Chapter 2.4, Problem 11E

(a)

To determine

To find: The slope of the curve $y=\frac{1}{x-1}$ at $x=2$.

Answer to Problem 11E

The slope of the curve $y=\frac{1}{x-1}$ at $x=2$ is $-1$.

Explanation of Solution

Given information:

The curve is $y=\frac{1}{x-1}$ and the point is $x=2$.

Calculation:

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

The slope of the curve $y=\frac{1}{x-1}$ at $x=2$ is:

  $\text{Slope}=\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}$

Substitute $2+h$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=\frac{1}{x-1}\\ f\left(2+h\right)=\frac{1}{2+h-1}\\ f\left(2+h\right)=\frac{1}{1+h}\end{array}$

Substitute $2$ for $x$ in the function.

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

Substitute $\frac{1}{1+h}$ for $f\left(2+h\right)$ and $1$ for $f\left(2\right)$ in the formula for slope.

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

Further simplify,

  $\begin{array}{c}\text{Slope}=\frac{-1}{1+0}\\ =-1\end{array}$

Therefore, the slope of the curve $y=\frac{1}{x-1}$ at $x=2$ is $-1$.

(b)

To determine

To find: The equation of the tangent line to the curve $y=\frac{1}{x-1}$ at $x=2$.

Answer to Problem 11E

The equation of the tangent line to the curve $y=\frac{1}{x-1}$ at $x=2$ is $y=-x+3$.

Explanation of Solution

Given information:

The curve is $y=\frac{1}{x-1}$ and the point is $x=2$.

Calculation:

The equation of the tangent line to $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ with slope at $x=a$ is:

  $y-f\left(a\right)=m\left(x-a\right)$

The function is $y=\frac{1}{x-1}$ and as calculated in part (a), the value of $f\left(2\right)$ is equal to $1$ . The value of slope of tangent to curve is $-1$.

Substitute $-1$ for $m$ , $2$ for $a$ and $1$ for $f\left(2\right)$ in the equation of tangent.

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

Therefore, the equation of the tangent line to the curve $y=\frac{1}{x-1}$ at $x=2$ is $y=-x+3$.

(c)

To determine

To find: The equation to the normal line of the curve $y=\frac{1}{x-1}$ at $x=2$.

Answer to Problem 11E

The equation of the normal line to the curve $y=\frac{1}{x-1}$ at $x=2$ is $y=x-1$.

Explanation of Solution

Given information:

The curve is $y=\frac{1}{x-1}$ and the point is $x=2$.

Calculation:

The equation of the normal line of $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ with slope at $x=a$ is:

  $y-f\left(a\right)=m\left(x-a\right)$

The function is $y=\frac{1}{x-1}$ and as calculated in part (a), the value of $f\left(2\right)$ is equal to $1$ . The value of slope of tangent to curve $y=\frac{1}{x-1}$ at $x=2$ is $-1$.

So, the slope of the normal to the curve $y=\frac{1}{x-1}$ at $x=2$ is equal to $1$.

Substitute $1$ for $m$ , $2$ for $a$ and $1$ for $f\left(2\right)$ in the equation of normal.

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

Therefore, the equation of the normal line to the curve $y=\frac{1}{x-1}$ at $x=2$ is $y=x-1$.

(d)

To determine

To graph: The curve $y=\frac{1}{x-1}$ , the tangent and normal line to the curve at $x=2$ in the same square viewing window.

Explanation of Solution

Given information:

The curve is $y=\frac{1}{x-1}$ and the point is $x=2$.

Graph:

As calculated in part (b), the equation of the tangent line to the curve $y=\frac{1}{x-1}$ at $x=2$ is equal to $y=-x+3$.

As calculated in part (c), the equation of normal line to the curve $y=\frac{1}{x-1}$ at $x=2$ is equal to $y=x-1$.

To graph the curve, tangent and normal to the curve, follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $y=\frac{1}{x-1}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{1}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{-}\boxed{1}\boxed{)}$ . Now, press the $\boxed{\text{ENTER}}$ button and enter right hand side of the equation $y=-x+3$ after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{(-)}\boxed{\text{X}}\boxed{+}\boxed{3}$ . Now, press the $\boxed{\text{ENTER}}$ button and enter right hand side of the equation $y=x-1$ after the symbol ${\text{Y}}_3$ . Enter the keystrokes $\boxed{\text{X}}\boxed{-}\boxed{1}$.

The display will show the equations,

  $\begin{array}{l}{\text{Y}}_{\text{1}}=1/\left(\text{X}-1\right)\\ {\text{Y}}_2=-\text{X}+3\\ {\text{Y}}_3=\text{X}-1\end{array}$

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below.

  

Figure (1)

Interpretation: From the graph it can be seen that the curve, the tangent and the normal line intersect the upper half of the curve at the point $\left(2,1\right)$.