`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`.

Textbook Question

Chapter 3, 3.2 - Problem 26 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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.

We’ve answered 318,916 questions. We can answer yours, too.

Ask a question