Is This Inequality Solvable Using Elementary Algebra Techniques?

[Newtime]

Here's a problem in my analysis book:

solve the following for x$$\in$$R:

|x³ - 3x + 1| < x³

Should be pretty simple right? After all, it's an inequality that might appear in my 14 year old sister's algebra book...so why can't I solve it? The obvious solution is that x > 1/3 setting the inside of the absolute value to positive. when you assume the inside of the absolute value is negative and work from there, that's when I get into a bind...I know there's the formula for solving cubics, but I find it hard to believe that's how we're supposed to be going about this...can someone please show me what I might be overlooking? By the way, the answer is ($$\frac{1}{3}$$, ($$\sqrt{3}$$-1) / 2 ) $$\cup$$ (1, $$\infty$$). Thanks.

 

[Petek]

Hint: Can you factor the cubic polynomial that's giving you trouble?

 

[Newtime]

Hint: Can you factor the cubic polynomial that's giving you trouble?

Well not that I can see. Here's my work:

x³ - 3x + 1 > -x³ (by definition of absolute value, and this is assuming that the polynomial inside the absolute value is less than zero, like I said in the OP, if you assume it is greater than zero then the x³ terms cancel and the answer is easy and obvious).

So from here I can either factor (...attempt to factor) x³-3x+1 or I can add x³ to both sides and attempt to factor 2x³-3x+1. Both of these are getting the better of me. I tried a few basic techniques to no avail. My guess is that I'm overlooking something very basic but maybe a little tricky. Any ideas?

Note: also, the reason I think that I'm overlooking something basic and that this is in fact an easy inequality is that it is in an analysis book where the focus is analysis (and in this chapter, the definition of inequalities), not how to factor polynomials.

 

[Petek]

The polynomial that I had in mind is $$2x^3-3x+1$$. Do you know the theorem that says (roughly): If a is a root of the polynomial f(x), then x - a divides f(x)?

 

[Newtime]

The polynomial that I had in mind is $$2x^3-3x+1$$. Do you know the theorem that says (roughly): If a is a root of the polynomial f(x), then x - a divides f(x)?

Wow, no I was not aware of that theorem, but it's a good one to know. And I think i should be able to get it from here. Thanks for the help.

 

[HallsofIvy]

If $|x^3- 3x+ 1|< x^3$ then either $x^3- 3x+ 1< x^3$ or $x^3- 3x+ 1< -x^3$. Now try solving the corresponding equalities to see where the sign can change.

If $x^3- 3x+ 1= x^3$ you can cancel the $x^3$ and the rest is easy. If $x^3- 3x+ 1= x^3$, then $2x^3- 3x+ 1= 0$ as Petek said. And that has at least one very obvious root.

(You may not have been aware of the "theorem", "if a is a root of the polynomial f(x), then x-a divides f(x)" but think about the very first method you learned to solve polynomial equations- factoring.

 

[Newtime]

If $|x^3- 3x+ 1|< x^3$ then either $x^3- 3x+ 1< x^3$ or $x^3- 3x+ 1< -x^3$. Now try solving the corresponding equalities to see where the sign can change.

If $x^3- 3x+ 1= x^3$ you can cancel the $x^3$ and the rest is easy. If $x^3- 3x+ 1= x^3$, then $2x^3- 3x+ 1= 0$ as Petek said. And that has at least one very obvious root.

(You may not have been aware of the "theorem", "if a is a root of the polynomial f(x), then x-a divides f(x)" but think about the very first method you learned to solve polynomial equations- factoring.

Thanks HallsofIvy, that's more or less what I've got written down right now and it's good to know it's what you got also. I suppose the thing I was missing was the obvious root: x=1, had I seen that, regardless of knowing it was a theorem, I would have divided by x-1.

 

[HallsofIvy]

A good "first step" for problems like these: Cross your fingers and hope there is a simple root! By the "rational root theorem" any rational root of [itex]ax^n+ bx^{n-1}+ \cdot\cdot\cdot+ yx+ z= 0[/must] with integer coefficients must be of the form m/n where m is an integer divisor of z and n is an integer divisor of a. Here, $x^3- 3x+ 1= 0$ a= 1 and z= 1 so the only possible rational roots are 1 and -1. It's certainly worth trying them to see![/itex]