Rolle's Theorem $x^3 - 3x +b$

Question

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.

Suppose that there are two distinct roots in the interval $[-1,1]$, apply Rolle's theorem, obtain a contradiction. — Antonio Vargas, Nov 05 '13 at 23:57

score 0 · Answer 1 · answered Nov 06 '13 at 00:00

0

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

answered Nov 06 '13 at 00:00

Martin Argerami

205,756

Rolle's Theorem $x^3 - 3x +b$

1 Answers1

Linked