College Algebra 7th Edition

$f^{-1}(x) = \frac{3 + x }{8x + 4}$
$f(x) = \frac{3 - 4x}{8x -1}$ $y= \frac{3 - 4x}{8x -1}$ y(8x - 1) = 3 - 4x 8xy - y = 3 - 4x 8xy + 4x = 3 + y x(8y + 4) = 3 + y $x = \frac{3 + y }{8y + 4}$ $f^{-1}(x) = \frac{3 + x }{8x + 4}$