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Local Max & Min

  • Thread starter Rossinole
  • Start date
20
0
1. Find the local max and min of x/(1+x^2)

3. 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

458
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
Good. Now stop.

[tex]\frac {(1-x)(1+x)}{(1+x^2)^2} \neq \frac {1-x}{(1+x)^2}[/tex]
 
20
0
Good. Now stop.

[tex]\frac {(1-x)(1+x)}{(1+x^2)^2} \neq \frac {1-x}{(1+x)^2}[/tex]
Would I factor the bottom into (1+x^2)(1+x^2)?
 
458
0
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.
 
1,341
3
[tex]f'(x)= \frac{(1-x^2)}{(1+x^2)^2} = 0[/tex]

Is 0 when the numerator is 0...

[tex]1-x^2 = 0[/tex]

[tex]x = \pm 1?[/tex]
 

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