Let $f:[a,b] \rightarrow \mathbb{R}$ be a twice differentiable function such that
$f(a) = f(b) = \frac{1}{b-a} \int_a^b f(x)\,dx$.
Show there exists $c \in (a,b)$ such that $f''(c) = 0$.
Attempts:
To me, this looks a lot like the intermediate value theorem on $f''(c)$. Assuming $f$ is Riemann integrable, the right is $\frac{F(b)-F(a)}{b-a}, F'(x) = f(x)$, which is the form of the mean value theorem.
I've gotten that $f'$ is Riemann integrable, and that $\int_a^b f'(x)\,dx = 0$ via the fundamental theorem of calculus.
However, I'm having trouble applying this information to the second derivative, which seems to lack a lot of information. $f''(x)$ has the intermediate value property, as all derivatives do. However, finding values of $f''(x)$ to apply the intermediate value property seems out of my grasp.
Answer
Let $F(x) = \int_{a}^{x}f(t)\,dt$. Then clearly $F$ is continuous on $[a, b]$ and differentiable in $[a, b]$ with $F'(x) = f(x)$. By Mean value Theorem $(F(b) - F(a))/(b - a) = F'(\xi) = f(\xi)$ for some $\xi \in (a, b)$. Thus we get $f(a) = f(b) = f(\xi)$ where $a < \xi < b$. Clearly by Rolle's Theorem we have $f'(\eta) = 0 = f'(\theta)$ for some $\eta \in (a, \xi)$ and some $\theta \in (\xi, b)$. And then again applying Rolle's theorem on $f'$ we get $f''(c) = 0$ for some $c \in (\eta, \theta) \subset (a, b)$.
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