your domain is all real values of x that will generate a real value of y.
in this equation, this will occur at all values of x except when the denominator is equal to 0.
to find that value of x, you set 3 + 4x = 0 and solve for x to get x = -3/4.
when x = -3/4, the value of y will be undefined because you are dividing by 0.
the range of your function is all real values of y except where these is a y-asymptote.
in this equation, this will occur for all real values of y except where there is a horizontal asymptote.
that occurs as x approaches plus or minus infinity.
here's how you analyze the function to find the horizontal asymptote, if any.
the equation is (6-x)/(3+4x)
divide numerator and denominator of the equation by the highest degree of x.
since the highest degree of x is x, then divide numerator and denominator by x.
you will get (6/x-x/x)/(3/x+4x/x)
simplify to get (6/x-1)/(3/x+4)
as x approaches infinity, the 6/x approaches 0 and the 3/x approaches 0 and the expression becomes -1/4.
that's your y-asymptote.
so, the domain of y = f(x) = (6-x)/(3+4x) is all values of x except at x = -3/4, and the range is y = -1/4 as x approaches infinity in the positive or negative direction.
here's a graph of the function.
you can see that the x-asymptote is at x = -3/4 and the y-asymptote is at y = -1/4.
now, when you have an inverse function, the domain of the function is equal to the range of the inverse function and the range of the function is equal to the domain of the inverse function.
therefore, you can expect that the domain of the inverse function will be all real values of x except at x = -1/4, and the range of the inverse function will be all real values of y except at y = -3/4.
you will see this shortly.
to find the inverse function, you do the following:
start with y = f(x) = (6-x)/(3+4x)
since y = f(x), disregard f(x) for now and just take y = (6-x)/(3+4x).
replace y with x and x with y to get x = (6-y)/(3+4y).
multiply both sides of the equation by (3+4y) to get x*(3+4y) = 6-y.
simplify to get 3x+4xy = 6-y
add y to both sides of the equation and subtract 3x from both sides of the equation to get y+4xy = 6-3x
factor out the y to get y*(1+4x) = 6-3x
divide both sides of the equation to get y = (6-3x)/(4x+1)
that's your inverse equation after you solved for y.
your inverse equation before you solved form y was x = (6-y)/(3+4y).
so your inverse equation after solving for y is y = (6-3x)/(4x+1)
your domain is all real values of x except when (4x+1) = 0.
that occurs when x = -1/4.
your range is all real values of y except when there is a y-asymptote.
that occurs when y = -3/4.
so, ...
the domain of y = f(x) = (6-x)/(3+4x) is all values of x except at x = -3/4, and the range is y = -1/4 as x approaches infinity in the positive or negative direction.
and, ...
the domain of y = f^-1(x) = (6-3x)/(4x+1) is all values of x except at x = -1/4, and the range is y = -3/4 as x approaches infinity in the positive or negative direction.
this is as it should be.
the domain of the function is the range of the inverse function.
the range of the function is the domain of the inverse function.
here's the graph of the inverse function.
here's the graph of both functions.
the function is red and the inverse function is blue
here's a reference on how to find asymptote4s of an equation.
http://www.purplemath.com/modules/asymtote4.htm