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 2 years 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
 2 years 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

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Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0Take the derivative of this function first

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

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0also 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

hiramoby
 2 years ago
Best ResponseYou've already chosen the best response.1so it is going to be e then

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0ok make a table dw:1351050419295:dw

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0check in between these numbers and right if they are positive or negative

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0you need to check to see if they are maximums or minimums

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0and I'm showing you how

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0Oh 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
 2 years ago
Best ResponseYou've already chosen the best response.0So 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
 2 years ago
Best ResponseYou've already chosen the best response.0then 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
 2 years ago
Best ResponseYou've already chosen the best response.0I mean minimum

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0we 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
 2 years ago
Best ResponseYou've already chosen the best response.0I hope this was helpful :S

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0oh 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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