use any method to find the range of the function f, where
1. f(x) = 1/(x-2)(x-6)
2. f(x) = x/(1-x)^2
i really dont know where to start here. i could graph these functions, but i just don't manage to break them down to recognisable patterns. otherwise i thought i could find the domain of the inverse and then use that as the range, but that doesnt work does it? anyway i would be needing some help...
ok i'm sure really heavy exchange was going on back there but i couldnt understand it so i cannot comment. however, the only post actually attempting to solve my problem was that by earboth which unfortunately:
1. utilized a method i am unacquainted with, which entails that a simpler solution must exist
2. was wrong, according to my textbook
so i am still awaiting help. thanks in advance.
Now consider first the case that : the largest value that g can assume for such an x is . If you let x tend toward either -2 or +2 in this intervall (-2,2) you get arbitrarily large negative values of g. Since g is continuous it assumes all values inbetween, therefore the image of (-2,2) under g is .
Next, consider the case that x<-2 or 2<x. Then, g(x)>0. You can get arbitrarily large values of g(x) for such an x close enough to -2 or +2. However, if you let x tend toward or , the values of g stay positive but get ever closer to 0. Thus for x<-2 and 2<x the values of g can take any value in .
To sum up: the range of g - and thus the range of f - is .
You agree, I hope, that assumes the lowest value at and thus the range of g is (g is just a quadratic function, its graph is a parabola open from above). I maintain that the range of g is actually the same as the range of f. Thus, the range of f is .