Chapter 1.4, Problem 35E

a.

To determine

To find: an equation for each tangent to the curve $y=\frac{1}{x-1}$ with slope $x=-1$ .

Answer to Problem 35E

The equation each tangent to the curve $y=\frac{1}{x-1}$ with slope $x=-1$ is:

For $x=0$ , the equation is $y=-x-1$ and for $x=2$ the equation is $y=-x+3$ .

Explanation of Solution

Given information:

The function is $y=\frac{1}{x-1}$ at $x=a$ , the slope is $-\frac{1}{{\left(a-1\right)}^2}$ .

Concept used:

For any function f , slope at any point $x=a$ can be calculated as: $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

The equation of a line passing through $\left(x_0,y_0\right)$ having slope m is given by:

  $y-y_0=m\left(x-x_0\right)$ .

The slope of the tangent is equal to the slope of the curve.

Calculation:

Let $f\left(x\right)=y$

For $y=\frac{1}{x-1}$ at $x=a$ the slope will be:

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

The slope of the function is $-\frac{1}{{\left(a-1\right)}^2}$ .

Given slope is $-1$ . So,

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

At $x=a=0$ .

  $\begin{array}{c}y=\frac{1}{\left(0\right)-1}\\ =-1\end{array}$

Now find $y-y_0=m\left(x-x_0\right)$ . Put $\left(x_0,y_0\right)=\left(0,-1\right)\text{ and }m=-1$ .

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

At $x=a=2$ .

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

Now find $y-y_0=m\left(x-x_0\right)$ . Put $\left(x_0,y_0\right)=\left(2,1\right)\text{ and }m=-1$ .

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

The equation each tangent to the curve $y=\frac{1}{x-1}$ with slope $x=-1$ is:

For $x=0$ , the equation is $y=-x+3$ and for $x=2$ the equation is $y=-x+3$ .

b.

To determine

To find: an equation for each normal to the curve $y=\frac{1}{x-1}$ with slope $x=1$ .

Answer to Problem 35E

The equation each normal to the curve $y=\frac{1}{x-1}$ with slope $x=1$ is:

For $x=0$ , the equation is $y=-1+x$ and for $x=2$ the equation is $y=x-1$ .

Explanation of Solution

Given information:

The function is $y=\frac{1}{x-1}$ at $x=a$ , the slope is $-\frac{1}{{\left(a-1\right)}^2}$ .

Concept used:

For any function f , slope at any point $x=a$ can be calculated as: $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

The equation of a line passing through $\left(x_0,y_0\right)$ having slope m is given by:

  $y-y_0=m\left(x-x_0\right)$ .

If the slope of a line is m, then the slope of the normal will be $-\frac{1}{m}$ . Also, the normal passes through the point $\left(x_0,y_0\right)$ . So, the equation of the normal will be:

  $y-y_0=-\frac{1}{m}\left(x-x_0\right)$ .

Calculation:

Let $f\left(x\right)=y$

For $y=\frac{1}{x-1}$ at $x=a$ the slope will be:

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

The slope of the function is $-\frac{1}{{\left(a-1\right)}^2}$ .

The slope of the normal is negative of the slope of the curve.

Given slope is $-1$ . So,

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

At $x=a=0$ .

  $\begin{array}{c}y=\frac{1}{\left(0\right)-1}\\ =-1\end{array}$

Now find $y-y_0=-\frac{1}{m}\left(x-x_0\right)$ . Put $\left(x_0,y_0\right)=\left(0,-1\right)\text{ and }m=1$ .

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

At $x=a=2$ .

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

Now find $y-y_0=-\frac{1}{m}\left(x-x_0\right)$ . Put $\left(x_0,y_0\right)=\left(2,1\right)\text{ and }m=1$ .

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

The equation each tangent to the curve $y=\frac{1}{x-1}$ with slope $x=1$ is:

For $x=0$ , the equation is $y=-1+x$ and for $x=2$ the equation is $y=x-1$ .