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

(a) $$f^{-1}(x)=\frac{x}{m}$$ (b) The graph of the inverse of $y=f(x)=mx$ is a line through the origin with a non-zero slope $\frac{1}{m}$
$$y=f(x)=mx\hspace{1cm}m\ne0$$ (a) To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=mx$$ $$x=\frac{y}{m}$$ 2) Interchange $x$ and $y$: $$y=\frac{x}{m}$$ Therefore, $$f^{-1}(x)=\frac{x}{m}$$ (b) As told in the exercise, $f(x)=mx$ in the graph is a line through the origin with a non-zero slope $m$. We found its inverse, which is $f^{-1}(x)=\frac{x}{m}$ If we replace $x=0$, we find that $f^{-1}(0)=\frac{0}{m}=0$. So the graph of the inverse also goes through the origin. Looking at the formula, we figure that the graph of the inverse is also a line, too, with a non-zero slope $\frac{1}{m}$ Therefore, we can conclude that the graph of the inverse of $y=f(x)=mx$ is a line through the origin with a non-zero slope $\frac{1}{m}$