Question: Let $f$ be a differentiable on $[1, 3]$ with $f(1) = 2$. Suppose $f'(x)$ is decreasing on $[1, 3]$ with $f'(1) = 3/2$ and $f'(3) = 0$. Why is it certain that $f(3) < 5$?

I know that $f''(x) < 0$ since $f'(x)$ is a decreasing function, but I'm not entirely sure how to proceed. I recently learned MVT and Rolle's Theorem, but don't see how it's applicable here.

Answer: 
If $f(3)>5$, then there exists $c\in(1,3)$ such that $f'(c)=\frac{f(3)-f(1)}{3-1}>\frac32$, which is not possible since $f'(1)=\frac32$ and $f'$ is decreasing on $[1,3]$.
As a consequence of MVT, there exists $c\in (1,3)$ such that
$$\frac{f(3)-f(1)}{3-1} = f'(c).$$
This is equivalent to
$$f(3) =2f'(c)+f(1) = 2f'(c) + 2.$$
But $f'(c)\leq f'(1)=\frac{3}{2}$ because $f'$ is decreasing in $[1,3]$. Thus,
$$f(3) \leq 2\frac{3}{2} + 2 = 5.$$
So, in order to complete the exercise, we must prove that $f(3)<5$. Assume that $f(3)=5$. Then,
$$\frac{f(3)-f(1)}{3-1} = \frac{5-2}{2} = \frac{3}{2}$$
Consequently, by the MVT there exists $c\in(1,3)$ such that $f'(c)=\frac{3}{2}$. But $f'$ is decreasing and $f'(1)=f'(c)$, thus
$$f'(x)=\frac{3}{2},\;\; \forall x\in [1,c].$$
It is not difficult to see that
$$f(x) = \frac{3}{2}x+k.$$
As $f(1)=2$ we can calculate $k$. So,
$$f(x) = \frac{3}{2}x+\frac{1}{2}.$$
As $f'(3)=0$, we assure the existence of certain value $c'$ such that $f'(x)<\frac{3}{2}$ for all $x>c'$ (this can be done using a supremum argument). Now, we can use the Mean Value Theorem again in the interval $[c',3]$ then, we have a $d\in(c',3)$ such that
$$f'(d) =\frac{f(3)-f(c)}{3-c} = \frac{5-\frac{3}{2}c-\frac{1}{2}}{3-c} = \frac{3}{2}.$$
But this last fact is impossible. Then, $f(3)<5$.