Question

How do you determine all values of c that satisfy the mean value theorem on the interval [-1, 1] for `f(x)=3sin(2πx)`?

Answer 1

`c=+-1/4`

Explanation:

`f'(x)=3cos(2pix)*2pi = 6pi cos(2pix)`
Mean Value Theorem
`f'(c)=(f(b)-f(a))/(b-a)`
`6pi cos (2pi c)=(f(1)-f(-1))/(1--1)`
`6pi cos (2pi c)=(3sin2pi-3sin(-2pi))/(1--1)`
`6pi cos (2pi c)=(3sin2pi+3sin(2pi))/(2)`
`6pi cos (2pi c)=(6sin(2pi))/(2)`
`6pi cos (2pi c)=0/2`

`6pi cos (2pi c)=0`

`cos(2pic)=0`

`2pic=cos^-1 0`
`2pic =+-pi/2`
`c=(+-pi/2 )/(2pi)= +-(pi/2) (1/(2pi))`
`c=+-1/4`