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I'm trying to solve this problem

Let $f:[a,b] \to \mathbb R$ a differentiable function with continuous derivative.

Suppose further that $f$ is twice differentiable on $(a,b)$.

Prove that if $f(a)=f(b)$ and f'(a)=f'(b)=0, then exist $x_1, x_2 \in (a,b)$ with $x_1 \neq x_2$ such that f''(x_1) = f''(x_2).

I'm trying to solve this problem graphically intuitively ; I tried, and likely the problem is real: if $f(a) = f(b)$ and f'(a) = f'(b) = 0, the points $a$ and $b$ are the points of maximum and minimum for the function. but how can you prove that there are $x_1, x_2 \in (a,b)$ with $x_1 \neq x_2$ such that f''(x_1) = f''(x_2)?

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    @Patrick: What we saw as `b` plus a space was in fact a compound of the letter `b` and of two invisible pseudo-characters attached to it and translated by the system as a space. That is, an encoding gone awry (but please do not ask me for more precise details...).2011-12-29

2 Answers 2

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If the function is constant, everything is easy.

If the function is not constant, it attains a maximum and a minimum value in our interval, and these values are distinct.

The max and min cannot both occur at endpoints, since $f(a)=f(b)$. So there is a local extremum in $(a,b)$, and therefore a point $c\in(a,b)$ such that f'(c)=0. Now you should be able to use Rolle's Theorem to show that there exist points $x_1\in(a,c)$ and $x_2\in(c,b)$ such that f''(x_1)=f''(x_2)=0.

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    @David Mitra: Of course you are right. I wanted to connect the finding of $c$ with something familiar, to make the geometry more concrete-seeming.2011-12-28
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Hint: Apply the Mean Value Theorem three times (or Rolle's Theorem if you prefer).

Applying it once to $f$ over $[a,b]$ gives you a point $c$ with $a where f'(c)=0.

Now apply the Mean Value Theorem to f' on each of the intervals $[a,c]$ and $[c,b]$.

(Note that the Mean Value Theorem gives you a point $c$ strictly between the endpoints of the interval that you're working over.)


It's not true that $a$ and $b$ necessarily give the minimum and maximum of the function. They could in fact give neither. For instance, the graph could resemble a $\sin$ wave over $[0,2\pi]$ that's been "smoothed out" at the endpoints (so that the derivative is zero at the endpoints).