Proving $f(3) < 5$ 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.
 A: 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 $[1,3]$.
A: 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, $f(3)<5$.
A: === original intuitive; big picture why;  my first answer-- I think it is correct but it involves a lot of handwaving and lack of rigor=======
If you extend a straight line through to point $(1,f(1))=(1,2)$ with the slope $f'(1)=\frac 32$ it will go through the point $(1+2, 2+\frac 2\times \frac 32)= (3,5)$.
$f(x)$ starts out increasing as fast as that line does but as $f'(x)$ is decreasing it does not mantain that rate of increase.  And $x=3$, $f'(3)=0$ so it has decreased and so because it increased at a rate less than the line did it can not have reached the heights that the line did.  As the line has $(3,5)$, the function $f$ must have $(3, k)$ where $k < 5$.
===== second formal argument: extreme reducto ad absurdum====
Case 1:  If all $x \in [1,3]$,   $f(x) = 2 + \frac 32(x-1)$.

Upshot:  The fails $f'(3) = 0$. This can not be true.

then $f'(3)=\lim_{h\to 0}\frac {f(x+h)-f(x)}{h}$ but the left side limit of that for $h \to 0; h< 0$ is equal to $\frac 23$ and not $f'(3) =0$ as stated.  So this is not the case.
So there is a $c \in [1,3]$ where $f(c) \ne 2 +\frac 23(c-1)$.
Case 2: $f(c) > 2+\frac 23(c-1)$.

Upshot: If we consider MVT when we  the mean value between $(1,f(1))$ and $(c,f(c))$ this contradicts that $f'(x)$ is decreasing. This can not be true.

According to the mean value theorem there is a point $d \in (1,c)$ where $f'(d)  = \frac {f(c) - f(1)}{c-1}> \frac {2+\frac 23(c-1) - 2}{c-1} = \frac 32$.  This contradicts that $f'(x)$ is decreasing and $f(1) =\frac 32$.
Case 3:  $f(c) < 2+\frac 32(c-1)$.

Upshot: this must be true as it is the only case left.  But if we apply the MVT to compare $(c,f(c))$ and $(3,f(3))$ and noting that $f'(x) \le \frac 32$ we will get a calculation that shows $f(3) < 5$.

By MVT there must be a point $d \in (c,3)$ there $f'(d) = \frac {f(3)- f(c)}{3-c} > \frac {f(3)-2-\frac 32(c-1)}{3-c}$ so $f(3) < f'(d)(3-c)+2+\frac 32(c-1)$.
Now $\frac 32 f'(1)\ge f(c) \ge f'(d)\ge f(3)=0$ and $1 < c < d < 3$. so
$f(3) < \frac 32(3-c) + 2 + \frac 32(c-1) = 5$.
=== My first formal argument but contains a fallacy that the derivative function is continuous and is a bit hard to follow (don't really like it but I leave it in for ... transparency) =====
Or.....
As $f'(1) = \frac 32$ and $f'(3) = 0$ there must be a value $c \in (1,3)$ there $f'(c) = 1$.
Suppose $f(3) \ge 5$.
If $f(c) > 2 + \frac 32(c-1)$ then there must be a point $d\in (1,c)$ there $f'(0) = \frac {f(c) - f(1)}{c-1} > \frac {2+\frac 32(c-1) - 2}{c-1} =\frac 32$ but $f'$ is decreasing so that is impposible
So $f(c) \le 2 + \frac 32(c-1)$.  No there must be a point $e$  where $f'(e) = \frac {f(3) - f(c)}{3-c} \ge \frac {f(3)-2 -\frac 32(c-1)}{3-c}$.  But $f'$ is decreasing so $f'(x) \le f'(c) =1$ so $1 \ge \frac {f(3) - f(c)}{3-c} \ge \frac {f(3)-2 -\frac 32(c-1)}{3-c}$ and $f(3)\le (3-c)+(2+\frac 32(c-1))= \frac 72 + \frac 12 c < \frac 72 + \frac 12\cdot 3 = 5$.
A: 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 $f(3) < 5$.
