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Math Help - Graphing using 1st and 2nd derivatives

  1. #1
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    Graphing using 1st and 2nd derivatives

    f(x) = 3 sin x + 3 cos x

    0 ≤ x ≤ 2π

    How do I use this to find local max & min, critical points, and inflection points?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    I'm not sure of what you have to do, but if you have to graph f(x), you can convert it into a simple form:

    3\sin x + 3 \cos x = R \sin(x + \alpha)

    R = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18}

    \alpha = \tan^{-1}\left(\dfrac33\right) = \dfrac{\pi}{4}

    3\sin x + 3 \cos x = \sqrt{18} \sin\left(x + \dfrac{\pi}{4}\right)

    The first and second derivative should also be easier to graph now, provided you know how to graph f(x).
    Last edited by Unknown008; November 2nd 2010 at 04:21 AM. Reason: corrected miscalculation
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  3. #3
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    Quote Originally Posted by looseenz2 View Post
    f(x) = 3 sin x + 3 cos x

    0 ≤ x ≤ 2π

    How do I use this to find local max & min, critical points, and inflection points?
    "Critical points" are, by definition, points where the first derivative is 0. "Local max and min" are points where the first derivative changes sign (and so the second derivative is not 0), and inflection points are where the second derivative is 0.

    f'(x)= 3cos(x)- 3 sin(x)

    That will equal 0 when cos(x)- sin(x)= 0 or cos(x)= sin(x). That happens when x= \pi/4 or x= 5\pi/4.

    f''(x)= -3 sin(x)- 3cos(x)

    That will equal 0 when sin(x)+ cos(x)= 0 or cos(x)=-sin(x). That happens when x= 3\pi/4 or x= 7\pi/4.

    ( \sqrt{3^2+ 3^2}= \sqrt{2(3^2)}= 2\sqrt{3}, not \sqrt{19}.)
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Oops... dang
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