$f(x)$ is continuous on $(a,b)$ such that $f(a)f(b) < 0$. Show the existence of $c$ element of $(a,b)$ such that $f(c) = 0$

Question

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?

Answer 1

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.

Answer 2

Hint: Use the intermediate value theorem.

Answer 3

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