# twice differentiable functions

Let $f:\mathbb R \to \mathbb R$ be a twice continuously differentiable function, with $f(0)=f(1)=f'(0) = 0$. Then

1. $f^{"}$ is the zero function.

2. $f^{"}(0)$ is zero.

3. $f^{"}(x)=0$ for some $x \in$ (0,1).

4. $f^{"}$ never vanishes.

I don't know where to start, tried rolles theorem but cant catch the point..please help me to start..

• I take it you're supposed to determine which of these things happens? More than one is possible. – Jakob Hansen Apr 15 '16 at 15:48
• There is one that will happen in all cases, though. – Mauro Apr 15 '16 at 15:50
• A fully satisfying answer should select the correct answer, AND show by way of counterexample or in any other manner why the other three answers are not correct too. – mathguy Apr 15 '16 at 15:56

Rolle's theorem says that because $f(0) = f(1)$, there is a point $t$ STRICTLY between 0 and 1 where $f'(t) = 0$. Now use this and $f'(0) = 0$ and apply Rolle's theorem to $f'$ over $[0,t]$. This will lead you to the correct choice.
Hoping someone may find it useful: Let $$f(x)=\cos2\pi x-1, f(0)=f(1)=0$$ $$f'(x)=2\pi\sin2\pi x,f'(0)=0$$
It's easy to counter 1,2 and 4 with $f$ ($f''$ vanishes at $x=\frac14$). And for 4 alone, $f(x)=0$ works.