Question

Let $f:R\to R$ be twice continuously differentiable (or $f^{\prime \prime }$ exists and is continuous) such that $f\left(0\right)=f\left(1\right)=f^{\prime }\left(0\right)=0$. Then

• A. $f^{\prime \prime }\left(c\right)=0$ for some $c\in R$
• B. there is no point for which $f^{\prime \prime }\left(x\right)=0$
• C. at all points $f^{\prime \prime }\left(x\right)>0$
• D. at all points $f^{\prime \prime }\left(x\right)<0$

Answer 1

The correct option is A $f^{\prime \prime }\left(c\right)=0$ for some $c\in R$

Consider $f\left(x\right)$ on $[0,1]$
Applying Rolle's theorem on the interval $[0,1]$,
$f^{\prime }\left(a\right)=0$ for some $a\in (0,1)$

Now, applying Rolle's theorem to $f^{\prime }\left(x\right)$on the interval $[0,a]$,
$f^{\prime \prime }\left(c\right)=0$ for some $c\in (0,a)$