5
$\begingroup$

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

enter image description here

$\endgroup$
  • 1
    $\begingroup$ 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 at 7:33
  • $\begingroup$ @stochasticboy321 I think you're right. Thanks $\endgroup$ – ahmed Nov 12 at 7:35
5
$\begingroup$

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 $f''(x)=2>0$.

$\endgroup$
  • 1
    $\begingroup$ 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 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 at 7:33
  • $\begingroup$ A comment on my first post confirms my remark actually $\endgroup$ – ahmed Nov 12 at 7:36
  • 1
    $\begingroup$ Yes, that would correct the problem. $\endgroup$ – José Carlos Santos Nov 12 at 7:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.