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Mean Value Theorem

  • Thread starter 1calculus1
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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?
PLEASE HELP.
 

Answers and Replies

  • #2
Dick
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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.
 
  • #3
Defennder
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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?
 
  • #4
Dick
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I thought the OP meant "equal the equation" to (f(b)-f(a))/(b-a). I may have been extrapolating on that.
 
  • #5
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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?
 
  • #6
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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!
 
  • #7
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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
 
  • #8
Dick
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Don't "equal it to 1". Equal it to (f(b)-f(a))/(b-a)=13. Read the mean value theorem again.
 
  • #9
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HAHA. Thanks. That's all I needed to know. You've solved one of my many problems, AGAIN! THANKS!
 
  • #10
Dick
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Gotta admit, you resolve your own problems quickly. Hope this is a short lived phase of confusion.
 

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