MVT Find bound A and B

Question

My friend threw this problem a few days ago, I don't recall the MVT ... could someone help me?

Suppose that $f'(x)=\frac{1}{2+x}$ and $f(0)=2$.

What bounds A and B, $A<f(4)<B$ come directly from the MVT?

(Answers should be written as fractions)

PD: I know I have been posting a lot of stuff, these questions have been sitting on my desk some months know and I am just excited to use this platform

Answer 1

Note that you can exactly get $f(4)$ without MVT here.

$f(x)=12x+\frac{x^2}{2}+c$ and since $f(0)=2$ then $c=2$

Thus $f(x)=\frac{x^2}{2}+12x+2$, so $f(4)=58$

Answer 2

By the Mean Value Theorem, $$ f(4) = f(0) + f'(t)\cdot 4 $$ where $t$ is some number in the open interval $(0,4)$. Since $f'(x) = \frac{1}{2+x}$, we know that for all $t\in (0,4)$, $$ \frac{1}{6} < f'(t) < \frac{1}{2} $$ Therefore, $$ 2 + \frac{1}{6} \cdot 4 < f(4) < 2 + \frac{1}{2} \cdot 4 \implies \frac{8}{3} < f(4) < 4 $$

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I kind of agree with Fareed; the MVT is not the right tool for this job since $f(x)$ can be found directly from $f'(x)$. But this is the technique used to estimate $f(x)$ when you know information about (e.g., upper and lower bounds of) $f'(x)$.