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Mean Value Theorem Calculus 1

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the function f(x)=2x^3−12x^2−72x+6 on the interval [−4,7] . Find the average or mean slope of the function on this interval.


    2. Relevant equations

    MEAN VALUE THEOREM
    [tex]f'(c) = \frac{f(b)-f(a)}{b-a}[/tex]

    3. The attempt at a solution

    When I set this problem up in the mean value theorem I found that

    6c^2 - 24c - 72 = (-400+26)/(11)
    Finally giving me

    6c^2 - 24c -38 = 0

    Solving with the quadriatic equation I got

    [tex]2\pm\frac{\sqrt93}{3}[/tex]


    Which I know is correct, but the program says it is not correct so maybe the question is not asking for me to do this.....?
     
  2. jcsd
  3. Nov 7, 2009 #2
    I am honestly not for sure what you are doing. You just have some equations there. What the mean value theorem says is that on a closed interval [a,b], there exists a point c such that f'(c)="mean slope of f over the interval [a,b]". What equation gives you the mean slope?
     
  4. Nov 7, 2009 #3
    I was finding the value of "c" in the mean value theorem, because I was overthinking this problem. After I read your post several times it finally sunk in that the mean slope is equal to (y1-y0)/(x1-x0). Which is -34 and the answer to the problem. Thank you for the help.
     
  5. Nov 8, 2009 #4
    Great! Yes, (f(b)-f(a))/(b-a) is the mean slope over the interval [a,b]. Just to reiterate, the mean value theorem guarantees there is at least one point c such that f'(c) is exactly equal to the mean slope over [a,b].
     
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