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A problem asks the following

$f$ 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).

What about this situation

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    $\begingroup$ 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. $\endgroup$ – stochasticboy321 Nov 12 '19 at 7:33
  • $\begingroup$ @stochasticboy321 I think you're right. Thanks $\endgroup$ – ahmed Nov 12 '19 at 7:35
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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 $f''(x)=2>0$.

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    $\begingroup$ Thanks. Can the problem statement be rectified if we assume $f'(a)f'(b)>0$ ? Graphically it seems to be the case. $\endgroup$ – ahmed Nov 12 '19 at 7:30
  • $\begingroup$ 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 $\endgroup$ – ahmed Nov 12 '19 at 7:33
  • $\begingroup$ A comment on my first post confirms my remark actually $\endgroup$ – ahmed Nov 12 '19 at 7:36
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    $\begingroup$ Yes, that would correct the problem. $\endgroup$ – José Carlos Santos Nov 12 '19 at 7:48

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