anonymous
  • anonymous
what are the critical numbers of f(x)= 4x/(x^2 +1)?
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
did u already calculate the derivative?
anonymous
  • anonymous
what's f ' (x) = ??
Goten77
  • Goten77
|dw:1358921318013:dw|

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anonymous
  • anonymous
do i simplify it further or just set it to 0 now?
Goten77
  • Goten77
to simplify makes setting it = to 0 much easier
anonymous
  • anonymous
ok i need help simplifying it…i don't think i am doing it right so far
Goten77
  • Goten77
k also the critical points are what makes the top part become 0 .. or what makes the bottom part not exist.. in this case all reals exist on the domain... so we just gotta worry bout setting the top = to 0
anonymous
  • anonymous
so i just completely forget the bottom part?
Goten77
  • Goten77
|dw:1358921549240:dw|
Goten77
  • Goten77
i wouldnt say forget bout the bottom part... but usually in the 1st derivative test its worthless.. unless theres a square root or something that might make it non existiant
Goten77
  • Goten77
hmm let me think... i gotta make sure i didnt say soemthing wrong
anonymous
  • anonymous
|dw:1358921732409:dw|
anonymous
  • anonymous
wait wait that was wrong i got it its 1
Goten77
  • Goten77
k i also need to add that critical points are when the derivative = 0 on the bottom or top.. and when its undefined... but it must all be in the domain
Goten77
  • Goten77
We say that x=c is a critical point of the function f(x) if exists and if either of the following are true. f'(c)=0 or doesnt exist
Goten77
  • Goten77
|dw:1358922243760:dw|
Goten77
  • Goten77
but ultimately the 1 and -1 are the critical points... sorry i kinda am lost in calculus *been a while since i did it* but making the top = to 0 and finding where stuff doesnt exist is where critical points are.... srry if i say something wrong but this info should be correct
anonymous
  • anonymous
thanks for the help!

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