How would I know if $f(x)=x^5-2x+10$ has a root at the interval $[-2, 2]$? Unsure on the procedure on this one and then how to explain it.  I don't think this function has any rational roots, right?
 A: Hint:
$$f(-2) = (-2)^5 -2(-2) + 10 = -32 + 4 + 10 = -18 < 0$$
while
$$f(2) = 2^5 - 2(2) + 10 = 38 > 0.$$
A: You can use Sturm's Theorem and Descartes's Rule of Signs.
A: 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. 





*

*For the proof, as mentioned in the comments by other users, do you know mean value theorem?

*For the third question, see rational root theorem. 
A: 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.
A: 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.
