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This problem is from Apostol's Mathematical Analysis:

Let $f$ be continuously differentiable in $[a, b]$. If $ f(a) = f(b)$ and if $f^{'}(a) = f^{'}(b)$, then prove that there exists $x_1$ and $x_2$ in $(a, b)$ such that $x_1\neq x_2$ but $f^{'}(x_1) = f^{'}(x_2)$.

My try:

By Rolle's Theorem $\exists x_0\in (a,b) $ such that $f^{'}(x_0)=0$. How to guarantee existence of $x_1,x_2 $ from here?

Can it be solved from a geometrical point of view?

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    $\begingroup$ you don't know that $f'(a)=f'(b)=0$ $\endgroup$ – tattwamasi amrutam Oct 2 '15 at 8:53
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    $\begingroup$ Geometrical point of view. At point a the function is going at a rate of f'. if that's positive the function is going up. At point b, f(a) = f(b) so the function had to have come down again. That means that somewhere f' was neg. But the means somewhere f' had to be zero. But right now at b f' is positive again. But that means it had to go from negative to positive. That means there was another point in between where f' was zero. Recap: function going up at a, peaks, f' = 0, goes down, plateaus, f' = 0 goes up again, reaches b. It's the only possibility. $\endgroup$ – fleablood Oct 2 '15 at 9:05
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    $\begingroup$ It will be nice to mention why you downvote and let people think and learn from your comments on their wrong answers. $\endgroup$ – Math-fun Oct 2 '15 at 9:41
  • $\begingroup$ thanks for your useful comments@fleablood $\endgroup$ – Learnmore Oct 3 '15 at 3:30
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As you said, by Rolle's Theorem, we are guaranteed existence of $c \in (a,b)$ such that $f'(c) = 0$. WLOG, let $f'(a) = f'(b) = k > 0$. By the continuity of $f'$ and the intermediate value theorem, we get existence of $c_1 \in (a,c)$ such that $f'(c_1) = k_1$ where $0<k_1<k$. Similarly, we get $c_2 \in (c,b)$ such that $f'(c_2) = k_1$.

Now, if $f'(a) = 0$, then let's look at some cases. If $f(c) \neq f(a)$, then the mean value theorem gives existence of $d \in (a,c)$ such that $\alpha = f'(d) = \frac{f(c) - f(a)}{c - a}\neq 0.$ Now, we can apply the above argument again.

If $f(c) = f(a) = f(b)$, then if $f'$ is nonzero at some point in the interval, the above argument works. Otherwise, $f'$ is $0$ and the result follows.

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(i) Let's suppose that $f'(a)>0$ (then we'll see the case $f'(a)=0$). Because of (i), you can find a $\eta > 0$ (as small as it is) that we have $f(a + \eta) > f(a)$ and $f(b - \eta) < f(b)$.

$f$ is continuous on $[a + \eta, b-\eta]$, therefore there exists $c \in [a + \eta; b - \eta]$ so that $f(c) = f(a) = f(b)$. ($f(c)$ in $[f(a+\eta], f(b-\eta)])$.

$f$ is continuous on both intervals $[a, c] $ and $[c, b]$ and reach its maximums in a $x_1$ and $x_2$ ($x_1$ in $(a, c)$ and $x_2$ in $(c, b)$ because $a, c$ and $b$ are not the maximum since $a+\eta$ and $b-\eta$).

We have $f'(x_1) = f'(x_2) = 0$

Opposite reasonning for $f'(a)<0$ leading to the same conclusion

(ii) if $f'(a) = 0$ because of Rolle's theorem you know that you have a $c$ in $(a,b)$ that $f'(c) = 0$. Assume that $f(c) = f(a)$ (then we reapply this reasonning on $a$ and $c$ instead of $a$ and $b$ until finding a $c_n$ that $f(c_n)$ is different from $f(a)$, if we can not find one then $f$ is constant and any $c_1<c_2$ in $(a, b)$ will do)

We have a $f(a) < f(c)$ and $f'$ continuous on $[a, c]$ and $f'(a) = f'(c) = 0$ but $f'$ not equal to $0$ and it exists a $c_1$ in $(a, c)$ so that $f'(c_1)>0$ because $f(a) < f(c)$.

Because of that and $f'$ continuous on this interval, $f'$ reaches its maximum in $d$ in $(a;c)$ and because of its continuity $f'$ reaches $f'(d)/2$ in two points $x_1,x_2$, where $a<x_1<d<x_2<c<b$.

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    $\begingroup$ i dont understand the down votes, because the arguments seems fine to me. $\endgroup$ – Lost1 Oct 3 '15 at 1:57
  • $\begingroup$ The format was messy. @Lost1 $\endgroup$ – user99914 Oct 3 '15 at 3:48

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