# Local Max & Min

1. Find the local max and min of x/(1+x^2)

## The Attempt at a Solution

Take the derivative of x/(1+x^2) and set it equal to zero.

f'(x)= (1-x^2)/(1+x^2)^2 = 0

This is where I am stuck. My first guess was to factor the top, resulting in (1-x)(1+x)/(1+x^2)^2 = 0, and then cancel the 1+x. But after that I'm still left with (1-x)/(1+x)^2 = 0. I know if you let x=0 you will get f'(x) = 0 because 0/anything = 0.

Really, I'm not sure what I can do after I take the derivative.

## Answers and Replies

This is where I am stuck. My first guess was to factor the top, resulting in (1-x)(1+x)/(1+x^2)^2 = 0

Good. Now stop.

$$\frac {(1-x)(1+x)}{(1+x^2)^2} \neq \frac {1-x}{(1+x)^2}$$

Good. Now stop.

$$\frac {(1-x)(1+x)}{(1+x^2)^2} \neq \frac {1-x}{(1+x)^2}$$

Would I factor the bottom into (1+x^2)(1+x^2)?

What's the use? That bugger is never going to equal 0 no matter what number you put in the denominator. Look at the numerator.

$$f'(x)= \frac{(1-x^2)}{(1+x^2)^2} = 0$$

Is 0 when the numerator is 0...

$$1-x^2 = 0$$

$$x = \pm 1?$$