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

(a) $$f^{-1}(x)=-x+1$$ The graphs are shown in the image below. The line $y=-x+1$ and the line $y=x$ intersect at $90^\circ$. (b) $$f^{-1}(x)=-x+b$$ The line $y=-x+b$ makes with the line $y=x$ an angle of $90^\circ$. (c) The inverses of those functions, whose graphs are perpendicular to the line $y=x$, are themselves.
(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. From the graphs, we can see that the line $y=-x+1$ and the line $y=x$ intersect at $90^\circ$. (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, $$f^{-1}(x)=-x+b$$ We can deduce from (a), or notice that the slope of $y=-x+b$ is $-1$ and the slope of $y=x$ is $1$ and since $-1\times1=-1$, these two lines would make an angle of $90^\circ$ with each other. (c) From (a) and (b), we notice that the formula of the inverse are exactly the same as the formula of the original functions. That means, their graphs must overlap each other, or in other words, the inverse of those functions, whose graphs are lines perpendicular to the line $y=x$, are themselves.