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Let $f:[-5,3]\to \mathbb{R}$ such that $f$ is differentiable, moreover $\int_{-5}^{-1}fdx=\int_{-1}^{3}fdx$
Prove: there is $x_0\in[-5,3]$ such that $f'(x_0)=0$

$f$ is differentiable $\Rightarrow$ continuous$\Rightarrow$ integrable, now, from the Mean Value Theorem for Integrals $\int_{-5}^{3}fdx=8\cdot f(c)$

I can not see how $\int_{-5}^{-1}fdx=\int_{-1}^{3}fdx$ can help me find that $f'(c)=0$

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    $\begingroup$ Perhaps this is a little brute force, but you could try proof with contrapositive. You can show that if $f'(c) \not = 0$ then the function is monotonic, and using riemann sums show that the definite integral on equal length intervals cannot be the same $\endgroup$ – Ovi Aug 1 '16 at 12:05
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A small modification of your Mean Value Theorem for Integrals idea works.

By the MVT for integrals, there is a $c$ strictly between $-5$ and $-1$ such that the first integral is $4f(c)$.

Similarly, there is a $d$ between $-1$ and $3$ such that the second integral is $4f(d)$.

But the two integrals are equal, so $f(c)=f(d)$. The result now follows from Rolle's Theorem.

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You need to show that $f$ has an extremum. If it doesn't then it must be strictly monotone and then the condition on integrals will fail (why?)

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  • $\begingroup$ Because than there can not be that $\int_{-5}^{-1}fdx=\int_{-1}^{3}fdx$? $\endgroup$ – gbox Aug 1 '16 at 12:11
  • $\begingroup$ Indeed! If e.g. $f$ is strictly increasing the latter must be strictly larger (since both are integrating over an interval of length 4). In fact: $\int_{-5}^{-1} f \;dx < 4 f(-1) < \int_{-1}^3 f \; dx$ $\endgroup$ – H. H. Rugh Aug 1 '16 at 13:21
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Using the mean value theorem, but now for the intervals $[-5, -1]$ and $[-1, 4]$, we get two constants $c \in [-5,-1]$ and $d \in [-1,3]$, such that $ 4 \cdot f(c) = \int_{-5} ^{-1} f dx = \int_{-1} ^{3}f dx = 4\cdot f(d).$ The statement now follows from Rolle's theorem, as $f(c)=f(d)$.

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You are given that the area under the curve in the interval $[-5,-1]$ is equal to that in the interval $[-1,3]$. If $f$ is strictly increasing in the interval $[-5,3]$, then the area under the curve in the interval $[-1,3]$ would be greater than that in the interval $[-5,-1]$. If $f$ is strictly decreasing in the interval $[-5,3]$, then the area under the curve in the interval $[-5,-1]$ would be greater than that in the interval $[-1,3]$.

Therefore, you know that the curve must change from increasing to decreasing or vice versa. Then by using the Mean Value Theorem, you know there must be a value $c$ in $[-5,3]$ such that $f'(c) = 0$.

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