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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 $f'(c) = 0$.

Im not exactly sure what to do because $f(1) = -2 + b$ and $f(-1) = 2 + b$, unless we are allowed to have different $b$ values.

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    $\begingroup$ Suppose that there are two distinct roots in the interval $[-1,1]$, apply Rolle's theorem, obtain a contradiction. $\endgroup$
    – Antonio Vargas
    Nov 5, 2013 at 23:57

1 Answer 1

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Rolle's Theorem tells you that between any two roots of $f(x)=x^3-3x+b$, there is a root of $f'(x)$.

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