Chapter 2.4, Problem 35E

(a)

To determine

To find: The equation of the tangents to the curve $y=\frac{1}{x-1}$ that has slope $-1$.

Answer to Problem 35E

The equation of the tangents to the curve $y=\frac{1}{x-1}$ that has slope $-1$ are $y=-x+3$ at $x=2$ and $y=-\left(x+1\right)$ at $x=0$.

Explanation of Solution

Given information:

The curve is $y=\frac{1}{x-1}$ and the slope of the tangents is $-1$.

Calculation:

The formula for the slope of the curve $f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$.

Substitute $a+h$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=\frac{1}{x-1}\\ f\left(a+h\right)=\frac{1}{a+h-1}\end{array}$

Substitute $a$ for $x$ in the function.

  $\begin{array}{l}f\left(x\right)=\frac{1}{x-1}\\ f\left(a\right)=\frac{1}{a-1}\end{array}$

Substitute $\frac{1}{a+h-1}$ for $f\left(a+h\right)$ and $\frac{1}{a-1}$ for $f\left(a\right)$ in the formula for slope.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0}{\lim }\frac{\frac{1}{a+h-1}-\frac{1}{a-1}}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{a-1-\left(a+h-1\right)}{\left(a-1\right)\left(a+h-1\right)}}{h}\\ =\underset{h\to 0}{\lim }\frac{-h}{h\left(a-1\right)\left(a+h-1\right)}\\ =\underset{h\to 0}{\lim }\left(\frac{-1}{\left(a-1\right)\left(a+h-1\right)}\right)\end{array}$

Further simplify.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\frac{-1}{\left(a-1\right)\left(a+0-1\right)}\\ =-\frac{1}{{\left(a-1\right)}^2}\end{array}$

Equate the slope to $-1$.

  $\begin{array}{c}-\frac{1}{{\left(a-1\right)}^2}=-1\\ {\left(a-1\right)}^2=1\\ a-1=\pm 1\\ a=2,0\end{array}$

So, the points at which the tangent has the slope equal to $-1$ are $x=2$ and $x=0$.

The equation of the tangent line at $x=2$.

  $\begin{array}{c}y-f\left(2\right)=m\left(x-2\right)\\ y-1=-1\left(x-2\right)\\ y-1=-x+2\\ y=-x+3\end{array}$

The equation of the tangent line at $x=0$.

  $\begin{array}{c}y-f\left(0\right)=m\left(x-0\right)\\ y-\left(-1\right)=-1\left(x-0\right)\\ y+1=-x\\ y=-x-1\end{array}$

Therefore, the equation of the tangents to the curve $y=\frac{1}{x-1}$ that has slope $-1$ are $y=-x+3$ at $x=2$ and $y=-\left(x+1\right)$ at $x=0$.

(b)

To determine

To find: The equation of the normal to the curve $y=\frac{1}{x-1}$ that has slope $1$.

Answer to Problem 35E

The equation of the normal to the curve $y=\frac{1}{x-1}$ that has slope $1$ is $y=x-1$ at both the points $x=2$ and $x=0$.

Explanation of Solution

Given information:

The curve is $y=\frac{1}{x-1}$ and slope of normal is $1$.

Calculation:

The points at which the tangent has the slope $-1$ are the same points where the normal has the slope equal to $-1$.

The equation of normal at $x=2$.

  $\begin{array}{c}y-f\left(2\right)=m\left(x-2\right)\\ y-1=1\left(x-2\right)\\ y-1=x-2\\ y=x-1\end{array}$

The equation of normal at $x=0$.

  $\begin{array}{c}y-f\left(0\right)=m\left(x-0\right)\\ y-\left(-1\right)=1\left(x-0\right)\\ y+1=x\\ y=x-1\end{array}$

Therefore, the equation of the normal to the curve $y=\frac{1}{x-1}$ that has slope $1$ is $y=x-1$ at both the points $x=2$ and $x=0$.