# Inverse Functions, think I might have this one. . .

• Nov 19th 2008, 11:05 AM
rtwilton
Inverse Functions, think I might have this one. . .
f(x) = (2x-3) / (x+4) is one-to-one. Find it's inverse, state the domain of f, and find the range using f-1.

So I did the work and got an inverse of 3-3x.

And range and domain always give me trouble. I know the range is the domain of the inverse, etc etc. . .

First of all, is my answer correct? LOL This is my new fav site.
• Nov 19th 2008, 02:00 PM
skeeter
$\displaystyle f(x) = \frac{2x-3}{x+4}$

domain of f(x) is all reals except x = -4

$\displaystyle y = \frac{2x-3}{x+4}$

swap variables ...

$\displaystyle x = \frac{2y-3}{y+4}$

solve for y ...

$\displaystyle x(y+4) = 2y-3$

$\displaystyle xy + 4x = 2y - 3$

$\displaystyle xy - 2y = -3x - 3$

$\displaystyle y(x - 2) = -3(x + 1)$

$\displaystyle y = -\frac{3(x+1)}{x-2}$

$\displaystyle f^{-1}(x) = -\frac{3(x+1)}{x-2}$

domain of the inverse is all reals except x = 2 ... so, the range of the original function is all reals except x = 2.
• Nov 19th 2008, 02:40 PM
rtwilton
I can see clearly now, thanks so much!!