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

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To solve the inequality x^3 + (1/x^3) ≥ 3, the expression is rearranged to x^6 - 3x^3 + 1 ≥ 0. This leads to a sixth-degree polynomial, which can be treated as a quadratic in terms of x^3. Using the quadratic formula, the real roots are found to be approximately 0.7256 and 1.378, with Descartes' rule of signs indicating the remaining roots are complex. The solution intervals are determined to be (-infinity, 0.7256) and (1.378, infinity), excluding the critical points.
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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.
 
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Move everything to the LHS, and then combine all terms into a rational expression...what do you then have?
 
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.
 
What you should have is:

$$\frac{x^6-3x^3+1}{x^3}\ge0$$

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?
 
"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?
 
RTCNTC said:
"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).

RTCNTC said:
"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.

RTCNTC said:
"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:

$$(x^3)^2-3x^3+1$$

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

$$x^3=\frac{3\pm\sqrt{5}}{2}\implies x=\sqrt[3]{\frac{3\pm\sqrt{5}}{2}}$$

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
 
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?
 
  • #10
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?
 
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