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

1. Nov 2, 2015

### Eclair_de_XII

1. The problem statement, all variables and given/known data
f(x) = (x3 - 1)/(x3 + 1)

2. Relevant equations
(d/dx)[f(x)/g(x)] = [f(x)⋅g(x) - f(x)⋅g(x)]/[g(x)]2

3. The attempt at a solution
f(x) = [3x2(x3 + 1) - 3x2(x3 - 1)]/[(x3 + 1)]2
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?

2. Nov 2, 2015

### andrewkirk

Your calculations look correct to me

3. Nov 2, 2015

### Staff: Mentor

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.

4. Nov 2, 2015

### Eclair_de_XII

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?

5. Nov 2, 2015

### Staff: Mentor

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.

6. Nov 3, 2015

### Ray Vickson

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 $y = x^3$.

Last edited by a moderator: Jan 28, 2016
7. Nov 3, 2015

### Staff: Mentor

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