# 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
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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