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

There are 2 tangent lines whose slope is $-1$: $y=-x-1$ and $y=-x+3$.
$$y=f(x)=\frac{1}{x-1}$$ The equation of the tangent lines whose slopes are $-1$ has this form: $$y=-x+m$$ To find $m$, we need to locate the tangent point of these lines, which is to find $x$ such that $$y'=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=-1\hspace{1cm}(1)$$ First, we examine $f(x+h)-f(x)$: $$f(x+h)-f(x)=\frac{1}{x+h-1}-\frac{1}{x-1}=\frac{x-1-x-h+1}{(x+h-1)(x-1)}$$ $$f(x+h)-f(x)=\frac{-h}{(x+h-1)(x-1)}$$ Therefore, $$\frac{f(x+h)-f(x)}{h}=\frac{-h}{h(x+h-1)(x-1)}=\frac{-1}{(x+h-1)(x-1)}$$ Replacing the result into $(1)$: $$\lim_{h\to0}\frac{-1}{(x+h-1)(x-1)}=-1$$ $$\frac{-1}{(x+0-1)(x-1)}=-1$$ $$\frac{1}{(x-1)^2}=1$$ $$(x-1)^2=1$$ $$x-1=\pm1$$ $$x=2\hspace{1cm}\text{or}\hspace{1cm}x=0$$ - For $x=0$, $y=f(0)=1/(0-1)=1/(-1)=-1$ Substitute $x$ and $y$ for $m$: $$0+m=-1$$ $$m=-1$$ The tangent line is $y=-x-1$. - For $x=2$, $y=f(2)=1/(2-1)=1/(1)=1$ Substitute $x$ and $y$ for $m$: $$-2+m=1$$ $$m=3$$ The tangent line is $y=-x+3$.