1. ## Rolle's Theorem

How do I do this problem?

Use Rolle's theorem to prove that, regardless of the value b, there is at most one point in the interval -1 <= x <= 1 for which x^3 - 3x + b = 0

2. Originally Posted by BrianW
How do I do this problem?

Use Rolle's theorem to prove that, regardless of the value b, there is at most one point in the interval -1 <= x <= 1 for which x^3 - 3x + b = 0
We need to understand this problem carefully, because when I was first reading it, it seemed as if it said "there is at least one point which is a zero".

We want to show that there is at most one point. We will argue by contradiction. Say there are more than two points on $\displaystyle [-1,1]$. Pick any two distinct ones $\displaystyle x_1,x_2\in [-1,1]$ with $\displaystyle x_1 < x_2$. Define the function $\displaystyle f(x) = x^3 - 3x + b$. Note that $\displaystyle f(x_1) = 0 \mbox{ and }f(x_2)=0$. Consider this function $\displaystyle f$ on the closed interval $\displaystyle [x_1,x_2]$. We notice that $\displaystyle f(x_1) = f(x_2)$, the function is continous on $\displaystyle [x_1,x_2]$, and differenciable on $\displaystyle (x_1,x_2)$. By Rolle's theorem we can find $\displaystyle x_0 \in (x_1,x_2)$ so that $\displaystyle f'(x_0)=0$. Meaning $\displaystyle 3x_0^2 - 3 = 0$. But that is impossible because $\displaystyle 3x_0^2 - 3 < 0$ since $\displaystyle - 1< x_0 < 1$. Hence a contradiction. That means there can be at most one zero.
Q.E.D.