# If \$f(a) leq 0, quad f(b) < 0 quad text{and} quad int_{a}^b f(x)text{dx} geq 0\$ for \$f: [a,b] to mathbb{R}\$, then \$f'(x_0) = 0\$.

Let $$a,b in mathbb{R}$$ with $$a < b$$ and $$f : (a,b) to mathbb{R}$$ be differentiable with
$$f(a) leq 0, quad f(b) < 0 quad text{and} quad int_{a}^b f(x)text{dx} geq 0.$$
I have to show that there exists a $$x_0 in (a,b)$$ with $$f'(x_0) = 0$$.

Here is my attempt: From the mean-value theorem I know that there exists a point $$c in (a,b)$$ such that
$$f'(c) = frac{f(b) – f(a)}{b-a}$$
and from the mean value theorem for definite integrals I know that there exists some point $$xi in (a,b)$$ such that
$$int_{a}^b f(x)text{dx} = f(xi)(b-a).$$
From here I don’t know what to do next. Can you give me a tip how I can continue? The case $$f (a) = f (b)$$ is clear, so that we can assume that $$f (a) neq f(b)$$.