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Math Help - Mean Value Theorem on indicated intervals

  1. #1
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    Mean Value Theorem on indicated intervals

    Just making sure if I got mean value theoroms,
    Apply the Mean Value Theorem to f on the indicated interval, and find all values of c in the interval (a, b) such that: f'(c)=(f(b)-f(a))/(b-a)

    So if I have something like f(x)=x2 and the intervals are [-2, 1]
    Answer is: f'(-1/2)=-1
    I would plug in the intervals first into the equation right?
    So f(a)=f(-2)=4 and f(b)=f(1)=1
    Then I just plugged in what I know back into that equation up there, which turns out to be (1-4)/(1+2)=(-3/3) = -1
    But I'm a bit confused, I got the -1 thing down, but how did they get the (-1/2) in there.
    So I thought it was the value of 'c', so I did f(c)=c2, then did f'(c), which is 2c=-1
    So c=-1/2, I think, if I did this right, but can someone verify if this is the way to do it.

    Another example is like f(x)=x2/3 with intervals [0, 1]
    Answer is: f'(8/27)=1
    I did the same thing,
    So f(a)=f(0)=0 and f(b)=f(1)=1
    Then I plugged it back into the equation (1-0)/(1-0)=1
    But I'm confused on where they got (8/27)
    So I thought f'(c)=(2/3)x-1/3=1, then x-1/3=(3/2), then I got lost on here, since I did both side to the third power, but that was a bit off o.o


    I'm not sure if that was a concidence that I get the ending result, but the thing inside the paranthesis inside the answer is what I'm lost at, which is the 'value of c' I believe.

    Can someone explain please? Thanks
    Last edited by Chaim; November 27th 2012 at 07:53 PM.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: Mean Value Theorem on indicated intervals

    For the first one, we then want to set:

    f'(x)=-1

    2x=-1

    x=-\frac{1}{2}

    So we know f'\left(-\frac{1}{2} \right)=-1

    See if you can use this technique to get the second one.
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  3. #3
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    Re: Mean Value Theorem on indicated intervals

    Quote Originally Posted by MarkFL2 View Post
    For the first one, we then want to set:

    f'(x)=-1

    2x=-1

    x=-\frac{1}{2}

    So we know f'\left(-\frac{1}{2} \right)=-1

    See if you can use this technique to get the second one.
    Ok, so for the 2nd one
    f(x)=x2/3 with intervals [0, 1]
    Answer is: f'(8/27)=1

    Lets see
    So first you would use that equation at the top right?
    So it's f(b) which is f(1)=1, and f(a)=0
    Then the equation is (1-0)/(1-0) = 1

    Then you took f(c) to the original equation, which then becomes c2/3
    Then the derivative of that is (2/3)c-1/3=1
    c-1/3=(3/2)
    So I multiplied the exponents both by (-3/1), I think, right? o.o
    Which makes it (3/2)(-3/1)

    So that means it'll be (3/2)-3, which then turns into (2/3)3, which equals 8/27!!!
    Thanks!!!
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    Re: Mean Value Theorem on indicated intervals

    Also just wondering, if I have something like x(x2-x-2) intervals [-1, 1]
    I got f(-1)=0
    f(1)=-2
    Then (-2-0)/(1-(-1)) = -2/2 = -1
    Then I did f(c)=c(c2-c-2)=-1
    Then I got c(c-1)(c-2)=-1
    I was a bit confused on this one, where there are more of the same variable, and the end is -1, so it's hard to find out since regularly it would be 0.

    So how would I find what c is here?
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: Mean Value Theorem on indicated intervals

    You want to set:

    f'(x)=-1

    3x^2-2x-2=-1

    3x^2-2x-1=0

    (3x+1)(x-1)=0

    Discard the root not within the open interval (-1,1) to get:

    c=-\frac{1}{3}
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  6. #6
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    Re: Mean Value Theorem on indicated intervals

    Quote Originally Posted by MarkFL2 View Post
    You want to set:

    f'(x)=-1

    3x^2-2x-2=-1

    3x^2-2x-1=0

    (3x+1)(x-1)=0

    Discard the root not within the open interval (-1,1) to get:

    c=-\frac{1}{3}
    Oh right!
    Forgot to take the derivative!
    Thanks!
    I'm starting to understand the Mean Value Theorom now
    I would add to your rep, but I gotta spread it xD.
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