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Polynomials of odd degree can have no critical point. 
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#1
Nov2312, 10:07 PM

P: 127

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. 


#2
Nov2312, 10:12 PM

P: 828

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 [itex]x^2 +1[/itex]. Do you see how to generalise? For 2) you want to use the Intermediate Value Theorem. 


#3
Nov2412, 03:20 PM

Sci Advisor
P: 6,076

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


#4
Nov2412, 05:17 PM

P: 828

Polynomials of odd degree can have no critical point.
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
Nov2512, 05:40 AM

P: 127

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
Nov2512, 06:19 AM

P: 606

All of them are true: (1) [itex]\,(x^2+1)^n\,[/itex] (2) The intermediate value theorem for continuous functions (3) [itex]\,\int (x^2+1)^ndx\,[/itex] DonAntonio 


#7
Nov2512, 09:34 AM

P: 828




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