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Math Help - Inflection Points

  1. #1
    Senior Member topher0805's Avatar
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    Inflection Points

    I already know my first derivative to be:

    f'(x) = 16\cos {x}(-\sin {x} - 1)

    So I solved for my second derivative and got:

    -16\sin {x}(-\sin {x} - 1) + 16\cos {x}(-\cos {x})

    = 16\sin {x}(\sin {x} + 1) - 16\cos^2 {x}

    I'm looking for where f is concave up and concave down, and I figured that inflection points were a good way to find that. So, I set my equation equal to zero:

    16\sin {x}(\sin {x} + 1) - 16\cos^2 {x} = 0

    16\sin {x}(\sin {x} + 1) = 16\cos^2 {x}

    \sin {x}(\sin {x} + 1) = \cos^2 {x}

    I am completely stuck here. How do I solve for x now?
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  2. #2
    o_O
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    Assuming everything you've done so far is correct:
    sinx(sinx+1) = cos^{2}x

    sin^{2}x + sinx = 1 - sin^{2}x

    2sin^{2}x  + sinx - 1 = 0

    Solve the quadratic.
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  3. #3
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    The sign of the second derivative will indicate which direction the function opens: positive second derivative means concave up, negative second derivative means concave down. So those intervals on which the function has a positive second derivative are the intervals in which the function is concave up, vice versa for points where the function has a negative second derivative.
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  4. #4
    Senior Member topher0805's Avatar
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    Quote Originally Posted by o_O View Post
    Assuming everything you've done so far is correct:
    sinx(sinx+1) = cos^{2}x

    sin^{2}x + sinx = 1 - sin^{2}x

    2sin^{2}x  + sinx - 1 = 0

    Solve the quadratic.
    How do you solve the quadratic when it involves trig?
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  5. #5
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    Treat it like a polynomial in \sin(x). If you like it better say \sin(x) = y.

    Then your quadratic becomes 2y^2+y-1 = 0

    once you solve for the possible values of y substitue \sin(x) back in and solve for x using \arcsin or some other technique.
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  6. #6
    Senior Member topher0805's Avatar
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    Quote Originally Posted by iknowone View Post
    Treat it like a polynomial in \sin(x). If you like it better say \sin(x) = y.

    Then your quadratic becomes 2y^2+y-1 = 0

    once you solve for the possible values of y substitue \sin(x) back in and solve for x using \arcsin or some other technique.
    So I would have:

    <br />
(2y-1)(y+1)=0

    (2\sin {x} - 1)(\sin {x} + 1) = 0

    So x is 0 when \sin {x} = -1, \frac {1}{2}

    Thanks guys I appreciate it.
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