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Math Help - 1st and 2nd derivative tests

  1. #1
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    1st and 2nd derivative tests

    Alright so I have a few problems using the derivative tests and I understand like if f' changes from positive to negative at c, the f has a local maximum at c. But then for like this problem cos^2x - 2sinx, 0 < x < 2pie those are less then or equal to don't know how to show that. And take the derivative -2sinx-2cosx so we have a sign change but whats c? So here are a couple problems i need help on.

    Find intervals at which f is increasing and decreasing, find local max and min values of f, and find the intervals of concavity and inflection points. for :

    f(x) = cos^2x - 2 sinx, 0 < x < 2pie (less then or equal to)


    Thanks for any help.
    Last edited by UMStudent; April 5th 2007 at 05:17 PM.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by UMStudent View Post
    Find intervals at which f is increasing and decreasing, find local max and min values of f, and find the intervals of concavity and inflection points. for :

    f(x) = cos^2x - 2 sinx, 0 < x < 2pie (less then or equal to)
    f(x) = cos^2x - 2sinx
    f'(x) = -2cosx*sinx - 2cosx
    f''(x) = -2cos^2x + 2sin^2x + 2sinx

    Set f'(x) = 0
    0 = -2cosx*sinx - 2cosx
    0 = -2cosx(sinx + 1)

    cosx = 0 --> x = pi/2, 3pi/2
    sinx + 1 = 0 --> sinx = -1 --> x = 3pi/2

    The roots of f'(x) are: x = pi/2 and x = 3pi/2

    Choose values between the following intervals to find where the graph is increasing or decreasing:
    For (0,pi/2), choose x = pi/4 --> f'(pi/4) = (negative)
    For (pi/2,3pi/2), choose x = pi --> f'(pi) = (positive)
    For (3pi/2,2pi), choose x = 7pi/4 --> f'(7pi/4) = (negative)

    Decreasing: (0,pi/2) U (3pi/2,2pi)
    Increasing: (pi/2,3pi/2)
    Minimum: x = pi/2, x = 2pi
    Maximum: x = 3pi/2, x = 0


    Set f''(x) = 0
    0 = -2cos^2x + 2sin^2x + 2sinx
    0 = -2(1 - sin^2x) + 2sin^2x + 2sinx = -2 + 4sin^2x + 2sinx
    0 = (sinx + 1)(4sinx - 2)

    sinx + 1 = 0 --> sinx = -1 --> x = 3pi/2
    4sinx - 2 = 0 --> sinx = 1/2 --> x = pi/6, 5pi/6

    The roots of f''(x) are: x = 3pi/2, x = pi/6, x = 5pi/6

    Choose values to find concavity:
    For (0,pi/6), choose x = pi/12 --> f''(pi/12) = (negative)
    For (pi/6,5pi/6), choose x = pi/2 --> f''(pi/2) = (positive)
    For (5pi/6,3pi/2), choose x = pi --> f''(pi) = (negative)
    For (3pi/2,2pi), choose x = 7pi/4 --> f''(7pi/4) = (negative)

    Concave Up: (pi/6,5pi/6)
    Concave Down: (0,pi/6) U (5pi/6,3pi/2) U (3pi/2,2pi)
    Inflextion: x = pi/6, x = 5pi/6

    Hopefully I made no mistakes in all of that. I suggest you check my work.
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