Here's the question you clicked on:
hiramoby
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
Take the derivative of this function first
First set f'(x) = 0 then solve for x, those will give you your critical points
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
so it is going to be e then
ok make a table |dw:1351050419295:dw|
check in between these numbers and right if they are positive or negative
you need to check to see if they are maximums or minimums
and I'm showing you how
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
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|
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
I mean minimum
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)
I hope this was helpful :S
oh crap made a mistake lol |dw:1351051431080:dw| it should be this you are right it is e maximum at (0,16)