# Thread: One Mean Value Thm Problem

1. ## One Mean Value Thm Problem

I'm new to M.V.T. and Rolle's Thm, so I guess I'm not quite comphrehending the concepts... Here's the problem:

Suppose that f is continuous on [0,4] and differentiable on (0,4). Also, suppose f(0)=1 and 3<f'(x)<6 for all x in (0,4). Which of the following must be true?

A) 13<f(4)<25

B) 3<f(4)<6

C) 12<f(4)<24

D) f(4) is undefined

Any help/explanation that would aid in my understanding of the concepts would be very appreciated!

2. Originally Posted by obsmith08
I'm new to M.V.T. and Rolle's Thm, so I guess I'm not quite comphrehending the concepts... Here's the problem:

Suppose that f is continuous on [0,4] and differentiable on (0,4). Also, suppose f(0)=1 and 3<f'(x)<6 for all x in (0,4). Which of the following must be true?

A) 13<f(4)<25

B) 3<f(4)<6

C) 12<f(4)<24

D) f(4) is undefined

Any help/explanation that would aid in my understanding of the concepts would be very appreciated!
the MVT says that for at least one value of x in the given interval ...

$f'(x) = \frac{f(4) - f(0)}{4 - 0}$

$f'(x) = \frac{f(4) - 1}{4}$

since $3 < f'(x) < 6$ ...

$3 < \frac{f(4) - 1}{4} < 6$

find the solution interval for $f(4)$

3. Originally Posted by skeeter
the MVT says that for at least one value of x in the given interval ...

$f'(x) = \frac{f(4) - f(0)}{4 - 0}$

$f'(x) = \frac{f(4) - 1}{4}$

since $3 < f'(x) < 6$ ...

$3 < \frac{f(4) - 1}{4} < 6$

find the solution interval for $f(4)$

Thank you so much! I can see now that I was making this problem out to be much more difficult than it actually is! (Math--especially Calculus--isn't really my forte though)