No differentiable function on $[-1,1]$ with derivative $\operatorname{sgn}(x)$ How can I prove for certain that there exists no differentiable function $g(x)\colon [-1,1]\rightarrow \mathbb{R}$ with
$$g'(x) = \operatorname{sgn}(x) = \begin{cases} +1,  & \text{for $x > 0$} \\
0, & \text{for $x=0$} \\ -1, & \text{for $x<0$} \end{cases} $$
I'm currently jumping to the absolute value function but that's not differentiable at $x = 0$ and that doesn't exactly disprove the existence of others.
Looking at the limits
\begin{align}
\lim_{h\to 0^{+}} g'(h) &=1\\
&= \lim_{h\to 0^{+}} \frac{g(0+h)-g(0)}{h}\\
&\neq  \lim_{h\to 0^{-}}\frac{g(0+h)-g(0)}{h}\\
&= -1\\
&= \lim_{h\to 0^{-}}g'(h)\\
&\neq 0\\
&= g'(0)
\end{align}
which implies discontinuity at $h = 0$.
Continuity and differentiability are topics that I still don't quite grasp properly. Can I use the contraposition from Darboux's Theorem for this specific case?

Assumed a g(x) exists.
1)Solution for $x>0:$ $g(x)=x +a;$
2)Solution for $x <0:$ $g(x)=-x+b;$
Since differentiability implies continuity: $a=b=0;$
$3)\lim_{x \rightarrow \pm 0} \dfrac{g(x)-0}{x}$ does not exist,
hence not differentiable at $0$.
