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

by RoRi

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

One comment

  1. Anonymous says:

    Yes, but with a direct proof, I am not sure how Rolle’s Theorem applies, since we do not have end point equality.

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