# `f(x) = (x^2 - 1)/x, [-1,1]` 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) = (x^2 - 1)/x, [-1,1]` 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 16 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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 polynomial functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval. Now, you need to check if f(-1) = f(1).

`f(-1) = ((-1)^2-1)/(-1)= 0`

`f(1) = (1^2-1)/(1)= = 0`

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

` f'(c)(b-a) = 0`

Replacing 1 for b and -1 for a, yields:

`f'(c)(1+1) = 0`

You need to evaluate f'(c), using quotient rule:

`f'(c) = ((c^2-1)/c)' => f'(c) =((c^2-1)'*c - (c^2-1)*c')/(c^2)`

`f'(c) = (2c^2 - c^2 + 1)/(c^2)`

`f'(c) = (c^2+1)/(c^2)`

Replacing the found values in equation `f'(c)(1+1) = 0`

`2(c^2+1)/(c^2)= 0 =>c^2 + 1 = 0 => c^2 = -1!in R`

Hence, in this case, there is no valid value of `c in(-1,1)` , for the Rolle's theorem to be applied.