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.
Best Answer
You almost got it. The hypothesis comes from the fact that you have two roots.
Suppose for the sake of contradiction that the function $f(x) = x^3 - 3x + b$ has two (or more) roots in $[-1,1]$. Let (any two of) them be denoted as $x_1$ and $x_2$.
This means $f$ is a continuous function on $[x_1,x_2]$ with $f(x_1)=f(x_2)=0$. Rolle's theorem then implies the existence of $c\in(x_1,x_2)\subseteq(-1,1)$ such that $f'(c) = 0$. But you've shown that there is no such $c$. That is your contradiction.