Is f(x) = (x) / (x^2 - 1) Always Decreasing?

[agv567]

Homework Statement

The equation is f(x) = (x) / (x^2 - 1)

Homework Equations

The equation is f(x) = (x) / (x^2 - 1)

The Attempt at a Solution

Well I first took the derivative, which was f'(x) = -(x^2 - 1) / (x^2 -1) ^2

I set it equal to zero to find the relative extremas, and I got -X^2 -1 = 0, which means X^2 = -1.

That isn't a real number, but so would that mean there are no intervals where it's increasing or decreasing? When I graphed it, it seemed like it was decreasing on all values, negative infinity to positive infinity...

 

[Poopsilon]

Where the derivative equals zero is where the function has a local minimum (bottoms out) or has a local maximum (peaks). Thus if it never does either of these things then it better always be doing either one of two things.

 

[SammyS]

Homework Statement

The equation is f(x) = (x) / (x^2 - 1)

Homework Equations

The equation is f(x) = (x) / (x^2 - 1)

The Attempt at a Solution

Well I first took the derivative, which was f'(x) = -(x^2 - 1) / (x^2 -1) ^2

I set it equal to zero to find the relative extremas, and I got -X^2 -1 = 0, which means X^2 = -1.

That isn't a real number, but so would that mean there are no intervals where it's increasing or decreasing? When I graphed it, it seemed like it was decreasing on all values, negative infinity to positive infinity...

If -(x² - 1) = 0 , then x² = 1, so that x = ±1 .

 

[Poopsilon]

His derivative is not correct, it should be: f'(x) = -(x^2 + 1) / (x^2 -1) ^2.

 

[SammyS]

His derivative is not correct, it should be: f'(x) = -(x^2 + 1) / (x^2 -1) ^2.

That makes sense considering that the function is decreasing over its entire domain.

 

[agv567]

Oh yes, I'm sorry. I had it written down correctly but I typed it incorrectly by adding a parentheses.

It should be -(x^2 + 1) / (X^2-1) ^2

Now when I set it equal to zero, I get X=-1, which does not exist. How would I tell that it was decreasing over the entire domain then without graphing?

 

[SammyS]

For what value of x is x² + 1 = 0 ?

That would require that x = -1. !

Added in Edit:

That was a typo

Should have said: That would require that x² = -1. !

 

[agv567]

But it's not X^2 + 1 = 0...

It's -X^2 -1 = 0, so X^2 = -1...which does not exist.

How would I tell that it was decreasing w/o a calculator?

 

[Poopsilon]

The fact that the derivative is never equal to zero tells you that it must either always be decreasing or always increasing. Thus you need only look at the derivative at any point where it exists and if it's negative you know the original function is decreasing at that point and thus must be decreasing everywhere.

Edit: Also X^2 + 1 = 0 and -X^2 - 1 = 0 are the exact same equation, just move the terms to the other side of the equals sign.

Remember the derivative tells you the slope of the line at any point at which it's defined, the reason its zeros are of particular interest is because when it comes to differentiable functions if the function is going to switch from decreasing to increasing or visa-versa the derivative will be zero right when that switch happens. For instance draw the graph of f(x) = x^2, and then start drawing some tangent lines at various points, notice that your tangent line right at the point where the graph stops going down and starts going up will be flat, and thus have slope equal to zero.