Let $f: [a, b] \to \Bbb R$ be a convex continuous function such that $f(a) < 0 < f (b)$. Prove that there is only one point $c \in (a, b)$ such that $f(c) = 0$.
TIP: The resolution to be made is similar to the Bolzano theorem used for the demonstration of the Intermediate Value Theorem. However, the idea is to use derivatives to solve this.
Preface: There's at least one such point by the Intermediate Value Theorem. And depending how you define "convexity", it usually implies that your function is at least twice differentiable.
Now suppose there are two values $c_1, c_2 \in (a, b)$ which meet your conditions. In other words, $f(c_1) = f(c_2) = 0$. The order is arbitrary, so define $c_1 < c_2$.
Since $c_1 > a$ and $f(c_1) > f(a)$, $\frac{f(c_1)-f(a)}{c_1 -a} > 0$. So by the Mean Value Theorem, there exists a point $x \in (a, c_1)$ such that $f'(x) > 0$.
Since $c_2 > c_1$ and $f(c_2) = f(c_1)$, $\frac{f(c_2)-f(c_1)}{c_2 - c_1} = 0$. So by the Mean Value Theorem, there exists a value $y \in (c_1, c_2)$ such that $f'(y) = 0$.
Since $b > c_2$ and $f(b) > f(c_2)$, $\frac{f(b)-f(c_2)}{b-c_2} > 0$. So by the Mean Value Theorem, there exists a value $z \in (c_2, b)$ such that $f'(z) > 0$.
This means that $f'$ has decreased between $x$ and $y$, but increased between $y$ and $z$. Thus the concavity of $f$ is not consistent. Since we assumed $f$ was convex, this is a contradiction.
So there can be only one such point. QED
