Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 3 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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.