Prove that 111x^111+11x^11+x+1 has only one real root

[AdrianZ]

Homework Statement

Prove that the function 111x^111+11x^11+x+1 has exactly one real root.

The Attempt at a Solution

Well, I know how to prove this. f is continuous because It's a polynomial. f(-1) equals -111-11-1+1 = -122 < 0 and f(0) equals 1. therefore, It satisfies all conditions of the intermediate value theorem and the theorem tells us that there must be a zero of f between -1 and 0. Now we can use Rolle's theorem to rule out any other possible real roots of the function. but the problem is the derivative of the function leads to an equation which can't be solved without a computer. so I can't show that the assumption that f has other real roots leads to a contradiction using Rolle's theorem. What can I do now?

 

[micromass]

Hi AdrianZ!

There is no need to calculate the possible roots, you just need to check that there is no root.

As derivative, you've probably found

12321x^{110}+121x^{10}+1

It suffices to show that there is no positive root (if there is a positive root a, then -a will be a negative root). Now, can you show that the derivative is always >0 for all positive numbers?

 

[AdrianZ]

How do you say that if a is a positive root then -a will be a negative root? the function is not even. I didn't understand that part.
well, I think I can show that this equation is always positive. what a stupid I am. for any real x: x^10 >= 0. if I multiply it by 121 I'll get 121x^10>=0 Now if I add 1 to both sides I'll get 121x^10+1>=1.
I can do the same for x^110.for any real x: x^110>=0. by summing the two equations I'll get f'(x) >= 1. which proves that It can't have a real root. so, the contradiction!

Thank you micromass xP You are always helpful xP

 

[micromass]

How do you say that if a is a positive root then -a will be a negative root? the function is not even. I didn't understand that part.

Well, it seems you figured it out without using evenness. But the function is even:

12321(-x)^{110}+121(-x)^{10}+1=12321x^{110}+121x^{10}+1

well, I think I can show that this equation is always positive. what a stupid I am. for any real x: x^10 >= 0. if I multiply it by 121 I'll get 121x^10>=0 Now if I add 1 to both sides I'll get 121x^10+1>=1.
I can do the same for x^110.for any real x: x^110>=0. by summing the two equations I'll get f'(x) >= 1. which proves that It can't have a real root. so, the contradiction!

That's good!

 

[AdrianZ]

Well, it seems you figured it out without using evenness. But the function is even:

12321(-x)^{110}+121(-x)^{10}+1=12321x^{110}+121x^{10}+1
That's good!

Oh! Again what a stupid I am. Of course the function is even lol Thanks. I need to rest for some hours. my brain is off.
Btw, How could you prove that the function has no real roots using evenness? You wanted to use reductio ad absurdum?

 

[gb7nash]

Oh! Again what a stupid I am. Of course the function is even lol Thanks. I need to rest for some hours. my brain is off.
Btw, How could you prove that the function has no real roots using evenness? You wanted to use reductio ad absurdum?

Think about it. I'll give you the idea in a nutshell:

Every x is raised to an even power in 12321x¹¹⁰ + 1232x¹⁰ + 1. No matter what value you plug into x, you get a positive value for x¹¹⁰ and x¹⁰. The smallest value you could get then is plugging in 0 for x, so you'll get 0 + 0 + 1 = 1.