(x)=\frac{1}{2}$.