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Calculus

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Given f(x)=sin(x)-2cos(x) on the interval [0,2pi]. Determine where the function is concave up and concave down.

  • Calculus -

    f'(x) = cosx + 2sinx
    f''(x) = -sinx + 2cosx

    points of inflection"
    f''(x) = 0
    -sinx + 2cosx = 0
    sinx = 2cosx
    tanx = 2
    x = 63.4 degrees or x = 243.4 degrees

    these last two are our important x values, since they are where the curve switches between concave up and concave down

    lets's pick values in between
    f''(x) = sin0 - 2cos0 = -2, so at 0 it is concave upwards
    f''(90) = sin90 - 2cos9- = 1, so at 90 degrees it is concave down
    f''(270) = sin270 - 2cos270 = -1 , so at 270 degrees it is concave upwards again.
    f''(360) = sin360 - 2cos360 = -2 , sure enough concave upwards

    so concave up: from 0 < x < 63.4
    concave down : 63.4 < x < 243.4
    concave up : 243.4 < x < 360

    Just noticed you probably wanted your answer in radians, my calculator was set to degrees.
    No big deal, just repeat my calculations with radian settings

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