# Mean Value Theorem

## Homework Statement

Given the function f(x)= x(x^2-8)-5 satisfies the hypothesis of the Mean Value Thereom on the interval [1,4], find a number C in the interval (1,4) which satisfies this thereom.

## Homework Equations

f'(c) = f(b)-f(a) / b-a

## The Attempt at a Solution

1) Expand the equation first
2) Find the first derivative.
3) Equal the equation to 1

Apparently, I got the wrong answer. What am I doing wrong?

Dick
Homework Helper
That's a great strategy. Impossible to tell you how you got the wrong answer until you tell us what you got for (f(b)-f(a))/(b-a) and C.

Defennder
Homework Helper
What do you mean by "equal the equation to 1"? Don't you have to find both f(1) and f(4) then use the Mean Value theorem to find c?

Dick
Homework Helper
I thought the OP meant "equal the equation" to (f(b)-f(a))/(b-a). I may have been extrapolating on that.

Hm.. I just realized it and I'm stuck. I don't know what to do or what I'm trying to get.. HAHA

On the bright side, I do have the C value and the value for (f(b)-f(a))/(a-b):

(f(b)-f(a))/(a-b)
( 27 + 12 )/(4-1) = 13

C value = x^3-8x-5
f' = 3x^2-8
3x^2-8 = 1
3x^2 = 9
9 / 3 ^1/2
= 3^1/2

So, what to do next? Or what the heck am I suppose to get?

I found out the tangent line is at (3^1/2, -13.66) which is parallel to the secant line through (1, -12) and (4, 27)

Now, I don't even know if that helps.. but there it is. Lol!

What do you mean by "equal the equation to 1"? Don't you have to find both f(1) and f(4) then use the Mean Value theorem to find c?
What I meant by equal the equation to one was that getting the derivative of the equation and equalling it to 1.

1 = 3x^2 -8

Dick
Homework Helper
Don't "equal it to 1". Equal it to (f(b)-f(a))/(b-a)=13. Read the mean value theorem again.

HAHA. Thanks. That's all I needed to know. You've solved one of my many problems, AGAIN! THANKS!

Dick