Show that a nonempty neighborhood exists around x for a given function that is 0 at x with nonzero derivative at the same point.

Question

Let $f : [0,1] \rightarrow \mathbb{R}$ and $x$ a fixed point in $(0,1)$. Assume $f(x)=0$ and $f$ is differentiable at $x$ with $f'(x)\neq 0$.

Prove: $\exists \delta \gt 0$ such that $f(t)\ne 0 \ \ \forall t$ with $0 \lt |{t-x}| \lt \delta$.

I'm pretty sure that I need to invoke continuity of $f$ in some way, but I don't see the immediate path forward. Any help is appreciated!

Answer 1

Suppose otherwise. Then for any $\;n\in\Bbb N\;$ there exists $\;t_n\in A_n:=\{t\in\Bbb R\;|\;|t-x|<\frac1n\}\;$ s.t. $\;f(t_n)=0\;$ , but then by definition

$$f'(x)=\lim_{t\to x}\frac{f(t)-f(x)}{t-x}\stackrel{\text{since the limit exists!}}=\lim_{n\to \infty\\t_n\in A_n}\frac{f(t_n)-f(x)}{t_n-x}=0$$

and we get a contradiction.