## Calculus: Early Transcendentals 8th Edition

$f^{-1}(x)=\frac{1}{3}x^2-\frac{2}{3}x-\frac{1}{3}, x\geq1$
$y=1+\sqrt {2+3x}$ $\because \sqrt {2+3x}\geq0, y\geq1+0=1$ $\sqrt {2+3x}=y-1$ $2+3x=(y-1)^2$ $3x=(y-1)^2-2=y^2-2y-1$ $x=\frac{1}{3}y^2-\frac{2}{3}y-\frac{1}{3}$ $x\rightleftharpoons y$ $f^{-1}(x)=\frac{1}{3}x^2-\frac{2}{3}x-\frac{1}{3}$ Domain of inverse: $x\geq1$