Rolle's Theorem

• June 7th 2007, 06:25 PM
BrianW
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
• June 7th 2007, 07:49 PM
ThePerfectHacker
Quote:

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