$f: [0,3] \rightarrow \mathbb R$ be a twice differentiable function function.
$f(0) = f(3) = 0 $ and $f'(0)f'(3)\geq0$
prove that that $\exists c \in (0,3)$ such that $f''(c) =0$
assume: $f'(0)>0$ and $f'(3)>0$ (the case where one is equal to zero is trivial)
Rolle's theorem: $\exists \hspace0.1cm \alpha \in(0,3) \hspace0.1cm f'(\alpha)=0$
lagrange's theorem:
$f'(3) – f'(\alpha) = f''(\beta)(3-\alpha)$ $\implies f''(\beta) > 0$
$f'(\alpha) – f'(0) = f''(\beta)(\alpha)$ $\implies f''(\beta) < 0$
but since the 2nd derivative is not necessary a continuous function I can't conclude $f''(\beta) = 0$
can this be fixed or am I missing something?
Best Answer
Suppose $f'(0)>0,f'(3)>0$. By the continuity of $f'(x)$ at $x=0,3$, there are $a,b\in(0,3), a<b$ such that $$ f(a)=f(0)+f'(c_1)a>0,f(b)=f(3)+f'(c_2)(x-3)<0 $$ for some $c_1\in(0,a),c_2\in(b,3)$. By IMVT, there is $c\in(a,b)$ such that $f(c)=0$. Now applying the Rolle's theorem in $[0,c]$ and $[c,3]$, there is $d_1\in(0,c),d_2\in(c,3)$ such that $$ f'(d_1)=f'(d_2)=0. $$ Applying Rolle's theorem in $[d_1,d_2]$, one concludes that there is $d\in(d_1,d_2)\subset(0,3)$ such that $f''(d)=0$.