Rolle's Theorem $x^3 - 3x +b$ Use Rolle's Theorem to prove that the equation $x^3 - 3x + b = 0$ has at most one root in the interval $[-1,1]$.
Rolle's Theorem : Suppose f is a continuous real-valued function on $[a,b]$ with $f(a) = f(b)$, and that f is differentiable on $(a,b)$. Then there exists c in $(a,b)$ such that $f'(c) = 0$.
Here is my proof:
Suppose there are 2 roots in $[-1,1]$. $f'(x) = 3(x^2-1)$, which is equal to zero at the endpoints -1 and 1. This results in a contradiction because there are are no roots in the interval $(-1,1)$. Therefore the original hypothesis that there are two roots in $[-1,1]$ is false, implying that there is at most 1 root.
I know that is incorrect because I never used the hypothesis, but I think its the right idea. Help would be more than appreciated.