g C1 in interval about $x_0, g(x_0) = 0$. if $g'(x_0) \neq 0$, there's a small interval around $x_0$ s.t. neither g nor g' vanish except $g(x_0)$

Question

The question:

Let $g$ be $C^1$ in an interval about $x_0$ and $g(x_0) = 0$. If $g'(x_0) \neq 0$, show there's a small interval about $x_0$ in which neither $g$ nor $g'$ vanish except for the root of $g$ at $x_0$.

I was able to show that $g$ doesn't vanish in some interval around $x_0$ except for at $x_0$. Any help for showing $g'$ doesn't vanish in the interval at all would be very helpful.

Answer 1

Say, $g'(x_{0})>0$, find some $\delta>0$ such that $g'(x)>\dfrac{g'(x_{0})}{2}$ for all $x\in(x_{0}-\delta,x_{0}+\delta)$.

For all $x\in(x_{0}-\delta,x_{0}+\delta)$, then $g(x)=g(x)-g(x_{0})=g'(\xi_{x})(x-x_{0})>\dfrac{g'(x_{0})}{2}(x-x_{0})$ if $x>x_{0}$ and $<\dfrac{g'(x_{0})}{2}(x-x_{0})$ if $x<x_{0}$, in either case, $g(x)\ne 0$ for such an $x$.