0
$\begingroup$

I believe I should be simply restating this question in with respect to Rolle's Theorem (if $f$ is continuous on $[a,b]$ and $f$ is differentiable on $(a,b)$, and $f(a) = f(b)$ then there exists $c \in (a,b)$ s.t. $f'(c) = 0$). BUT the problem states that it is continuous on open interval, not closed, and with $f(a)f(b)$ being a negative number, I can't figure $f(a)$ equaling $f(b)$...AND what clue tells me it IS differentiable other than it is continuous so starting to run in circles Perhaps just staying in Mean Value Theorem? Or is there an epsilon/delta method of proving this?

$\endgroup$
  • 2
    $\begingroup$ If $$f(x) = \begin{cases}1;& x \in [a,b)\\ -1;& x =b\end{cases}\text{,}$$ then there is no such $c\in (a,b)$ that $f(c) = 0$. $\endgroup$ – Antoine Oct 12 '14 at 20:07
  • $\begingroup$ @Antoine that ain't continuous. $\endgroup$ – JP McCarthy Oct 12 '14 at 20:17
  • 2
    $\begingroup$ @JpMcCarthy it's continuous on (a,b) $\endgroup$ – David Peterson Oct 12 '14 at 20:20
  • $\begingroup$ Apologies and in fact yours should be an answer. $\endgroup$ – JP McCarthy Oct 12 '14 at 20:40
  • 1
    $\begingroup$ @MathRaider the problem as stated is flawed $\endgroup$ – David Peterson Oct 12 '14 at 21:14
3
$\begingroup$

It's just the Intermediate Value Theorem: $f(a)f(b)<0$ so either $f(a)>0>f(b)$ or $f(a)<0<f(b)$. In either case, there must be a $c$ such that $f(c)=0$.

Edit: the argument here needs $f$ to be continuous on $[a,b]$. Continuity on $(a,b)$ isn't enough, as Antoine has shown in his comment.

$\endgroup$
  • $\begingroup$ unless Antoine's example above? $\endgroup$ – Math Raider Oct 12 '14 at 21:06
  • $\begingroup$ that's what I am thinking and others here are belabored by this on my behalf and I can't wait to argue this in class tomorrow, so thanks for the input $\endgroup$ – Math Raider Oct 12 '14 at 23:25
2
$\begingroup$

Hint: Use the intermediate value theorem.

$\endgroup$
  • $\begingroup$ DOH! I forgot that in my fog! $\endgroup$ – Math Raider Oct 12 '14 at 20:11
1
$\begingroup$

Maybe he's trying to prove the intermediate value theorem:

f does'nt have to be differentiable, but the product f(a)*f(b)<0 implies: f(a) and f(b) are of opposite sign. So if there isn't such a c as f(c)=0 on (a,b), then f is of constant sign on it because of its continuity, and there is an absurdity

$\endgroup$
  • $\begingroup$ by contradiction then? $\endgroup$ – Math Raider Oct 12 '14 at 20:14
  • $\begingroup$ that's a way to do it. Prove that f has to be of constant sign and you're good $\endgroup$ – mvggz Oct 12 '14 at 20:17

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.