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

$$f^{-1}(x)=\frac{2x+3}{x-1}$$ - Domain: $(-\infty,1)\cup(1,\infty)$ - Range: $(-\infty,2)\cup(2,\infty)$
$$y=f(x)=\frac{x+3}{x-2}\hspace{1cm}$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=\frac{x+3}{x-2}$$ $$y = \frac{(x-2)+5}{x-2}$$ $$y =1+\frac{5}{x-2}$$ $$\frac{5}{x-2}=y-1$$ $$x-2=\frac{5}{y-1}$$ $$x=\frac{5}{y-1}+2=\frac{5+2y-2}{y-1}=\frac{2y+3}{y-1}$$ 2) Interchange $x$ and $y$: $$y=\frac{2x+3}{x-1}$$ Therefore, $$f^{-1}(x)=\frac{2x+3}{x-1}$$ - 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: $$f^{-1}(x)=\frac{2x+3}{x-1}=\frac{2x-2+5}{x-1}=2+\frac{5}{x-1}$$ We know that the range of $(x-1)$ is $(-\infty,0)\cup(0,\infty)$. So the range of $\frac{5}{x-1}$ would also be $(-\infty,0)\cup(0,\infty)$ ($\frac{5}{x-1}$ cannot be $0$). Therefore, the range of $f^{-1}(x)=2+\frac{5}{x-1}$ is $(-\infty,2)\cup(2,\infty)$ *Check: $$f(f^{-1}(x))=\frac{\frac{2x+3}{x-1}+3}{\frac{2x+3}{x-1}-2}=\frac{\frac{2x+3+3x-3}{x-1}}{\frac{2x+3-2x+2}{x-1}}=\frac{5x}{5}=x$$ $$f^{-1}(f(x))=\frac{2\times\frac{x+3}{x-2}+3}{\frac{x+3}{x-2}-1}=\frac{\frac{2x+6+3x-6}{x-2}}{\frac{x+3-x+2}{x-2}}=\frac{5x}{5}=x$$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$