# What are the intervals where the function is increasing or decreasing(if any)?

1. Jan 7, 2012

### agv567

1. The problem statement, all variables and given/known data

function is (X) / (X^2 - 1)

The derivative(as far as I know) is (-X^2-1) / (X^2-1)^2

3. The attempt at a solution

So I set it equal to zero, and I get -X^2 -1 = 0, which means X^2 = -1

This does not exist, so what would I say for the intervals? When I graph it, the function is decreasing on all, but there are asymptotes for X = +-1.

Last edited: Jan 7, 2012
2. Jan 7, 2012

### Staff: Mentor

It looks like there's a term missing here. What is the correct formula for the function?
Since I don't know what you started with, there's no way to tell if this is right.

3. Jan 7, 2012

### SammyS

Staff Emeritus
I assume that you mean:
$\displaystyle f(x)=\frac{x}{x^2-1}\,.$​
In that case, your derivative is correct.

So you have found that the derivative is never equal to zero.

It is discontinuous for two values of x. So it is continuous over three intervals. Check the sign of the derivative in each of the three intervals.

4. Jan 7, 2012

### agv567

Well by graphing it, all of them are negative.

How would I know that you would get 2 valus for X algebraically when the derivative is never equal to zero?

5. Jan 8, 2012

### Staff: Mentor

Your function is f(x) = x/(x2 - 1). For which x values if this function undefined? Those values determine the intervals that Sammy was talking about.

6. Jan 8, 2012

### agv567

The values are +-1
When I check the sign, all of them are negative

So would the answer look like this?

f(x) is decreasing on (negative infinity, -1) U (-1, 1) U (1, infinity)?

U meaning union

7. Jan 8, 2012

### SammyS

Staff Emeritus
Yes.