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anonymous
 5 years ago
can someone help me find the absolute and local extrema in this equation (3x4)/(x^2+1)
anonymous
 5 years ago
can someone help me find the absolute and local extrema in this equation (3x4)/(x^2+1)

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1(x^2+1)(3)  (3x4)(2x) = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0on the intervals [2,2]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.13x^2 +3 6x^2 +8x = 0 3x^2 +3 +8x = 0 0 = 3x^2 8x 3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1(x3) (x+ 1/3) = x = x and x = 1/3 are the max and min

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x = 3 and x = 1/3......

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1f(2) = ? f(1/3) = ? f(2) = ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what do u do with the denominator

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1nothing; the top directs the =0 part; not the bottom

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1put these x values into the original equationand determine which is higher

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0awesome thanks a lot man the denominator was throwing me off

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1there is no vertical asypmtote; so you aint gotta worry about bad values :)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0also x^2+1=0 has no real solution so this does not give you a critcal number anyways

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0right what amistre said no vertical asym

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0awesome thanks a lot guys
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