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Math Help - Mean Value Theorem

  1. #1
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    Mean Value Theorem

    Apply the mean value theorem to the function f(x)= x^2 +2x on the interval [0,2] to determine any valid conclusion the theorem would make.

    I know the formula is f(b)-f(a) / b-a = f'(c)

    I ended up with (4+2)-(2) / (2-0) = 2

    So do I take that 2 and plug it into the derivative of the equation. I did that and got 2(2)+2=6

    I am not sure if I did that right, or what it says about the equation?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by KarlosK View Post
    Apply the mean value theorem to the function f(x)= x^2 +2x on the interval [0,2] to determine any valid conclusion the theorem would make.

    I know the formula is f(b)-f(a) / b-a = f'(c)

    I ended up with (4+2)-(2) / (2-0) = 2

    So do I take that 2 and plug it into the derivative of the equation. I did that and got 2(2)+2=6

    I am not sure if I did that right, or what it says about the equation?
    Note f(2)=8 and f(0)=0.

    Now, \frac{f(2)-f(0)}{2-0}=\frac{8-0}{2-0}=4.

    Also note that f^{\prime}(x)=2x+2\implies f^{\prime}(c)=2c+2

    Therefore, by MVT, \exists\,c\in[0,2]: 2c+2=4.

    So what is c?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Note f(2)=8 and f(0)=0.

    Now, \frac{f(2)-f(0)}{2-0}=\frac{8-0}{2-0}=4.

    Also note that f^{\prime}(x)=2x+2\implies f^{\prime}(c)=2c+2

    Therefore, by MVT, \exists\,c\in[0,2]: 2c+2=4.

    So what is c?
    C=1

    So you plug the first numbers into the original function... and only plug the value in the first equation into the derivative.
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