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Math Help - Simple Factoring

  1. #1
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    Simple Factoring

    So I have this equation
    f" (x) = -2sinx - 4sin2x
    My question is what can I factor out of this equation? my guess is - 2sin or is it just -2?? I'm confused....
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  2. #2
    Senior Member MacstersUndead's Avatar
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    Re: Simple Factoring

    Recall that sin(2x) = 2sin(x)cos(x)
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    Re: Simple Factoring

    So I'm asked to find the points of inflection and to discuss the concavity of the graph of the function
    In order to do that I know i need to take the first and second derivative... so here's the work that I have so far

    - Original Function
    f(x) = 2sinx + sin2x

    - First Derivative
    f' (x) = 2cosx + 2cos2x

    - Second Derivative
    f" (x) = -2sinx - 4sin2x

    I only needed to take up to the second derivative.... if I take the sin2x from 4sin2x and turn it into 2sin(x)cos(x) wouldn't I be taking the third derivative when i only needed to take the second derivative??
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    Senior Member MacstersUndead's Avatar
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    Re: Simple Factoring

    In the case of substituting sin(2x) with 2sin(x)cos(x), all we would be using is a trigonometric identity to simplify.

    f''(x) = -2sin(x) - 4sin(2x) = -2sin(x) - 4*(2sin(x)cos(x) = -2sin(x) - 8sin(x)cos(x) = -2sin(x) [ 1 + 4cos(x) ]
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    Re: Simple Factoring

    So to find the inflection points i need to set x equal to zero... heres what i got

    -2sinx = 0
    sinx = -2

    and the other one is...

    1 + 4cos(x) = 0
    4cos(x) = 1
    cos(x) = -1/4

    So when i set both x equal to zero i got sinx = -2 and cos(x) = -1/4
    NOW at this point I'm lost.... I'm not familiar with this type of situation... usually i'm used to seeing sinx = 1/2, and i know that equals to Pi/6 and 5Pi/6 in the unit circle and just plug it in a number line and check for concavity... but now i'm confused, what do i do now?
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    Senior Member MacstersUndead's Avatar
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    Re: Simple Factoring

    Simple arithmetic error
    -2sin(x) = 0
    sin(x) = 0 [divide by -2 on both sides]

    which makes sense since the zeroes wouldn't change if you were to apply a vertical stretch by a factor of -2.

    For cos(x) = -1/4, we know that a solution exists since -1/4 is in [-1,1] so apply cos^-1 to both sides to get your solution.
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    Re: Simple Factoring

    okay so since sinx = 0, then that corresponds to Pi and 0 in the unit circle
    and cosx is equal to -1 and 1? if it is then cos of -1 corresponds to Pi while cos of 1 corresponds to 0... am i right?
    Last edited by asilvester635; November 17th 2012 at 04:27 PM.
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    Re: Simple Factoring

    okay so since sinx = 0, then that corresponds to Pi and 0 in the unit circle
    and cosx is equal to -1 and 1? if it is then cos of -1 corresponds to Pi while cos of 1 corresponds to 0... am i right?
    Last edited by asilvester635; November 17th 2012 at 04:27 PM.
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  9. #9
    Senior Member MacstersUndead's Avatar
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    Re: Simple Factoring

    No. [-1,1] is the interval in which cos^-1 is formally defined. For example, if we had cos^-1 = -5, then the solution would not exist. but since -1/4 is in the interval [-1,1], then the solution does exist.

    cos^-1(-1/4) is approx. 1.82 radians
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    Re: Simple Factoring

    thank you so much.... but i am right with the "sinx = 0, then that corresponds to Pi and 0 in the unit circle" part right?
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  11. #11
    Senior Member MacstersUndead's Avatar
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    Re: Simple Factoring

    Yes.
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