Does the equation $x^{3} - x - p = 0$ always has a solution in $\mathbb{R}$

Question

Does the equation $x^{3} - x - p = 0$ always has a solution in $\mathbb{R}$ for all $p \in \mathbb{R}$.

Is there a real solution for $x$ for each real number $p$?

I am new to the theory of cubic equations.

At least one, but why? Do you have any thoughts? – mickep Mar 22 '15 at 14:25 — mickep, Mar 22 '15 at 14:25
Any odd degree polynomial has ... – Hagen von Eitzen Mar 22 '15 at 14:26 — Hagen von Eitzen, Mar 22 '15 at 14:26

score 1 · Accepted Answer · answered Mar 22 '15 at 14:26

1

Yes. The function $x^3 - x - p$ is continuous, it's negative for large negative values of $x$ and positive for large positive values of $x$, so it must be zero somewhere in between, by the intermediate value theorem.

answered Mar 22 '15 at 14:26

bubba

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Does the equation $x^{3} - x - p = 0$ always has a solution in $\mathbb{R}$

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