# AP CALC

Find the POSITIVE value of x that satisfies the mean value theorem for
f(x)=sin(x) on the closed interval
[-3pi/2, 3pi/2].

please help I have no idea how to solve this problem

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1. sin 3 pi/2 = sin 270 = -1
sin - 3 pi/2 = +1

change in sin x = -1 - +1 = -2
change in x = 3 pi/2+3pi/2 = 3 pi
so
slope from left to right = -2 / 3pi
= -.2122

so where does the derivative of sin x = -.2122 ?

d/dx (sin x) = cos x
so cos x = -.2122
x = 102.25 degrees = 1.78 radians

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