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

$$f^{-1}(x)=\frac{2x+b}{x-1}$$ - Domain: $(-\infty,1)\cup(1,\infty)$. - Range: $(-\infty,2)\cup(2,\infty)$.
$$y=f(x)=\frac{x+b}{x-2}\hspace{1cm}b\gt-2$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=\frac{x+b}{x-2}$$ $$y=\frac{x-2+(b+2)}{x-2}$$ $$y=1+\frac{b+2}{x-2}$$ $$\frac{b+2}{x-2}=y-1$$ $$x-2=\frac{b+2}{y-1}$$ $$x=\frac{b+2}{y-1}+2=\frac{b+2+2y-2}{y-1}=\frac{2y+b}{y-1}$$ 2) Interchange $x$ and $y$: $$y=\frac{2x+b}{x-1}$$ Therefore, $$f^{-1}(x)=\frac{2x+b}{x-1}$$ - Domain: $x$ is defined where $(x-1)\ne0$, or $x\ne1$. So the domain of $f^{-1}$ is $(-\infty,1)\cup(1,\infty)$. - Range: $$f^{-1}(x)=\frac{2x+b}{x-1}=\frac{2x-2+(b+2)}{x-1}=2+\frac{b+2}{x-1}$$ Since $b\gt-2$, which means $b+2\gt0$, in the domain for $x$ above, $\frac{b+2}{x-1}$ range from $-\infty$ to $\infty$, except that $\frac{b+2}{x-1}\ne0$ Thus, $\frac{b+2}{x-1}+2$ also ranges from $-\infty$ to $\infty$, except that $\frac{b+2}{x-1}+2\ne2$ In other words, the range of $f^{-1}$ is $(-\infty,2)\cup(2,\infty)$ *Check: $$f(f^{-1}(x))=\frac{\frac{2x+b}{x-1}+b}{\frac{2x+b}{x-1}-2}=\frac{\frac{2x+b+bx-b}{x-1}}{\frac{2x+b-2x+2}{x-1}}=\frac{2x+bx}{2+b}=x$$ $$f^{-1}(f(x))=\frac{2\times\frac{x+b}{x-2}+b}{\frac{x+b}{x-2}-1}=\frac{\frac{2x+2b+bx-2b}{x-2}}{\frac{x+b-x+2}{x-2}}=\frac{2x+bx}{2+b}=x$$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$