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March 28, 2017

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Determine the concavity and find the points of inflection.
y=2cosx + sin2x 0≤x≤2pi
y'=-2sinx + 2cos2x
y"=-2cosx-4sinx

How do I find the IP(s)?

  • Calculus - ,

    Your second derivative's last term should be -4sin2x

    set y'' = 0 for points of inflection
    -2cosx - 4sin2x = 0
    cosx + 2sin2x = 0
    cosx + 4sinxcosx = 0
    cosx(1 + 4cosx) = 0
    cosx = 0 or cosx = -1/4

    You seem to know what you are doing, and can probably take it from here.

    Concavity in short:
    If for some f(x), the second derivative exists,
    the curve is concave upwards if f''(x) > 0
    the curve is concave downwards if f''(x) < 0
    if f''(x) = 0 it is neither and you have a point of inflection as seen above.

  • correction - Calculus - ,

    from cosx + 4sinxcosx = 0
    it should have been
    cosx(1 + 4sinx) = 0
    cosx = 0 or sinx = -1/4

  • Calculus - ,

    I got-4sin2x for the second term, I made a typo.

    How do you translate sinx=-1/4 to decimal numbers?

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