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.

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

share|improve this question
3  
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
add comment

5 Answers 5

Hint:

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

share|improve this answer
add comment

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

share|improve this answer
2  
Intermediate value theorem, not mean value theorem, surely? –  Peter Taylor Dec 15 '11 at 10:22
3  
I think saying "roots of polynomials" is very common, actually. –  Cam McLeman Dec 15 '11 at 14:04
add comment

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.

share|improve this answer
    
Thanks this helped a lot! –  Max Dec 15 '11 at 4:57
    
Thanks that helps a lot! –  Max Dec 15 '11 at 4:57
2  
@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
add comment

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.

share|improve this answer
add comment

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.