Let f be a function continuous on [0, 1] and twice differentiable on (0, 1).

a) Suppose that f(0) = f(1) =0 and f(c) > 0 for some c ∈ (0,1). Prove that there exists $x_0$ ∈ (0,1) such that f′′($x_0$) < 0.)

b) Suppose that $$\int_{0}^{1}f(x)\,\mathrm dx=f(0) = f(1) = 0.$$

Prove that there exists a number x$_0$ ∈ (0,1) such that f′′(x$_0$) = 0.

How do I solve the above two questions? I have tried using Mean Value Theorem but it gives me zero when I differentiate for the first time. Not sure how I can get the second derivative to be less than zero.

Any help is much appreciated! Thanks!

  • $\begingroup$ Hint: The function reaches a maximum. $\endgroup$ – André Nicolas Oct 25 '15 at 17:49
  • $\begingroup$ Maximum implies $f''<0$ $\endgroup$ – Zelos Malum Oct 25 '15 at 17:57
  • $\begingroup$ @ZelosMalum: It is slightly more complicated. The second derivative can be $0$ at a local maximum. $\endgroup$ – André Nicolas Oct 25 '15 at 18:04
  • $\begingroup$ That is true, I forgot about that, been to long, thanks! $\endgroup$ – Zelos Malum Oct 25 '15 at 18:05

For the second problem, we note that if $f$ is constant, then any point satisfies the requirement, so we can suppose $f$ is not constant.

If we can show that we can split $[0,1]$ into two intervals of non-zero length $[0, t^*], [t^* ,1]$ such that $f(t^*) = 0$, and there exists $t_1 \in (0, t^*)$ and $t_2 \in (t^*,1)$ such that $f'(t_1)=0, f'(t_2) = 0$, we can apply the mean value theorem to find some $t_3 \in (t_1,t_2)$ such that $f''(t_3 ) = 0$.


Let $\phi(t) = \int_0^t f(x) dx$ and note that $\phi(0) = \phi(1) = 0$. Since $f$ is not constant, and $\phi'(t) = f(t)$, we see that $\phi$ is not constant. Then $\phi$ must have a maximum or minimum at $t^* \in (0,1)$, and we see that $\phi'(t^*) = f(t^*) = 0$.

  • $\begingroup$ Whoa, please don't switch answers! I am happy to help, but not to usurp. $\endgroup$ – copper.hat Oct 25 '15 at 18:38

We solve the first problem only. The function $f$ reaches a (positive) maximum at some point $p$ strictly between $0$ and $1$. At any such $p$, we have $f'(p)=0$.

Since $f(1)=0$, by the Mean Value Theorem there is a $q$ strictly between $p$ and $1$ such that $f'(q)\lt 0$.

Thus by the Mean Value Theorem there is a point $r$ strictly between $p$ and $q$ such that $f''(r)\lt 0$.

  • $\begingroup$ Thanks for your help! How did you use the Mean Value Theorem from f'(q)<0 to f''(r)<0? I don't think it is possible to use the Mean Value Theorem for the second time because f'(q) is now is only continuous on (0,1) instead of [0,1]. Do correct me if I'm wrong! $\endgroup$ – DanaS Oct 25 '15 at 18:08
  • $\begingroup$ We are strictly between $0$ and $1$, for $0\lt p\lt 1$ and $p\lt q\lt 1$. Since $f$ is twice differentiable on $(0,1)$, the function $f'$ is continuous on $[p,q]$. $\endgroup$ – André Nicolas Oct 25 '15 at 18:11
  • $\begingroup$ Oh I see. How about the second part to this question? What I have gotten so far is since the integral is 0, f(x) must be a straight line on the x-axis and that's why f''(x)=0. Is this right? $\endgroup$ – DanaS Oct 25 '15 at 18:18
  • $\begingroup$ For the second question, it is not that simple. I would prefer not to solve the problem for you. But as a hint, if the function is identically $0$ there is nothing to do. Else there is a place strictly between $0$ and $1$ at which $f$ attains a max, and a place strictly between $0$ and $1$ at which it attains a min. $\endgroup$ – André Nicolas Oct 25 '15 at 18:29
  • $\begingroup$ If there is a max and min, how can f''(x) = 0? Since the function will be increasing and decreasing then it is not possible for f''(x) to be zero. $\endgroup$ – DanaS Oct 25 '15 at 18:34

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.