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

(a) $$f^{-1}(x)=x-1$$ The graphs are in the image below. (b) $$f^{-1}(x)=x-b$$ The graphs of $f$ and $f^{-1}$ are parallel to each other. (c) We can conclude that the inverse of those functions whose graphs are parallel to the line $y=x$: - Their graphs are also parallel to the line $y=x$ and to the graphs of the original functions. - They lie across the line $y=x$ from the original graphs, and their distance to the line $y=x$ is equal to that of the original graphs.
(a) $$y=f(x)=x+1\hspace{1cm}$$ - To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=x+1$$ $$x=y-1$$ 2) Interchange $x$ and $y$: $$y=x-1$$ Therefore, $$f^{-1}(x)=x-1$$ The graphs are shown in the image below. (b) $$y=f(x)=x+b\hspace{1cm}$$ - To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=x+b$$ $$x=y-b$$ 2) Interchange $x$ and $y$: $$y=x-b$$ Therefore, $$f^{-1}(x)=x-b$$ Looking at the formula of $f$ and $f^{-1}$, since both have slope $1$, their graphs would be parallel with each other. (c) We see in (a) that the graph of $f^{-1}$ is across the other side of line $y=x$ from the graph of $f$. And in fact, in (b), the same thing would happen. They are not just on opposite sides, but the distance from them to line $y=x$ are equal to the other. So we can conclude that the inverse of those functions whose graphs are parallel to the line $y=x$: - Their graphs are also parallel to the line $y=x$ and to the graphs of the original functions. - They lie across the line $y=x$ from the original graphs, and their distance to the line $y=x$ is equal to that of the original graphs.