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Find all relative extrema of the function F(X)=16/(X^2+1). Use the SecondDerivative 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
 one year ago
 one year ago
Find all relative extrema of the function F(X)=16/(X^2+1). Use the SecondDerivative 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
 one year ago
 one year ago

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AustralopithecusBest ResponseYou've already chosen the best response.0
Take the derivative of this function first
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
First set f'(x) = 0 then solve for x, those will give you your critical points
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
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
 one year ago

hiramobyBest ResponseYou've already chosen the best response.1
so it is going to be e then
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
ok make a table dw:1351050419295:dw
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
check in between these numbers and right if they are positive or negative
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
you need to check to see if they are maximums or minimums
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
and I'm showing you how
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
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
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
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
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
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
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
I mean minimum
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
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)
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
I hope this was helpful :S
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
oh crap made a mistake lol dw:1351051431080:dw it should be this you are right it is e maximum at (0,16)
 one year ago
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