# Rolle's theorem for second derivative

$$f$$ is a twice-differentiable function on some segment $$[a,b]$$ such that $$f(a)=f(b)$$ and $$f'(a)f'(b)<0$$. it asks to prove that the second derivative of $$f$$ vanishes at some point between $$a$$ and $$b$$ (strictly).

• This might be a typo - if we change the question to $f'(a)f'(b)>0$ then the statement is true. Wlog assume $f(a) =f(b) = 0$ and $f'(a) >0$. Then the function must have a third $0$ crossing in the middle since it must be $<0$ close to $b$ in order to have $f(b) = 0$ and $f'(b) > 0,$ and is $>0$ close to $a$. This is enough because it gives two zeros of $f'$ by using Rolle's theorem twice, which induce a zero of $f''$ by another application of Rolle's theorem. – stochasticboy321 Nov 12 '19 at 7:33
You are right: the statement is false and you have the right idea. An example would be $$f(x)=x^2$$, $$a=-1$$, and $$b=1$$. Then $$f'(a)f'(b)=-4<0$$, but you always have $$f''(x)=2>0$$.
• Thanks. Can the problem statement be rectified if we assume $f'(a)f'(b)>0$ ? Graphically it seems to be the case. – ahmed Nov 12 '19 at 7:30
• My idea then is to prove first that f' vanishes at least twice on $(a,b)$ then apply Rolle on it between the two points where it vanishes – ahmed Nov 12 '19 at 7:33