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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!

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  • $\begingroup$ 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
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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 $f''(t_3 ) = 0$.

Hint:

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 $\phi'(t^*) = f(t^*) = 0$.

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  • $\begingroup$ Whoa, please don't switch answers! I am happy to help, but not to usurp. $\endgroup$ – copper.hat Oct 25 '15 at 18:38
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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 $f''(r)\lt 0$.

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  • $\begingroup$ 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

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