Asked by Anonymous
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
f(x)=sin(x) on the closed interval
[-3pi/2, 3pi/2].
please help I have no idea how to solve this problem
Answers
Answered by
Damon
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
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