0
$\begingroup$

I was solving Stephen Abbot exercises and I came across this one I'm struggling with:

Assume $g$ is differentiable at point $c \in (a,b)$. If $g'(c) \neq 0$ show that there exists $\delta $ neighborhood around $c \in (a,b)$ for which $g(x)\neq g(c), \forall x$ in that neighborhood .

Although I think I understand intuition behind this, I still don't now how to approach this problem. I think I have to prove it by contradiction, assuming that for all $\delta$ neighborhoods around $c$, $g(x) = g(c)$ for at least one $x$, but I don't know where to go starting from here. I think, at some point I'll have to end up stating that in that case $g'(c)$ has to be equal to zero, which would be the desired contradiction.

I also tried to build sequences in $(a,b)$ that converge to $c$, but apparently that didn't help at all.

$\endgroup$
  • 1
    $\begingroup$ You have all the right ideas to solve the problem written in your question. I think you can do it. $\endgroup$ – Lorenzo Najt May 6 '18 at 19:24
1
$\begingroup$

Your approach works indeed: suppose for sake of contradiction that for all $\delta$-neighbourhoods of $c$, there is an $x \ne c$ in said neighbourhood such that $g(x) = g(c)$. In particular, we can find for all integers $n$ big enough an $x_n \in (c - 1/n, c + 1/n)$ such that $x_n \ne c$ and $g(x_n) = g(c)$. Here, big enough means such that $(c - 1/n, c + 1/n) \subseteq (a, b)$. Now, what can you say about the sequence $(x_n)_n$?

$\endgroup$
  • 1
    $\begingroup$ It seems that this sequence converges to c. So it is enough to state that $g'(c) = \lim_{x \to \infty }\frac{g(x_{n})-g(c)}{x_{n}-c}$ = 0 ? $\endgroup$ – Arci May 6 '18 at 20:15
  • $\begingroup$ @Anton Indeed, that's correct. $\endgroup$ – P. Senden May 6 '18 at 20:26
0
$\begingroup$

You know that$$\lim_{x\to c}\frac{g(x)-g(c)}{x-c}=g'(c)\neq0.$$Take $\delta>0$ such that$$|x-c|<\delta\implies\left|\frac{g(x)-g(c)}{x-c}-g'(c)\right|<\bigl|g'(c)\bigr|.$$Then$$|x-c|<\delta\implies\frac{g(x)-g(c)}{x-c}\neq0\implies g(x)\neq g(c).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.