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I am given a function $f(x) = x^3 + 3x - 1$, and I am asked to prove that $f(x)$ has exactly one real root using the Intermediate Value Theorem and Rolle's theorem. So far, I managed to prove the existence of at least one real root using IVT.

Note that $f(x)$ is continuous and differentiable for all $x \in \Bbb R$. By inspection, since $f(-1) = -5$ and $f(1) = 3$, by the IVT, there exists a $c \in [-1, 1] $ such that $f(c) = 0$. Therefore, $f(x)$ must have at least one root. I am not sure if this is the best way to prove that it has at least one root - if there is a better way, please let me know. I don't know how I can use Rolle's theorem to prove that there is only one root. I had an idea of integrating the polynomial and then proving that it has just one maximum using Rolle's theorem but that did not work out so well.

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up vote 4 down vote accepted

Between any two distinct real roots, there is, by Rolle's Theorem, a root of the derivative. But the derivative has no roots.

There is a perhaps somewhat better way to use IVT to show the existence of a root. Don't bother to find explicit $a$ and $b$ such that our function is negative at $a$ and positive at $b$. It is obvious that when $x$ is large enough negative, our function is negative, and that when $x$ is large enough positive, our function is positive. Thus any cubic has a real root. By the same argument, any polynomial of odd degree has a real root.

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Ah, perfect - this is just what I needed! – hesson Dec 15 '12 at 0:46

IVT shows the existence of the root and you can use the first derivative to see your function is increasing, which means it is the unique point which makes $f(x)$ zero.

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