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Unsure on the procedure on this one and then how to explain it. I don't think this function has any rational roots, right?

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It depends on what you're allowed to use. If you evaluate $f(-2)$ and $f(2)$, what do you get? If you're allowed something like the intermediate value theorem and you know that polynomials are continuous, then these two values get you what you want. At the level of precalculus I don't know offhand. – Dylan Moreland Dec 15 '11 at 3:34


$$f(-2) = (-2)^5 -2(-2) + 10 = -32 + 4 + 10 = -18 < 0$$ while $$f(2) = 2^5 - 2(2) + 10 = 38 > 0.$$

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You can use Sturm's Theorem and Descartes's Rule of Signs.

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As I understand, you actually have three questions:

  • Does $f(x)=x^5-2x+10$ has zeros on the interval $[-2,2]$? (Notice that you can either say the root of $f(x)=0$ or zeros of $f(x)$. I don't think people would say "the root of $f(x)$".)
  • How to prove the existence of non-existence above?
  • If the root of $f(x)=0$ exists, is it rational?

Here are my answers:

  • First, try it on Mathematica. You can see the answer from the picture.

enter image description here

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Intermediate value theorem, not mean value theorem, surely? – Peter Taylor Dec 15 '11 at 10:22
I think saying "roots of polynomials" is very common, actually. – Cam McLeman Dec 15 '11 at 14:04

Calculate $f(2)$ and $f(-2)$. In your case they are 38 and -18 respectively. Since $f(x)$ changes its sign as one decreases $x$ from 2 to -2, $f(x)$ must have crossed the $f(x)=0$ line at some $x$. This proves that $f(x)$ has a root somewhere in the interval $[-2,2]$.

PS : Given the fact that $f(x)$ is continuous. See a comment below.

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Thanks this helped a lot! – Max Dec 15 '11 at 4:57
Thanks that helps a lot! – Max Dec 15 '11 at 4:57
@Dilawar When using this argument, you must remember to mention that $f$ is continuous. – Julián Aguirre Dec 15 '11 at 9:16
Yes, I totally agree. Thanks! Sorry for leaving a gap! – Dilawar Dec 15 '11 at 9:42

If you want to know if a polinomial has rational roots, you use the Rational Roots Theorem. For a polynomial $$a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$$ All rational roots must be of the form $$\pm \frac{p}{q}$$ where $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$. No other rational roots may exist. In this case, you have $p=\{1,2,5,10\}$ and $q=1$, so your possible rational roots are $\{\pm1,\pm2,\pm5,\pm10\}$. If you evaluate the function at these points (don't do it by hand), you'll see none of them equal zero, so your function has no rational roots.

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