`f(x) = (x/2) - sin((pix)/6), [-1,0]` Determine whether Rolle's Theorem can be applied to `f` on the interval and, if so, find all values of `c` in the open interval `(a,b)` such that `f'(c) = 0`.
The given function is continuous and differentiable over the given interval, as all trigonometric functions are. For Rolle's Theorem to be applied, you also need to test if `f(-1) = f(0).`
`f(-1) = -1/2 - sin(pi/12)`
You need to evaluate `sin(pi/12)` as `sin (alpha)/2 = sqrt((1 - cos alpha)/2):`
`sin((pi/6)/2) = sqrt((1 - cos(pi/6))/2) => sin((pi/6)/2) = (sqrt(2-sqrt3))/2`
`f(-1) = (-1 - sqrt(2-sqrt3))/2 `
`f(0) = 0 - sin 0 = 0 - 0 = 0`
Notice that `f(-1) != f(0), ` hence, since this condition is not satisfied, Rolle's theorem cannot be applied.