Find the value of c which satisfies Mean Value Theorem for the function f(x)=sin(x) on the closed interval (-3ð/2,3ð/2).

y = sin x

y' = cos x

I guess that is -3 pi/2 to + 3 pi/2
sin 3 pi/2 = -1
sin -3 pi/2 = 1

-1 - 1 = -2

-1/(3 pi) = -.1061

cos^-1 -.1061 = 96.1 degrees
= 1.677 radians
cos is odd function so also -1.677 radians

To find the value of c that satisfies the Mean Value Theorem for the function f(x) = sin(x) on the closed interval (-3π/2, 3π/2), we need to consider the conditions of the Mean Value Theorem.

The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in (a, b) such that the derivative of the function at c, denoted by f'(c), is equal to the average rate of change of f(x) over the interval [a, b].

In this case, the function f(x) = sin(x) is continuous on the closed interval (-3π/2, 3π/2) and differentiable on the open interval (-3π/2, 3π/2). Therefore, the conditions of the Mean Value Theorem are satisfied.

Now, let's find the average rate of change of f(x) over the interval (-3π/2, 3π/2). Since f(x) = sin(x), the average rate of change can be calculated using the formula:

Average rate of change = [f(3π/2) - f(-3π/2)] / (3π/2 - (-3π/2))

To simplify, we substitute the values:

Average rate of change = [sin(3π/2) - sin(-3π/2)] / (3π/2 - (-3π/2))

Now, sin(3π/2) = -1 and sin(-3π/2) = -1 , so we have:

Average rate of change = [-1 - (-1)] / (3π/2 - (-3π/2))
= [-1 + 1] / (3π/2 + 3π/2)
= 0 / (6π/2)
= 0 / 3π
= 0

The average rate of change is 0.

According to the Mean Value Theorem, there exists at least one value c in (-3π/2, 3π/2) such that f'(c) = 0.

Since f'(x) is the derivative of f(x), we can find the derivative of f(x) = sin(x):

f'(x) = cos(x)

Now, we need to find the value of c such that f'(c) = 0.

cos(c) = 0

To find the values of c, we need to consider the unit circle and find the angles where cos(x) = 0.

In the unit circle, cos(x) = 0 when x = π/2 or x = 3π/2.

Therefore, the value of c that satisfies the Mean Value Theorem for the function f(x) = sin(x) on the closed interval (-3π/2, 3π/2) can be π/2 or 3π/2.