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