Rolles theorem application

Question

How do I solve this question:

$g(1)=g(0)=g'(0)=0 $ Show that there is a number $t ∈ (0, 1)$ such that $g"(t)=0$. How do I apply Rolle's theorem to this?

Answer 1

By Rolle's theorem you will get $a \in (0,1)$ s.t $g'(a)=0$ Now apply Rolle's theorem on $g'$ in $[0,a]$.

Answer 2

Rephrasing:

Assuming $g$ twice differentiable.

MVT:

Given : $g(0)= g(1) = 0.$

$\dfrac{g(1)-g(0)}{1}= g'(t)$, $t \in (0,1).$

$g'$ has $2$ distinct zeroes at $x=0$ (given) , and at $x= t.$

Rolle's theorem:

There is a $s\in (0,t)$ with $g''(s)=0.$