anonymous
  • anonymous
Find all relative extrema of the function F(X)=16/(X^2+1). Use the Second-Derivative Test when applicable. Answer a. The relative maximum is .1,0 b. The relative minimum is .0,1 c. The relative maximum is .0,16 d. The relative minimum is .0,16 e. The relative maximum is .16,0
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Australopithecus
  • Australopithecus
Take the derivative of this function first
anonymous
  • anonymous
here we go
Australopithecus
  • Australopithecus
First set f'(x) = 0 then solve for x, those will give you your critical points

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anonymous
  • anonymous
I got 0
anonymous
  • anonymous
-32
Australopithecus
  • Australopithecus
also make sure to check the domain of f'(x) and the domain of f(x) if something is not in the domain of f'(x) but is in the domain of f(x) then it is a critical point if a point is not in the domain of f(x) but is in the domain of f'(x) then it is not a critical point
anonymous
  • anonymous
so it is going to be e then
Australopithecus
  • Australopithecus
ok make a table |dw:1351050419295:dw|
Australopithecus
  • Australopithecus
check in between these numbers and right if they are positive or negative
anonymous
  • anonymous
no b is the answer
Australopithecus
  • Australopithecus
you need to check to see if they are maximums or minimums
Australopithecus
  • Australopithecus
and I'm showing you how
Australopithecus
  • Australopithecus
Oh I made a mistake the table should be |dw:1351050692097:dw| so the general rule is 1. If f'(x) > 0 on the Interval than f(x) is increasing on the interval 2. If f'(x) < 0 on the Interval than f(x) is decreasing on the interval
Australopithecus
  • Australopithecus
So to show you an example of this -32x/(x^4 + 2x^2 + 1) sub in a number in the interval (-infinity, 0) so I'm going with -1 -32(-1)/((-1)^4 + 2(-1)^2 + 1) = a positive number so we add in the table |dw:1351050909010:dw|
Australopithecus
  • Australopithecus
then we do it again for the next interval (0, -32) so I will pick 1 f'(1) = -32(1)/((1)^4 + 2(1)^2 + 1) = a negative number |dw:1351051038432:dw| so we see from this table that 0 is a maximum
Australopithecus
  • Australopithecus
I mean minimum
Australopithecus
  • Australopithecus
we can insert 0 into f(x) to see where this minimum is f(0) = 16/(X^2+1) f(0) = 16/1 f(0) = 16 so we have a minimum at (0,16)
Australopithecus
  • Australopithecus
I hope this was helpful :S
Australopithecus
  • Australopithecus
oh crap made a mistake lol |dw:1351051431080:dw| it should be this you are right it is e maximum at (0,16)

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