Find the range of the following function.
f(x) = (4+x)/(3-2x)
Looking for something like this
y= x^2
x = +/- sqrt(y)
y>0
range of $\displaystyle f(x)$ will be the domain of $\displaystyle f^{-1}(x)$
$\displaystyle y = \frac{4+x}{3-2x}$
swap variables ...
$\displaystyle x = \frac{4+y}{3-2y}$
solve for $\displaystyle y$ ...
$\displaystyle 3x - 2xy = 4+y$
$\displaystyle 3x - 4 = y + 2xy$
$\displaystyle 3x-4 = y(1 + 2x)$
$\displaystyle \frac{3x-4}{1+2x} = y = f^{-1}(x)$
domain of the inverse function is all reals except $\displaystyle x = -\frac{1}{2}$
so ... range of $\displaystyle f(x)$ is all reals except $\displaystyle y = -\frac{1}{2}$