Prove: there is a point where the derivative is zero 
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 $f'(c)=0$
 A: 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?)
A: 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.
A: 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 $f(c)=f(d)$.
A: 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 $f'(c) = 0$.
