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Mean Value Theroem

  1. Apr 4, 2004 #1
    The Mean Value Theroem there is a number c in the interval (a,b) such that
    f'(c)=f(b)-f(a)/b-a or f(b)-f(a)=f'(c)(b-a)

    Okay, say a function has a derivative of zero. I'm supposed to explain what the Mean Value Theroem states about a function whose derivative is zero.

    Any takers?
  2. jcsd
  3. Apr 4, 2004 #2


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    This isn't right; you forgot some parentheses.

    Anyways, you're told that the derivatve is zero, right? Well, plug in zero for the derivative and what do you get?
  4. Apr 4, 2004 #3
    i plug in 0 for c it would be 0. so what?
  5. Apr 4, 2004 #4
    If the derivative of a function is zero at all points, then the total change in value of the function is going to be zero. So now, does it really matter which point you pick? It's going to be zero either way!

  6. Apr 4, 2004 #5


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    You were told the derivative of f is zero, right? Is c the derivative of f? (No) So why plug 0 in for c? Plug 0 in for the derivative of f.
    Last edited: Apr 4, 2004
  7. Apr 5, 2004 #6
    plug 0 in where?
  8. Apr 5, 2004 #7
    f'(c) = 0

    so we get

    0 = (f(b) - f(a))/(b-a)

    Now solve for c.

  9. Apr 5, 2004 #8
    how do i know what the functino is or a and b? do i just solve for it in terms of f a and b?
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