1. ## Find the range

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

2. Originally Posted by glover_m
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 $\displaystyle y = -\frac{1}{2}$

3. thanks so much