Finding Solutions for x^3 + (1/x^3) ≥ 3

[mathdad]

Section 2.6
Question 82

Solve: x^3 + (1/x^3) ≥ 3. (Use a calculator to approximate the key numbers.)

I need someone to get me started.

 

[MarkFL]

Move everything to the LHS, and then combine all terms into a rational expression...what do you then have?

 

[mathdad]

x^3 + (1/x^3) - 3 ≥ 0

x^3(x^3) + [(1/x^3)](x^3)/1 -3x^3 ≥ 0

x^6 + 1 - 3x^3 ≥ 0

x^6 - 3x^3 + 1 ≥ 0

I am stuck here.

 

[MarkFL]

What you should have is:

\(\displaystyle \frac{x^6-3x^3+1}{x^3}\ge0\)

Recall, you don't want to multiply an inequality by an expression having an unknown sign.

We have a 6th degree polynomial in the numerator, and so the fundamental theorem of algebra tells us that we should expect to find 6 roots...what does Descartes' rule of signs tell us about these roots? Do you recognize that you can treat the numerator like a quadratic?

 

[mathdad]

"Recall, you don't want to multiply an inequality by an expression having an unknown sign."

Yes, I forgot that important rule.

"We have a 6th degree polynomial in the numerator, and so the fundamental theorem of algebra tells us that we should expect to find 6 roots."

What is the Fund. Theo. of Algebra?"What does Descartes' rule of signs tell us about these roots?"

I am not there in the textbook and thus, do not know this rule.

"Do you recognize that you can treat the numerator like a quadratic?"

Can the numerator by written as (x^2)^3 - 3x^3 + 1?

 

[MarkFL]

"Recall, you don't want to multiply an inequality by an expression having an unknown sign."

Yes, I forgot that important rule.

"We have a 6th degree polynomial in the numerator, and so the fundamental theorem of algebra tells us that we should expect to find 6 roots."

What is the Fund. Theo. of Algebra?

One of the proven assertions of the FTOA is:

A polynomial of degree n will have exactly n roots (counting multiplicity).

"What does Descartes' rule of signs tell us about these roots?"

I am not there in the textbook and thus, do not know this rule.

That's why I provided a link...it is a useful rule when studying polynomials, and its usage is within the grasp of the Pre-Calc student. I advise you to read the information to which I linked, and let me know what this rule says about the roots of the polynomial under consideration here.

"Do you recognize that you can treat the numerator like a quadratic?"

Can the numerator by written as (x^2)^3 - 3x^3 + 1?

No, this is what I mean:

\(\displaystyle (x^3)^2-3x^3+1\)

We have a quadratic in $x^3$. This will give you two of the roots, and you should be able to use Descartes' rule of signs to determine if there are any other real roots. :D

 

[mathdad]

The expression (x^3)^2 - 3x^3 + 1 cannot be factored.

 

[MarkFL]

The expression (x^3)^2 - 3x^3 + 1 cannot be factored.

It has no rational roots, but we can use the quadratic formula to state:

\(\displaystyle x^3=\frac{3\pm\sqrt{5}}{2}\implies x=\sqrt[3]{\frac{3\pm\sqrt{5}}{2}}\)

Descartes' rule of signs indicates that the remaining roots must be complex, and so these two real roots are all we need to be concerned about. So, you have 3 critical values, all of which are roots of odd multiplicity, so you know what to do from here. :D

 

[mathdad]

Thanks. I will need to use my calculator. The critical values are complicated. They belong on the number line and the rest remains the same steps, right?

 

[mathdad]

Let CR = CUBE ROOT

Let SR = SQUARE ROOTCR{(3 - SR{5})/2} = x which is about 0.7256

CR{(3 + SR{5})/2} = x which is about 1.378

<----------(0.7256)----------(1.378)----------->

For (-infinity, 0.7256), let x = 0. True statement.

For (0.7256, 1.378), let x = 1.2. False statement.

For (1.378, infinity), let x = 4. True statement.

We exclude the critical points.

Solution:

(-infinity, 0.7256) U (1.378, infinity)

Correct?