Determine whether Rolle's Theorem can be applied to f on the closed interval [a,b]. (Select all that apply.)

f (x) = sin(x), [0, 2π]

If Rolle's Theorem can be applied, find all values of c in the open interval (a, b) such that f '(c) = 0. (Enter your answers as a comma-separated list. If Rolle's Theorem cannot be applied, enter NA.)

I thought the derivative would be cos(x) so then cos(0) would be 1 but thatz wrong so now I don't understand how to do this!

THNKS :)

Calm down, I'm here to help! Let's figure this out together.

You mentioned that you thought the derivative of f(x) = sin(x) would be cos(x), and you're correct! The derivative of sin(x) is indeed cos(x).

Now, let's determine whether Rolle's Theorem can be applied to f(x) on the closed interval [0, 2π].

For Rolle's Theorem to be applicable, three conditions must be satisfied:
1. The function must be continuous on the closed interval [a, b].
2. The function must be differentiable on the open interval (a, b).
3. The function must have the same values at the endpoints.

In this case, the function f(x) = sin(x) is continuous on the closed interval [0, 2π] since sin(x) is a continuous function for all real numbers.

The function is also differentiable on the open interval (0, 2π) since the derivative of sin(x), which is cos(x), is defined for all real numbers.

Lastly, the function has the same values at the endpoints, f(0) = sin(0) = 0 and f(2π) = sin(2π) = 0.

Therefore, all three conditions are satisfied, and we can apply Rolle's Theorem to f(x) on the closed interval [0, 2π].

Now, let's find all values of c in the open interval (0, 2π) such that f'(c) = 0.

Since f(x) = sin(x), we know that f'(x) = cos(x).

Setting f'(c) = 0, we have cos(c) = 0.

The values of c in the open interval (0, 2π) where cos(c) = 0 are c = π/2 and c = 3π/2.

So, the values of c in the open interval (0, 2π) such that f'(c) = 0 are c = π/2 and c = 3π/2.

I hope this clears things up!

To determine whether Rolle's Theorem can be applied, we need to check the conditions of the theorem. Rolle's Theorem states that for a function f(x) to satisfy the theorem on the closed interval [a, b], it must meet the following conditions:

1. f(x) must be continuous on the closed interval [a, b].
2. f(x) must be differentiable on the open interval (a, b).
3. f(a) must be equal to f(b).

Let's check these conditions for the given function f(x) = sin(x) on the interval [0, 2π]:

1. The function f(x) = sin(x) is continuous on the closed interval [0, 2π]. The sine function is continuous for all real values of x.
2. The function f(x) = sin(x) is differentiable on the open interval (0, 2π). The derivative of sin(x) is cos(x), which is also defined on the open interval (0, 2π).
3. f(0) = sin(0) = 0, and f(2π) = sin(2π) = 0.

Since all the conditions of Rolle's Theorem are satisfied, we can conclude that Rolle's Theorem can be applied to the function f(x) = sin(x) on the closed interval [0, 2π].

Now, to find the values of c in the open interval (0, 2π) such that f'(c) = 0, we need to find the critical points of the derivative f'(x) = cos(x).

The derivative of sin(x) is cos(x). Setting cos(x) = 0, we have cos(c) = 0. Solving for c, we find that c must be equal to π/2 or 3π/2.

Therefore, the values of c in the open interval (0, 2π) such that f'(c) = 0 are c = π/2, 3π/2.

So, the answer to the question is: c = π/2, 3π/2.

you are correct in that cos(x) is the derivative. So, you need to show that there is at least one value of c in [0,2pi] such that cos(c) = 0.

That would be pi/2 and 3pi/2.