How to find the interval of decrease on (x^3 - 1)/(x^3 + 1)?

[Eclair_de_XII]

Homework Statement

f(x) = (x³ - 1)/(x³ + 1)

Homework Equations

(d/dx)[f(x)/g(x)] = [f`(x)⋅g(x) - f(x)⋅g`(x)]/[g(x)]²

The Attempt at a Solution

f`(x) = [3x<sup>2</sup>(x<sup>3</sup> + 1) - 3x<sup>2</sup>(x<sup>3</sup> - 1)]/[(x<sup>3</sup> + 1)]<sup>2</sup>
f`(x) = 3x²[(x³ + 1) - (x³ - 1)]/[(x³ + 1)]²
f`(x) = 3x²(2)/[(x³ + 1)]²

f`(x) = 6x²/(x³ + 1)²

When I made a sign chart for this derivative, I couldn't find any x that could make the value of f`(x) negative. It's always positive, because both terms are squared. So I can't find any interval of decrease, and in turn, any local minimum or maximum. I'm asked to also write the intervals of concave up/down and any inflection points. I highly doubt my professor would assign a problem where I would just write N/A for six of these values. Am I doing something wrong? What x would make this derivative negative? Or perhaps, am I not deriving this correctly?

 

[andrewkirk]

Your calculations look correct to me

 

[Mark44]

Homework Statement

f(x) = (x³ - 1)/(x³ + 1)

Homework Equations

(d/dx)[f(x)/g(x)] = [f`(x)⋅g(x) - f(x)⋅g`(x)]/[g(x)]²

The Attempt at a Solution

f`(x) = [3x<sup>2</sup>(x<sup>3</sup> + 1) - 3x<sup>2</sup>(x<sup>3</sup> - 1)]/[(x<sup>3</sup> + 1)]<sup>2</sup>
f`(x) = 3x²[(x³ + 1) - (x³ - 1)]/[(x³ + 1)]²
f`(x) = 3x²(2)/[(x³ + 1)]²

f`(x) = 6x²/(x³ + 1)²

When I made a sign chart for this derivative, I couldn't find any x that could make the value of f`(x) negative. It's always positive, because both terms are squared. So I can't find any interval of decrease, and in turn, any local minimum or maximum. I'm asked to also write the intervals of concave up/down and any inflection points. I highly doubt my professor would assign a problem where I would just write N/A for six of these values. Am I doing something wrong? What x would make this derivative negative? Or perhaps, am I not deriving this correctly?

Technically, f' is not always positive -- it's undefined for x = -1. Otherwise, f' > 0 for any $x \ne -1$.
As far as concavity, the 2nd derivative changes sign at a number of points, so writing N/A for these parts would be incorrect.

 

[Eclair_de_XII]

As far as concavity, the 2nd derivative changes sign at a number of points, so writing N/A for these parts would be incorrect.

You mean f``(x)? I looked into local minima/maxima a bit more just now, and it looks like it depends on the second derivative. Other than that, there is no interval of decrease, is there?

 

[Mark44]

You mean f``(x)?

Yes, f'' determines the concavity. On any interval where f''(x) > 0, the graph of f is concave up, and on any interval where f''(x) < 0, the graph of f is concave down.

I looked into local minima/maxima a bit more just now, and it looks like it depends on the second derivative.

No. Since f'(x) is never 0, which you showed, there are no local maxima or minima.

Other than that, there is no interval of decrease, is there?

No, there isn't.

 

[Ray Vickson]

Yes, f'' determines the concavity. On any interval where f''(x) > 0, the graph of f is concave up, and on any interval where f''(x) < 0, the graph of f is concave down.
No. Since f'(x) is never 0, which you showed, there are no local maxima or minima.
No, there isn't.

Actually, $f'(0) = 0$, so $f'(x)$ can be zero in places. However, $x=0$ is a stationary inflection point, rather like in the graph of $y = x^3$.

 

[Mark44]

Actually, $f'(0) = 0$, so $f'(x)$ can be zero in places. However, $x=0$ is a stationary inflection point, rather like in the graph of $y = x^3$.

Yes, you're correct about f'(0), Ray. I forgot that the numerator could be zero, and misspoke.