Opps, posting the full question completely slipped my mind, it
is pretty long actually 
f = 3x^2 - 6x - 1 = 3(x - 1)^2 - 4
Rf = [-4, infin.) ~ (by the way, why is the -4 inclusive? wont
it y=k cut at 2 points at y=-4?)
for f^-1 to exist, x must be equals to or greater than 1.
f^-1 = sqrt [(x+4)/3]+1 ---x equals to or greater than -4
(same question as abovementioned point)
function g = sqrt(x+4), x equals to or greater than -4
Determine whether the composite function gf^-1 exists. Find the
range of gf^-1, if it exists.
Hi,
Is the domain of f given in the question?
Assuming that the domain of f is the entire real set of
numbers, the range you stated makes sense. At y = -4, the
horizontal line is tangent to the curve of f.
Yes, you are correct that x >= 1 for the inverse of f
to exist. Another possible maximal domain is given by the set of x
such that x <= 1.
To check for the existence of the composite function g
f^(-1), we consider range of f^(-1) and domain of g:
Range of f^(-1) = domain of f = (1, inf);
Domain of g = (-4, inf), which has been given.
Now (1, inf) is indeed a subset of (-4, inf), so the
composite function g f^(-1) does exist.
To find the range of g f^(-1), we take these systematic
steps:
1. Sketch the graph of y = g(x), a C-shaped
half-parabola.
2. Map the range of f^(-1) onto the x-values in the
graph.
3. Find the set of y-values corresponding to point 2.
So the range of g f^(-1) = ( sqrt(5), inf ).
Thanks.
P.S. Do give the question "as it is" in future. Thanks.
Cheers,
Wen Shih