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 $[1,3]$.
This proof is for $f^\prime$ **monotonically** decreasing, which means **not increasing** (but possibly constant).

The case for $f(3) > 5$ is easily handled with MVT as others have shown. We need only prove $f(3) \neq 5$.

Let's assume $f(3)= 5$.

We'll show that the only $f$ that is monotonically decreasing and meets all the other requirements, and that passes through both $(1,2)$ and $(3,5)$ is the line $f(x) = 3/2(x-1) + 2$, and then we'll have a contradiction because the slope of this line at $x = 3$ is not $0$.

First, let's assume $f$ contains a point $(x\_1,f(x\_1))$ above the line. We have

$f(x\_1) > 3/2(x\_1 - 1) + 2$

Using MVT between the point $(1,2)$ and this point $(x\_1, f(x\_1))$ above the line, shows that there is some $y$ in the interval $(1,x\_1)$ s.t.

$f^\prime(y) = \frac{f(x\_1) - 2} {x\_1 - 1} > \frac{3/2(x\_1 - 1) + 2 - 2}{x\_1 - 1}$

$f^\prime(y) > 3/2$

which is a contradiction.

Similarly, we can show that $f$ must not contain any points below the line:

If $f(x\_1) < 3/2(x\_1 - 1) + 2$, then using MVT on the interval $[x\_1, 3]$ we find that there must be a $y$ in $(x\_1, 3)$ s.t. $f^\prime(y) > 3/2$.

$f^\prime(y) = \frac{5 - f(x\_1)} {3 - x\_1} > \frac{5 - 3/2(x\_1 - 1) - 2}{3 - x\_1}$

$f^\prime(y) > \frac{3 - 3/2(x\_1 - 1)} {3 - x\_1}$

$f^\prime(y) > \frac{9/2 - 3/2x\_1}{3 - x\_1}$

$f^\prime(y) > 3/2$.

So, we've shown that $f$ must not contain any points above or below the line connecting the points $(1,2)$ and $(3,5)$. Therefore, $f$ *is* that line.

However, we also require $f$ to be differentiable on the interval $[1,3]$. Our $f$ has $f^\prime(3) = 3/2$, and also $0$, another contradiction.

Therefore, our beginning assumption, that $f(3) = 5$ must be wrong.

As we've already ruled out $f(3) > 5$, we are left with $f(3) < 5$.