Proving the Mean Value Theorem with 3 ≤ f'(x) ≤ 5: A Homework Help Guide

[Loppyfoot]

Homework Statement

Let us suppose that, 3≤ f '(x) ≤5 for all x values. Show that 18≤ f(8) - f(2) ≤30.

The Attempt at a Solution

Alright folks... I am unsure where to start, or where to apply the MVT or the Rolle's Theorem.

Thanks

 

[rasmhop]

Well the mean-value theorem states that you can find some c in the interval (2,8) such that:
f'(c) = \frac{f(8)-f(2)}{8-2}
Now just note:
3 \leq f'(c)=\frac{f(8)-f(2)}{8-2} \leq 5

 

[Loppyfoot]

Alright, I understand that that is the equation of the secant line. How do I prove that it is ≤18 and ≤30?

 

[rasmhop]

You have:
3 \leq \frac{f(8)-f(2)}{6}\leq 5
by my previous post. Multiplying by 6 you get:
3\times 6 \leq f(8)-f(2)\leq 5\times 6

 

[Loppyfoot]

Oh... silly me. You multiply the six out of the bottom.
Thanks man, that really helped. Much love brah.