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

There are two points $(5/2,9/4)$ and $(3/2,-1/4)$ where the slope of the curve is $-3/2$.
$$y=\frac{x}{2}+\frac{1}{2x-4}$$ 1) First, find the derivative of $y$: $$y'=\frac{1}{2}+\frac{(1)'\times(2x-4)-1\times(2x-4)'}{(2x-4)^2}$$ $$y'=\frac{1}{2}+\frac{0\times(2x-4)-1\times2}{(2x-4)^2}$$ $$y'=\frac{1}{2}+\frac{(-2)}{(2x-4)^2}$$ $$y'=\frac{1}{2}-\frac{2}{(2x-4)^2}$$ 2) To find if there are any points where the slope is $-3/2$, we find if there are any $x$ so that $y'=-3/2$ or not. $$y'=-\frac{3}{2}$$ $$\frac{1}{2}-\frac{2}{(2x-4)^2}=-\frac{3}{2}$$ $$\frac{2}{(2x-4)^2}=\frac{1}{2}+\frac{3}{2}=2$$ $$(2x-4)^2=\frac{2}{2}=1$$ $$2x-4=\pm1$$ - For $2x-4=1$, then $x=\frac{1+4}{2}=\frac{5}{2}$. Therefore, $$y=\frac{\frac{5}{2}}{2}+\frac{1}{2\times\frac{5}{2}-4}=\frac{5}{4}+\frac{1}{1}=\frac{9}{4}$$ - For $2x-4=-1$, $x=\frac{-1+4}{2}=\frac{3}{2}$. Therefore, $$y=\frac{\frac{3}{2}}{2}+\frac{1}{2\times\frac{3}{2}-4}=\frac{3}{4}+\frac{1}{(-1)}=-\frac{1}{4}$$ So there are two points $(5/2,9/4)$ and $(3/2,-1/4)$ where the slope of the curve is $-3/2$.