anonymous
  • anonymous
can someone help me find the absolute and local extrema in this equation (3x-4)/(x^2+1)
Mathematics
chestercat
  • chestercat
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amistre64
  • amistre64
sure
amistre64
  • amistre64
(x^2+1)(3) - (3x-4)(2x) = 0
anonymous
  • anonymous
on the intervals [-2,2]

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amistre64
  • amistre64
3x^2 +3 -6x^2 +8x = 0 -3x^2 +3 +8x = 0 0 = 3x^2 -8x -3
amistre64
  • amistre64
(x-3) (x+ 1/3) = x = x and x = -1/3 are the max and min
amistre64
  • amistre64
x = 3 and x = -1/3......
amistre64
  • amistre64
f(-2) = ? f(-1/3) = ? f(2) = ?
anonymous
  • anonymous
so what do u do with the denominator
amistre64
  • amistre64
nothing; the top directs the =0 part; not the bottom
amistre64
  • amistre64
put these x values into the original equationand determine which is higher
anonymous
  • anonymous
awesome thanks a lot man the denominator was throwing me off
amistre64
  • amistre64
there is no vertical asypmtote; so you aint gotta worry about bad values :)
myininaya
  • myininaya
also x^2+1=0 has no real solution so this does not give you a critcal number anyways
myininaya
  • myininaya
right what amistre said no vertical asym
anonymous
  • anonymous
awesome thanks a lot guys

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