Let $f$ be a function continuous on $[0,1]$ and twice differentiable on $(0,1)$. Suppose that $\int_0^1 f(x)\, dx=f(0)=f(1)=0$ and $f(c)>0$ for some $c\in(0,1)$.

How do i prove that there exists $x_0\in(0,1)$ such that $f''(x_0)=0$? I'm stuck and I do not know how to proceed.

  • 3
    $\begingroup$ You could possibly show that there must be a point $\xi\in(0,1)$ such that $f(\xi)=0$ and so you would have $3$ zeros of $f$ in $[0,1]$. By repeated application of Rolle's Theorem, there would exist a $x_0\in(0,1)$ such that $f''(x_0)=0$. $\endgroup$ – Guest Oct 23 '15 at 10:48

Hints. (1) If we had $f \ge 0$, then as $f$ is continuous and $f(c) > 0$, we would have $\int_0^1 f(x)\, dx > 0$. Hence $f(d) < 0$ for some $d \in (0,1)$.

(2) Apply the intermediate value theorem to find $e \in [\min\{c,d\}, \max\{c,d\}] \subseteq (0,1)$ such that $f(e) = 0$.

(3) Now apply the mean value theorem twice to $f$ and then once to $f'$.


Sketch: Show that there is $d\in(0,1)$ with $f(d)<0$ (you'll need the assumption on the integral for that). It follows that both maximum and minimum of $f$ are in $(0,1)$. The derivative $f'$ vanishes at each one them. Apply Role's theorem.


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