Let $f$ be a differentiable function with $|f'(x)|\leq1$ and $f(-3)=-3, f(3)=3$. Then find $f(0)$. 
Let $f$ be a differentiable function with $|f'(x)|\leq1$ and $f(-3)=-3, f(3)=3$. Then find $f(0)$.

I think this is mean value theorem problem. But I can't solve... help me please. 
 A: MVT states that there exists a unspecified c between a and b such that
$$f'(c) = \frac{f(b)-f(a)}{b-a} $$


*

*Apply MVT between endpoints -3 and 0:
There exists c1, between -3 and 0, ie $ -3 \leq c1 \leq 0 $, such that


$$ f'(c1) = \frac{f(0)-f(-3)}{0-(-3)} $$
$$ f'(c1) = \frac{f(0)+3}3  $$
but $ |f'(c1)|\leq 1$ 
$$ \left|\frac{f(0)+3}3\right| \leq 1 $$
$$ |f(0)+3| \leq 3 $$
There are two cases to this: whether or not f(0)+3 is negative or positive.
$$ \text{case a)} \quad f(0)+3 \geq0 \quad\quad\quad\quad\quad\quad \text{case b)} \quad f(0)+3 \leq 0 $$
$$ (f(0) +3) \leq 3 \quad\quad\quad\quad\quad\quad -(f(0) +3) \geq -3$$
$$ (f(0) +3) \leq 3 \quad\quad\quad\quad\quad\quad -f(0) -3 \geq -3 $$
$$ f(0) \leq 0 \quad\quad\quad\quad\quad\quad\quad\quad -f(0) \geq 0 $$
$$ f(0) \leq 0 \quad\quad\quad\quad\quad\quad\quad\quad f(0) \leq 0 $$
$$ f(0) \leq 0 $$


*Now apply MVT between endpoints 0 and 3:
There exists c2, between 0 and 3, ie $ 0 \leq c2 \leq 3 $, such that


$$ f'(c2) = \frac{f(3)-f(0)}{3-0} $$
$$ f'(c2) = \frac{3 - f(0)}3  $$
$$ \left|\frac{3 - f(0)}3\right| \leq 1 $$
$$ |3 - f(0)| \leq 3 $$
There are also two cases to this: whether or not 3 - f(0) is negative or positive.
$$ \text{case a)} \quad 3 - f(0) \geq0 \quad\quad\quad\quad\quad\quad \text{case b)} \quad 3 - f(0) \leq 0 $$
$$ (3 - f(0)) \leq 3 \quad\quad\quad\quad\quad\quad -(3 - f(0)) \geq -3$$
$$ -f(0) \leq 0 \quad\quad\quad\quad\quad\quad -3 +f(0) \geq -3 $$
$$ f(0) \geq 0 \quad\quad\quad\quad\quad\quad\quad f(0) \geq 0 $$
$$ f(0) \geq 0 $$
Combining  $ f(0) \leq 0 \text{ and } f(0) \geq 0 $
$$ 0 \leq f(0) \leq 0 $$
Then by the squeeze theorem $ f(0) = 0 $.
A: Since $|f'(x)|\leq 1$ then we take $f'(x)\leq 1$
1) integrate both sides from $-3$ to $u$ 
$\int_{-3}^uf'(x)dx\leq \int_{-3}^udx$
gives
$f(u)-f(-3)\leq u+3$
then
$f(u)\leq u$
2)integrate both sides from $u$ to $3$ 
$\int_{u}^3f'(x)dx\leq \int_{u}^3dx$
gives
$f(3)-f(u)\leq 3-u$
then
$-f(u)\leq -u$
which simply gives
$f(u)\geq u$
..
combine step 1 and 2 
$u\leq f(u)\leq u$
implies that $f(u)=u$ and $f(0)=0$
