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

Question

Elijah Abram · Accepted Answer

#c=+-1/4# #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#

Christopher Allers · Answer

undefined To determine all values of c that satisfy the Mean Value Theorem on the interval [-1, 1] for the function f(x) = 3sin(2πx), we first need to find the average rate of change of the function over that interval, which is given by the slope of the secant line between the endpoints of the interval.

The average rate of change (slope of the secant line) is given by [f(1) - f(-1)] / (1 - (-1)).

Next, we find the derivative of f(x), which is f'(x) = 6πcos(2πx).

Then, we set the derivative equal to the average rate of change and solve for c:

f'(c) = [f(1) - f(-1)] / (1 - (-1)).

Finally, we solve for c:

6πcos(2πc) = [3sin(2π) - 3sin(-2π)] / 2.

After solving for c, we find all values of c within the interval [-1, 1] that satisfy the Mean Value Theorem for the function f(x) = 3sin(2πx).