Proving that a nth degree polynomial is > (or < etc.) some constant.

QuarkCharmer
Messages
1,049
Reaction score
3

Homework Statement


Let x\inR,
Prove that if x>2 then x^4 - 8x^3+24x^2-32x+16

Homework Equations

The Attempt at a Solution


So far I have only learned proofs involving even and odd numbers, that sort of thing. I'm not really sure how to approach this one. I was thinking that a proof by cases would suffice, so:

Proof:
Case I: Assume that x>2, and let S = (0,\infty). If follows that,
x = (m+2) for some m\inS.

And so,
x^4 - 8x^3+24x^2-32x+16 = (m+2)^4 - 8(m+2)^3+24(m+2)^2-32(m+2)+16

but now I see the problem becoming too difficult. We haven't learned about epsilon-delta et al. yet. I can see that the conclusion is ALWAYS true, so really the proof is trivial, but I am not sure how to say that.
 
Physics news on Phys.org
QuarkCharmer said:

Homework Statement


Let x\inR,
Prove that if x>2 then x^4 - 8x^3+24x^2-32x+16.

Your question looks incomplete. Is that supposed to be x^4 - 8x^3+24x^2-32x+16 &gt; 0?

What's ##(x-2)^4##? How does that help in your question?
 
Thank you for the reply,

Sorry, yes the whole equation should be greater than zero, or equal to/less than something, I just made the question up as an example.

I see that (x-2)^4 is in fact equal to that polynomial. Therefor, for it is always going to be greater than zero because it is to an even power (and there is the case where x is 2, and then it IS equal to zero). I don't know how I am supposed to write this though? I'm aware that the conclusion is true for all x, but how would I even prove that:

(x-2)^4 &gt; 0, for some x\inR

Edit:

Oh wait, I think I see what you are getting at...
 
Last edited:
QuarkCharmer said:
Thank you for the reply,

Sorry, yes the whole equation should be greater than zero, or equal to/less than something, I just made the question up as an example.

I see that (x-2)^4 is in fact equal to that polynomial. Therefor, for it is always going to be greater than zero because it is to an even power (and there is the case where x is 2, and then it IS equal to zero). I don't know how I am supposed to write this though? I'm aware that the conclusion is true for all x, but how would I even prove that:

(x-2)^4 &gt; 0, for some x\inR

That should be ##(x-2)^4 \geq 0## for some ##x \in \mathbb{R}##. Equality occurs at x = 2. These little details are important.

You don't even need to bother with the case ##x \leq 2## since the question didn't ask for it.

Sketch the curve, and just make the appropriate arguments.

To prove the function is always increasing beyond x = 2, what can you say about the derivative for any x greater than 2?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top