Taking q to the limit

[bonanova]

Find the real numbers a and b for which

 

a < q < b  if and only if  x ³ + 1/x ³ = q

 

is not satisfied for any real x.

[DejMar]

x

³ + 1/x³ = q
multiplying both sides by x^3, and restating
x⁶ - qx³ + 1 = 0
let a = x³, then
a² - qa + 1 = 0
using the quadratic formula, we find the discriminant as (q² - 4)
The discriminant shows a,i.e., x³, is complex and not real if q² < 4, i.e.,
-2 < q < 2.

[k-man]

a=-2; b=2

f(x) = x³+1/x³ is undefined for x=0 and has a limit of infinity for all positive x and negative infinity for all negative x. f(x) also has a limit of infinity as x approaches 0 from positive side and negative infinity for negative side.

This means that f(x) must have a local minimum for some positive value of x and a local maximum for some negative x.

 

Taking a derivative and assigning it to 0 will find these extremums.

f'(x) = 3x²-3/x⁴

 

Solving 3x²-3/x⁴=0 results in x⁶=1, which has 2 solutions x=-1 and x=1. Plugging these values of x into f(x) we get the minimum of -2 and the maximum of 2.

[bonanova]

Do the extrema relate to the OP conditions?

[plasmid]

I think I'm mis-translating from Math to English. In my mind, this says:

Find a and b such that there is no real x for which

a < q < b if and only if q = x³ + 1/x³, for all values of q

But the thing on the left of the "if and only if" is true for an open range of values of q while the thing on the right of the "if and only if" is true for exactly one value of q (except when x=0). So the "if and only if" can never be satisfied.

Do you have a different interpretation of the OP in mind?

[bonanova]

Find the real numbers a and b for which,

when (and only when) a < q < b,

x³ + 1/x³ = q does not hold for any real value of x (has no real solutions in x.)

[k-man]

Do the extrema relate to the OP conditions?

 

The way I interpret the OP and your later clarification, they do.

x³ + 1/x³ = q has no real solution for x when -2 < q < 2. For example, x³ + 1/x³ = 1 has no real solutions.

[plasmid]

Nevermind, misread

Edited December 11, 2014 by plasmid

[DejMar]

Multiplying both sides of the equation, x

³ + 1/x³ = q, by x³, and rearranging the equation in standard polynomial form, results in a sextic (aka hexic) polynomial:
x⁶ - qx³ + 1 = 0
The sextic polynomial has six roots, some of which may be multiple roots and degenerate roots. As the polynomial is with even degree, it has both a global maximum and global minimum in addition to local maximum(s) and local minimum(s).
By substituting a = x³, the discrimanant of the quadratic equation for the roots of a can be used to denote the inflection points of the polynomial. (One may also find these inflection points by taking the derivative of the function). These occur at -2 and 2.
If x = 0, we see the equation is undefined (1 = 0).
As x : x < -2 → ^-∞, q → ^-∞,
as x : x > 2 → ⁺∞, q → ⁺∞,
as x : x < -2 → 0^-, q → ^-∞, and
as x : x > 2 → 0⁺, q → ⁺∞.

 

Restated,
x³ + 1/x³ = q is discontinuous where x=0 (thus, not satisfied for that real x),
For a < q < b : q = 0,
a = lim [x→0^-] x³+1/x³ = ^-∞
b = lim [x→0⁺] x³+1/x³ = ⁺∞

Edited December 11, 2014 by DejMar

[bonanova]

k-man (post #2) finds the region between local extrema and states (post 7) that in that region there is no real solution.

dejMar (post #6) finds the region where the discriminant is negative.

Both approaches give the right answer.

Post #6 is what I was looking for.

But before I mark it, I'll ask for comments regarding whether post #2 an equivalent condition for real solutions.