Mean Value Theorem Help

1. The problem statement, all variables and given/known data
Let us suppose that, 3≤ f '(x) ≤5 for all x values. Show that 18≤ f(8) - f(2) ≤30.

3. The attempt at a solution
Alright folks... I am unsure where to start, or where to apply the MVT or the Rolle's Theorem.

Well the mean-value theorem states that you can find some c in the interval (2,8) such that:
[tex]f'(c) = \frac{f(8)-f(2)}{8-2}[/tex]
Now just note:
[tex]3 \leq f'(c)=\frac{f(8)-f(2)}{8-2} \leq 5[/tex]
Alright, I understand that that is the equation of the secant line. How do I prove that it is ≤18 and ≤30?
You have:
[tex]3 \leq \frac{f(8)-f(2)}{6}\leq 5[/tex]
by my previous post. Multiplying by 6 you get:
[tex]3\times 6 \leq f(8)-f(2)\leq 5\times 6[/tex]
Oh... silly me. You multiply the six out of the bottom.
Thanks man, that really helped. Much love brah.

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