Calculus

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Determin the absolute extreme values of the function on the given interval.

1.)y= 3 sin x + 4 cos x, xE[0, 2pie]

  • Calculus -

    dy/dx = 3 cos x - 4 sin x
    =0 at max or min
    3 cos x = 4 sin x
    so
    tan x = 3/4 at extreme
    3,4,5 triangle.
    tan is 3/4 in quadrants 1 and 3
    In quadrant 1
    sin x = 3/5
    cos x = 4/5
    3 sin x + 4 cos x = 9/5 + 16/5 = 5 (maximum probably)

    do similar computation in quadrant 3 and I suspect you will find the minimum.

  • Calculus -

    y'=3cosx-4sinx=0
    3cosx=4sinx
    tanx=3/4

    solve for x. You will get two angles, 180degrees apart.

    One is a max, one is a min.

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