## University Calculus: Early Transcendentals (3rd Edition)

$$f^{-1}(x)=\frac{9x^2}{(x-1)^2}$$ - Domain: $(-\infty,1)\cup(1,\infty)$ - Range: $[0,9) \cup (9,\infty)$
$$y=f(x)=\frac{\sqrt x}{\sqrt x-3}\hspace{1cm}$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=\frac{\sqrt x}{\sqrt x-3}$$ $$y=\frac{\sqrt x-3+3}{\sqrt x-3}$$ $$y=1+\frac{3}{\sqrt x-3}$$ $$\frac{3}{\sqrt x-3}=y-1$$ $$\sqrt x-3=\frac{3}{y-1}$$ $$\sqrt x=\frac{3}{y-1}+3=\frac{3+3y-3}{y-1}=\frac{3y}{y-1}$$ $$x=\frac{9y^2}{(y-1)^2}$$ 2) Interchange $x$ and $y$: $$y=\frac{9x^2}{(x-1)^2}$$ Therefore, $$f^{-1}(x)=\frac{9x^2}{(x-1)^2}$$ - Domain: $x$ is defined in $R$ except where $x-1=0$ or $x=1$. So the domain of $f^{-1}$ is $(-\infty,1)\cup(1,\infty)$. - Range: It would be much more easier if we use a graphing calculator and have it graph the function $f^{-1}(x)=\frac{9x^2}{(x-1)^2}$ Looking at the graph, we see that the range of $f^{-1}(x)$ is: - Range: $[0,9) \cup (9,\infty)$ *Check: $$f(f^{-1}(x))=\frac{\sqrt{\frac{9x^2}{(x-1)^2}}}{\sqrt{\frac{9x^2}{(x-1)^2}}-3}=\frac{\frac{3x}{x-1}}{\frac{3x}{x-1}-3}=\frac{\frac{3x}{x-1}}{\frac{3x-3x+3}{x-1}}=\frac{3x}{3}=x$$ $$f^{-1}(f(x))=\frac{9\times\Big(\frac{\sqrt x}{\sqrt x-3}\Big)^2}{\Big(\frac{\sqrt x}{\sqrt x-3}-1\Big)^2}=\frac{9\times\frac{x}{(\sqrt x-3)^2}}{\Big(\frac{\sqrt x-\sqrt x+3}{\sqrt x-3}\Big)^2}=\frac{\frac{9x}{(\sqrt x-3)^2}}{\Big(\frac{3}{\sqrt x-3}\Big)^2}=\frac{\frac{9x}{(\sqrt x-3)^2}}{\frac{9}{(\sqrt x-3)^2}}=\frac{9x}{9}=x$$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$