guessing the value based on derivative

Question

My third question today, but don't get me wrong, I did put a lot of thoughts before asking.

In this question, I assume that $f'$ may be a convex function, similar to $x^2$, so I suppose $f'(1)=f'(3)=0$ and $f'(2)=-1$, so $f(0)<f(1)$ and $f(1)>f(3)$ but with the longer length, then I guess the solution is D, which is right.

However, I would need a concrete answer rather than just a lucky guess. Can you help me?

Thank you

Answer 1

The hypotheses for the mean value theorem are conspicuously satisfied on the interval $(-1,4)$, so $$ \frac{f(3)-f(0)}{3}=f'(c)\geq -1\implies f(0)\leq 8 $$

Answer 2

Let apply MVT

$$\frac{f(3)-f(0)}{3-0}=f'(c)\implies f(0)=f(3)-3f'(c)\implies f(0)_{max}=8$$

Answer 3

$f(3)=f(0)+ \int_0^3 f'(t) dt \ge f(0) + \int_0^3 (-1) dt = f(0) -3$ and so $f(0) \le f(3)+3 = 8$.