## Calculus 8th Edition

The inverse function is: $$f^{-1}(x)=\frac{(x-1)^2-2}{3}.$$
For the function $f$ to be defined $2+3x\geq0\Rightarrow 3x\geq-2\Rightarrow x\geq-2/3.$ Now let $y=f(x)=1+\sqrt{2+3x}$. We need to express $x$ in terms of $y$ because then $x=f^{-1}(y)$. $$y=1+\sqrt{2+3x}\Rightarrow y-1=\sqrt{2+3x}$$ squaring the last equality $$(y-1)^2=2+3x\Rightarrow 3x=(y-1)^2-2$$ and finally $$x=\frac{(y-1)^2-2}{3}$$ which means $$f^{-1}(y)=\frac{(y-1)^2-2}{3}.$$ Renaming $y$ back to $x$ we get $$f^{-1}(x)=\frac{(x-1)^2-2}{3}.$$