# Polynomials of odd degree can have no critical point.

1. Nov 23, 2012

### batballbat

Are these assertions true?
I am referring to polynomials with real coefficients.
1. There exists of polynomial of any even degree such that it has no real roots.
2. Polynomials of odd degree have atleast one real root
which implies that polynomial of even degree has atleast one critical point.
3. Polynomials of odd degree can have no critical point.

Last edited by a moderator: Feb 6, 2013
2. Nov 23, 2012

### Robert1986

Re: polynomials

The first two are. What do you mean by "critical point"? If you mean a place where the derivative is zero, then this is not right.

For 1) a polynomial of degree two that has no real roots is $x^2 +1$. Do you see how to generalise? For 2) you want to use the Intermediate Value Theorem.

3. Nov 24, 2012

### mathman

Re: polynomials

2) not true: example y=x3 + x. It has a real root at x=0, but the derivative is never zero (for real x).
3) not true: example y=x3/3 - x. Critical points at x = 1, -1.

4. Nov 24, 2012

### Robert1986

Re: polynomials

Did you edit #2 from the original post to add the second part? Because I didn't mean to include that part in my response. I was only saying that the first line of #2 is correct.

5. Nov 25, 2012

### batballbat

Re: polynomials

i dont get it.
Shouldnt a polynomial of even degree have atleast one critical point?

in question 3. i meant for any odd degree we can find a polynomial which has no critical point.

6. Nov 25, 2012

### DonAntonio

Re: polynomials

All of them are true:

(1) $\,(x^2+1)^n\,$

(2) The intermediate value theorem for continuous functions

(3) $\,\int (x^2+1)^ndx\,$

DonAntonio

7. Nov 25, 2012

### Robert1986

Re: polynomials

Yes, it does, apparently I have misread your post twice now. If all odd polys. have a real root, then this means that the derivative of all even degree polys. have a zero, which means all even degree polys. have a critical point.

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