What are the stationary points for this function?

gva0324 · #1 $Old$ May 20th, 2009, 09:49 AM

Hello, I need help deriving this and finding the stationary points for the following:

$y= 3x / (x^2 + 1)$

If somebody could please show me the method of how to do this. I compared my answer with the result in a calculator and I do not know what I am doing wrong.

TheAbstractionist · #2 $Old$ May 20th, 2009, 09:59 AM

Quote:

Originally Posted by gva0324 $View Post$

Hello, I need help deriving this and finding the stationary points for the following:

$y= 3x / (x^2 + 1)$

If somebody could please show me the method of how to do this. I compared my answer with the result in a calculator and I do not know what I am doing wrong.

Hi gva0324.

Use the quotient rule:

$\mbox{If}\ y=\frac uv,\ \frac{dy}{dx}=\frac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}$

The stationary points are where $\frac{dy}{dx}=0.$

gva0324 · #3 $Old$ May 20th, 2009, 10:25 AM

Yes.

I obtain $3 (x-1)^2 / (x^2 +1)$

Which seems to be wrong as the answer is

$3(1-x^2) / (x^2 + 1)$

How do I obtain the answer above?

and using the answer what would be the stationary points?

there are none?

TheAbstractionist · #4 $Old$ May 20th, 2009, 10:35 AM

Quote:

Originally Posted by gva0324 $View Post$

Yes.

I obtain $3 (x-1)^2 / (x^2 +1)$

Which seems to be wrong as the answer is

$3(1-x^2) / (x^2 + 1)$

How do I obtain the answer above?

and using the answer what would be the stationary points?

there are none?

The answer I got is $\frac{3(1-x^2)}{(x^2+1)^2}.$ The stationary points are when this is 0 (there are two of them).