If $f$ is continuously differentiable at $a$ with $f(a)=0$ and $f'(a)\ne0$, then $a$ is an isolated root of $f$

Question

Let $f:\mathbb R\to\mathbb R$ be continuously differentiable at $a\in\mathbb R$ with $f(a)=0$ and $f'(a)\ne0$.

It's intuitively trivial, but how do we formally prove that $a$ is an isolated point of $\left\{x\in\mathbb R:f(x)=0\right\}$, i.e. there is a $\delta>0$ with $f(x)\ne0$ for all $x\in B_\delta(a)\setminus\left\{a\right\}$?

It might be crucial to note that, since $f'$ is continuous at $a$, there is a $\delta>0$ with $$\operatorname{sgn}f'(x)=\operatorname{sgn}f'(a)\;\;\;\text{for all }x\in B_\delta(a)\tag1.$$ By this observation it's obvious that the claim is true, but how do we show it formally?

Remark: Maybe we need to slightly strengthen the assumption and assume that $f$ is continuously differentiable in a neighborhood of $a$.

Answer 1

Note that, since $f(a)=0$,\begin{align}\lim_{h\to0}\frac{f(a+h)-f(a)}h=f'(a)&\iff\lim_{h\to0}\frac{f(a+h)}h=f'(a)\\&\iff\lim_{h\to0}\frac{f(a+h)-hf'(a)}h=0.\end{align}So, there is some $\delta>0$ such that$$\lvert h\rvert<\delta\implies\bigl\lvert f(a+h)-hf'(a)\bigr\rvert<\bigl\lvert hf'(a)\bigr\rvert.$$But then $\lvert h\rvert<\delta\implies f(a+h)\neq0$.

Answer 2

By definition, we have $$f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x-a}.$$

Now assume that $a$ is not isolated. Then we can find a sequence $(x_n)_{n \in \mathbb{N}}$ that converges to $a$ such that $f(x_n) = 0$ for all $n$.
Using the definition of a limit, we should get the value $f'(a) \neq 0$ for all sequences. But the sequence $(x_n)$ when put into the equation would give $0$ as limit, a contradiction. Thus, $a$ must be an isolated point.

Answer 3

1) Assume $f'(a)>0$.

Since $f'$ continuos there is a $\delta$ neighbourhood of $a$, s.t.

$|x -a| \lt \delta$ implies $f'(x)>0$.

For $x \in B_{\delta}(a)$

$\dfrac{f(x)-f(a)}{x-a}= f'(t)$, $t \in (\min(a,x),\max(a,x)).$

$f(x)= f(a)+f'(t)(x-a) = f'(t)(x-a) \not =0.$

2) Likewise for $f'(a)<0$.