# `f(x) = cos(x), [0,2pi]` Determine whether Rolle’s Theorem can be applied to `f` on the closed interval `[a, b]`. If Rolle’s Theorem can be applied, find all values of `c` in the open...

`f(x) = cos(x), [0,2pi]` Determine whether Rolle’s Theorem can be applied to `f` on the closed interval `[a, b]`. If Rolle’s Theorem can be applied, find all values of `c` in the open interval

### Textbook Question

Chapter 3, 3.2 - Problem 18 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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The Rolle's theorem is applicable to the given function, only if the function is continuous and differentiable over the interval, and f(a) = f(b). Since all trigonometric functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval. Now, you need to check if f(0) = f(2pi).

`f(0) = cos 0 = 1`

`f(2pi) = cos (2pi) = 1`

Since all the three conditions are valid, you may apply Rolle's theorem:

`f'(c)(b-a) = 0`

Replacing `2pi` for b and 0 for a, yields:

`f'(c)(2pi-0) = 0`

You need to evaluate f'(c):

`f'(c) = (cos c)' => f'(c) = - sin c`

Replacing the found values in equation `f'(c)(2pi-0) = 0`

`-2pi*sin c = 0 => sin c = 0 => c = 0, c = pi, c = 2pi`

Since `c = {0,2pi} !in (0,2pi), ` the only valid value for c is `pi` .

Hence, in this case, the Rolle's theorem may be applied for `c = pi` .