## College Algebra 7th Edition

$f^{-1}(x) = \frac{x + 2}{4 - 3x}$
$f(x) = \frac{4x - 2}{3x + 1}$ $y = \frac{4x - 2}{3x + 1}$ y(3x + 1) = 4x - 2 3xy + y = 4x - 2 y + 2 = 4x - 3xy y + 2 = x(4 - 3y) $x = \frac{y + 2}{4 - 3y}$ $f^{-1}(x) = \frac{x + 2}{4 - 3x}$