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Use the Intermediate Value Theorem to show that the equation $x^3+x+1=0$ has a solution.

How to do this? :S

Thank you very much!

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Define $f(x) = x^3 + x + 1$. How does $f$ behave when $x \rightarrow \infty$? When $x \rightarrow -\infty$? What can you deduce? –  levap Oct 1 '12 at 0:03
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Or even, like, $f(0)$ and $f(-1).$ –  user17794 Oct 1 '12 at 0:04

2 Answers 2

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Let $f(x) = x^3 + x + 1$. Then $f$ is a polynomial, so its continuous. Then we notice that

$$f(-1) = -1, \text{ and } f(0) = 1.$$

But then by the intermediate value theorem, this means that for all $y\in[-1,1]$ we have some $x\in [-1,0]$ such that $f(x) = y$. Therefore we have some $z$ such that $f(z) = 0$.

Since $f(z) = 0$ we know that $z^3 + z + 1 = 0$ and so the equation $$x^3 + x + 1 = 0$$ has a solution, namely $z$.

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If you want to use the Intermediate Value Theorem, first you're going to need the statement of the Intermediate Value Theorem. Once you look at it, it tells you exactly what to do. From Calculus by Varberg, Purcell, and Rigdon:

Intermediate Value Theorem: Let $f$ be a function defined on $[a, b]$ and let $W$ be a number between $f(a)$ and $f(b)$. If $f$ is continuous on $[a, b]$, then there is at least one number $c$ between $a$ and $b$ such that $f(c) = W$.

Now, a very common use for this theorem is to show that a function is 0 somewhere. How that works is as follows. Say out of $f(a)$ and $f(b)$, one is positive and one is negative. Then $0$ is in between $f(a)$ and $f(b)$. If the function in question is continuous on the closed interval $[a, b]$, then all the conditions of the Intermediate Value Theorem apply. In that case, we can conclude there is some $c$ between $a$ and $b$ such that $f(c) = 0$. Even if don't know what $c$ is, and even if we have no idea how to ever figure out what $c$ is, we know there is some $c$ such that $f(c) = 0$.

In your specific example, $f(x) = x^3 + x + 1$ is continuous everywhere, in particular on any closed interval. So, we have no trouble on that issue. All we need to know is $f(x)$ is positive somewhere and negative somewhere else. Just try a few values of $x$ and see if you can find one where $f(x)$ is positive and one where $f(x)$ is negative. Start by trying the easy values. In this case, Deven already has shown the easiest ones to try, and it turns out $f(0) = 1$ is positive and $f(-1) = -1$ is negative, so that we know the Intermediate Value Theorem applies and tells us there is some $c$ between $-1$ and $0$ such that $f(c) = 0$.

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