anonymous
  • anonymous
Find the critical numbers for 8x-12x^(2/3)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
To find a critical number you must first find the derivative of the function which is in this case:\[f'(x) = 8 - 8(x)^(-1/3)\] then you take the 2 conditions for the critical numbers : 1) f'(x) = 0 2) f'(x) is undefined, hence we have Vertical Asymptotes
anonymous
  • anonymous
give it a try now ^_^
amistre64
  • amistre64
8 - (8/cbrt(x)) 8cbrt(x) - 8 ---------- = 0 : when x = 1 cbrt(x) x = 1 is a critical number.... But is it a max or min? or even an inflection? test with the second derivative...

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amistre64
  • amistre64
If I did it right, its a minimum....
anonymous
  • anonymous
to find the critical points we derive once and to find the inflection points we derive twice ^_^
anonymous
  • anonymous
she just wants to find the critical numbers lol :)
anonymous
  • anonymous
so the one you have found is a critical number :)
amistre64
  • amistre64
actually, there are some instances where an inflection point has a 0 slope, so it is premature to say that y' = 0 is NOT an inflection.
anonymous
  • anonymous
all she wanted was the critical numbers ^_^
amistre64
  • amistre64
yep; and I think its 1 right? if I did it right :)
anonymous
  • anonymous
yes ^_^
amistre64
  • amistre64
whew!! .... :)
anonymous
  • anonymous
:D

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