How do you determine whether Rolle's theorem can be applied to #f(x) = (x^2 - 1) / x# on the closed interval [-1,1]?

Question

How do you determine whether Rolle's theorem can be applied to #f(x)  = (x^2 - 1) / x# on the closed interval [-1,1]?

Charles Arocha · Accepted Answer

Rolles theorem states: "Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). If f(a) = f(b), then there is at least one point c in (a, b) where f '(c) = 0."

The first criteria is NOT met. Specifically, f is NOT continuous on the interval [-1, 1] at the point x = 0. At x = 0, the function is undefined. Hope that helps.

Emma Aldridge · Answer

undefined Rolle's theorem can be applied to a function $ f(x) $ on a closed interval $[a, b]$ if the following conditions are met:

1. $ f(x) $ is continuous on $[a, b]$
2. $ f(x) $ is differentiable on $(a, b)$
3. $ f(a) = f(b) $

For the function $ f(x) = \frac{x^2 - 1}{x} $ on the closed interval $[-1, 1]$, we need to check these conditions:

1. $ f(x) $ is continuous on $[-1, 1]$ since it is a rational function and is continuous everywhere except at $ x = 0 $, which is not in the interval $[-1, 1]$.
2. $ f(x) $ is differentiable on $(-1, 1)$ since it is a rational function and is differentiable everywhere except at $ x = 0 $, which is not in the interval $(-1, 1)$.
3. $ f(-1) = \frac{(-1)^2 - 1}{-1} = \frac{1 - 1}{-1} = 0 $ and $ f(1) = \frac{1^2 - 1}{1} = \frac{1 - 1}{1} = 0 $.

Since all three conditions are met, Rolle's theorem can be applied to the function $ f(x) = \frac{x^2 - 1}{x} $ on the closed interval $[-1, 1]$.