Show that there is a $c$ such that $f'(c)=0$ Let $f:[a,b]\to\mathbb{R}$ be continuous and differentiable on $(a,b)$. Assume that $f(a)=0$ and $f(b)=-1$, and $\int_a^bf(x)dx=0$. Prove that there is a $c\in (a,b)$ such that $f'(c)=0$.
My attempt: Since $\frac{1}{b-a}\int_a^bf(x)dx=0$, there is a $\theta\in (a,b)$ such that $f(\theta)=0$ (since $f$ is continuous). Since $f$ is differentiable on $(a,b)$, Rolle's theorem gives that there is a $c\in (a,\theta)$ such that $f'(c)=0$.
I think there is something wrong with this though because I haven't used the fact that $f(b)=-1$.
 A: It looks good, you used the following result:
If a continuous function $f:[a,b]$ is only zero at $a$ then its integral is not zero.
Proof: Suppose $f(b)>0$, then by the intermediate value theorem the whole function is non-negative, but we know that the only non-negative continuous function with integral $0$ is the constant $0$ function (all of the other ones have positive integral), so the integral cannot be $0$. Analogously when $f(b)<0$.
A: Your proof looks good.
An alternate proof using IVT
We find that if there exists no $c$ such that $f'(c) = 0$, then $f$ must be strictly increasing or decreasing. However, $f$ may not be strictly increasing, because we require that there exists an $a \in (0,1)$ for every $x \in [-1,0]$ such that $f(a) = x$ by the intermediate value theorem; and thus must decrease from its initial value of $0$ to satisfy for all $-1 \leq x < 0$.
Thus, $f$ must be strictly decreasing. Now try to show that the integral must be negative for strictly decreasing $f$, forming a contradiction.
A: There must be some point $s \in (a,b)$ such that $f(s) >0$. Otherwise, continuity and the fact that $f(b) <0$ would result in $\int_a^b f(x)dx < 0$.
Hence there is some point $c \in (a,b)$ at which $f$ attains a maximum, and
we have $f'(c) = 0$.
