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Math Help - integral..find local max

  1. #1
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    integral..find local max

     f(x) = \int_0^x \frac{ t^2 - 36 }{ 1+\cos^2(t)} dt

    At what value of  x does the local max of  f(x) occur?

     x = ?


     f '(x) = \int_0^x \frac{ x^2 - 36 }{ 1+\cos^2(x)}
    im stuck on this part, not sure what to do after.
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  2. #2
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    Quote Originally Posted by viet View Post
     f(x) = \int_0^x \frac{ t^2 - 36 }{ 1+\cos^2(t)} dt

    At what value of  x does the local max of  f(x) occur?
    If a real function is differenciable then the local max (relative maximum) must occurs when f'(x)=0

    Using the Fundamental theorem,
    f'(x)=\frac{x^2-36}{1+\cos^2 x}
    Thus,
    \frac{x^2-36}{1+\cos^2 x}=0
    Thus,
    x^2-36=0, x=\pm 6
    Those are the possible points.
    To determine what status they have we will use the first-derivative test.

    Consider the three intervals,
    x<-6----> f'(x)<0----> decreasing.
    -6<x<6----> f'(x)>0----> increasing.
    6<x----> f'(x)<0----> decreasing.

    Thus, the relative maximum occurs at x=6
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by viet View Post
     f(x) = \int_0^x \frac{ t^2 - 36 }{ 1+\cos^2(t)} dt

    At what value of  x does the local max of  f(x) occur?

     x = ?


     f '(x) = \int_0^x \frac{ x^2 - 36 }{ 1+\cos^2(x)}
    im stuck on this part, not sure what to do after.
    If:

    <br />
f(x) = \int_0^x \frac{ t^2 - 36 }{ 1+\cos^2(t)} dt<br />

    then (fundamental theorem of calculus)

    <br />
f'(x) = \frac{ x^2 - 36 }{ 1+\cos^2(x)}<br />

    since { 1+\cos^2(x)} \ne 0 the stationary points of f(x) are the roots of x^2-36=0, which are x=\pm 6.

    To determine which is the maximum and which the minimum find f''(x) at x=\pm 6, and check the sign negative and its a maximum and positive its a minimum (the minimum is at -6 but you had better check)

    RonL
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