Proving that a certain closed interval exists

[Usagi]

Let $a$, $b$ be reals and $f: (a,b) \rightarrow \mathbb{R}$ be twice continuously differentiable. Assume that there exists $c \in (a,b)$ such that $f(c) = 0$ and that for any $x \in (a,b)$, $f'(x) \neq 0$. Define $g: (a,b) \rightarrow \mathbb{R}$ by $\displaystyle g(x) = x -\frac{f(x)}{f'(x)}$ and note that $g(c) = c$.

Deduce that there exists $\delta_1 >0$ such that $I_1 = [c -\delta_1, c+\delta_1]$ is contained in $I = (a,b)$ and for any $x$, $|g'(x)| \le 1/2$.

Query 1: Is the question missing a phrase after "for any $x$", should it say "for any $x \in I_1$"?

Query 2: My (incomplete) attempt so far: I've shown that $g(x)$ is differentiable on $I = (a,b)$ and has a continuous derivative on $I$ with $\displaystyle g'(x) = \frac{f(x) f''(x)}{[f'(x)]^2}$, also $g'(c) = 0$. Also, since we know that $(a,b)$ is an open set, then for any $c \in (a,b)$, let $\epsilon = \min\{c-a, b-c\}$ then clearly $(c-\epsilon, c+\epsilon) \subset I$. Now pick $\delta_1$ such that $0<\delta_1 <\epsilon$ and consider the closed interval $I_1 = [c-\delta_1, c+\delta_1]$. So $I_1 \subset (c-\epsilon, c+\epsilon)$ and hence $I_1$ is contained in $I$.

I am just not sure how to show that for any $x \in I_1$ (?), $|g'(x)| \le 1/2$. Maybe something to do with the Extreme Value Theorem?

 

[Opalg]

Query 1: Yes, I think that the question must refer to $I_1$.

Query 2: Your "incomplete attempt" is practically complete. You have shown that $g$ is continuously differentiable and that $g'( c ) = 0.$ If a continuous function is $0$ at $c$ then it must be small in some neighbourhood of $c$ ...

 

[Usagi]

Thanks! However, how do you that this neighbourhood is contained in (a,b)?

 

[Evgeny.Makarov]

However, how do you that this neighbourhood is contained in (a,b)?

If it is not, make it smaller.

 

[Usagi]

I understand this intuitively, but how can I make it rigorous using $\epsilon-\delta$ arguments?

 

[Evgeny.Makarov]

If a continuous function is $0$ at $c$ then it must be small in some neighbourhood of $c$ ...

I understand this intuitively, but how can I make it rigorous using $\epsilon-\delta$ arguments?

It's a general fact that
\[
(\forall x\in A.\;P(x))\land B\subseteq A\implies\forall x\in B.\; P(x).
\]
You agreed that there exists a $\delta>0$ such that
\[
\forall x\in(c-\delta,c+\delta).\;|g'(x)|\le 1/2.
\]
There exists a $0<\delta_1<\min(\delta,c-a,b-c)\le\delta$ such that $[c-\delta_1,c+\delta_1]\subseteq (c-\delta,c+\delta)$. Therefore,
\[
\forall x\in[c-\delta_1,c+\delta_1].\;|g'(x)|\le 1/2.
\]

 

[Usagi]

Ah yup, makes perfect sense, thank you!