Prove there is only one zero of x3 - 3x + b in [-1,1]

Prove, using Rolle’s theorem, that for any value of $b$ there is at most one $x \in [-1,1]$ such that

$x^3 - 3x + b = 0.$

Proof. The argument is by contradiction. Suppose there are two or more points in $[-1,1]$ for which $x^3 - 3x + b = 0$ . Let $x_1, x_2$ be two such points with $x_1 < x_2$ . Since $x_1, x_2$ are both in $[-1,1]$ we have

$-1 \leq x_1 < x_2 \leq 1 \qquad \text{and} \qquad f(x_1) = f(x_2) = 0.$

Since $f(x) = x^3 - 3x + b$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$ (all polynomials are continuous and differentiable everywhere), we may apply Rolle’s theorem on the interval $[x_1, x_2]$ to conclude that there is some $c \in (x_1, x_2)$ such that

$f'(c) = 0 \quad \implies \quad 3c^2 - 3 = 0 \quad \implies \quad c = \pm 1.$

But this contradicts that $c \in (-1,1)$ (since $c \in (x_1, x_2)$ and $-1 \leq x_1 < x_2 \leq 1$ ). Hence, there can be at most one point $x \in [-1,1]$ such that $f(x) = 0. \qquad \blacksquare$

Stumbling Robot

Prove there is only one zero of x³ – 3x + b in [-1,1]

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