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

At 2 points $(4,2)$ and $(0,0)$, $g(x)$ is perpendicular with the given line.
$$y=g(x)=\frac{x}{x-2}$$ 1) Find $g'(x)$: $$g'(x)=\frac{x'(x-2)-x(x-2)'}{(x-2)^2}=\frac{1\times(x-2)-x\times1}{(x-2)^2}$$ $$g'(x)=\frac{x-2-x}{(x-2)^2}=\frac{-2}{(x-2)^2}$$ 2) The slope of the tangent line to $g(x)$ at $x=a$ is $g'(a)=\frac{-2}{(a-2)^2}$. We know that the product of the slope of 2 perpendicular lines equals $-1$. Therefore, all the tangent lines to $g(x)$ perpendicular to $y=2x+3$ must have $2g'(a)=-1$. In other words, in order to find the required points, we now have to find $a$ such that $2g'(a)=-1$: $$2\times\frac{-2}{(a-2)^2}=-1$$ $$\frac{-4}{(a-2)^2}=-1$$ $$(a-2)^2=\frac{-4}{-1}=4$$ $$a-2=\pm2$$ $$a=4\hspace{1cm}\text{or}\hspace{1cm}a=0$$ - For $a=4$: $$g(a)=\frac{4}{4-2}=\frac{4}{2}=2$$ - For $a=0$: $$g(a)=\frac{0}{0-2}=0$$ Therefore, at 2 points $(4,2)$ and $(0,0)$, $g(x)$ is perpendicular with the given line.