## College Algebra 7th Edition

$g^{-1}(5)$ = 1
$g(x) = x^{2} + 4x$, $x\geq2$ Plug in 5 for g(x) (5 is the x value for $g^{-1}(5)$, so it is the y value for g(x)) $5 = x^{2} + 4x$ $0 = x^{2} + 4x - 5$ 0 = (x - 1)(x + 5) x = 1, -5 x = -5 because it is not in the domain of g(x) x value of g(x) equals the y value of $g^{-1}(x)$ $g^{-1}(5)$ = 1