Rolle's Theorem not applicable?

Use Rolle's Theorem to show that the equation $\displaystyle x^3+x-1=0$ has exactly one root in the interval [-1,1].

So, I noticed that it is continuous on the closed interval and differentiable on the open interval; however, f(-1) \neq f(1). Doesn't that mean Rolle's Theorem is not applicable?

How would I answer this question? Do I just state that it does not apply and leave it?

Re: Rolle's Theorem not applicable?

Quote:

Originally Posted by

**terrorsquid** Use Rolle's Theorem to show that the equation $\displaystyle x^3+x-1=0$ has exactly one root in the interval [-1,1].

So, I noticed that it is continuous on the closed interval and differentiable on the open interval; however, f(-1) \neq f(1). Doesn't that mean Rolle's Theorem is not applicable?

How would I answer this question? Do I just state that it does not apply and leave it?

First show that there is a sign change.

$\displaystyle \displaystyle f(-1) = (-1)^3 + (-1) - 1 = -3$ and $\displaystyle \displaystyle f(1) = 1^3 + 1 - 1 = 1$.

So the function definitely crosses the x axis.

If there aren't any turning points, then the function only has one root.

$\displaystyle \displaystyle \begin{align*} f(x) &= x^3 + x - 1 \\ f'(x) &= 3x^2 + 1 \\ 0 &= 3x^2 + 1 \\ -1 &= 3x^2 \\ -\frac{1}{3} &= x^2 \end{align*}$

This equation is not solvable, so there are not any turning points. That means there is exactly one root.

Re: Rolle's Theorem not applicable?

Isn't that using the intermediate value theorem though and not Rolle's? Would you first say that Rolle's does not apply and then use the intermediate value theorem? Maybe that's what the question is supposed to be?

Re: Rolle's Theorem not applicable?

Quote:

Originally Posted by

**terrorsquid** Isn't that using the intermediate value theorem though and not Rolle's? Would you first say that Rolle's does not apply and then use the intermediate value theorem? Maybe that's is what the question is supposed to be?

In this case, that is all one can do.