Chapter 12.3, Problem 50E

a.

To determine

ToEvaluate: The slope of the graph of $f\left(x\right)=\frac{1}{x-3}$ at the point $\left(4,1\right)$

Answer to Problem 50E

The slope of the graph of $f\left(x\right)=\frac{1}{x-3}$ at the point $\left(4,1\right)$ is $m=-1$

Explanation of Solution

Given:

The function $f\left(x\right)=\frac{1}{x-3}$ at the point $\left(4,1\right)$

Concept Used:

The slope $m$ of the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ is equal to the slope of its tangent line at $\left(x,f\left(x\right)\right)$ , and is given by

  $m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Calculation:

Given the function $f\left(x\right)=\frac{1}{x-3}$

The slope $m$ of the graph of $f$ at the point $\left(x,f\left(x\right)\right)$ is given by

  $\begin{array}{c}m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{\frac{1}{x+h-3}-\frac{1}{x-3}}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\frac{\left(x-3\right)-\left(x+h-3\right)}{\left(x+h-3\right)\left(x-3\right)}}{h}\right)\end{array}$

  $\begin{array}{c}=\underset{h\to 0}{\lim }\left(\frac{x-3-x-h+3}{h\left(x+h-3\right)\left(x-3\right)}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{-h}{h\left(x+h-3\right)\left(x-3\right)}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{-1}{\left(x+h-3\right)\left(x-3\right)}\right)\\ =\frac{-1}{\left(x+0-3\right)\left(x-3\right)}\text{ Substituting limit}\end{array}$

  $=\frac{-1}{{\left(x-3\right)}^2}$

At $\left(4,1\right)$ , the slope is

  $\begin{array}{c}m=\frac{-1}{{\left(4-3\right)}^2}\\ =\frac{-1}{{\left(1\right)}^2}\\ =-1\end{array}$

Therefore, slope of the graph of $f\left(x\right)=\frac{1}{x-3}$ at the point $\left(4,1\right)$ is $m=-1$

Conclusion:

Theslope of the graph of $f\left(x\right)=\frac{1}{x-3}$ at the point $\left(4,1\right)$ is $m=-1$

b.

To determine

ToEvaluate: The equation of the tangent line to the graph of $f\left(x\right)=\frac{1}{x-3}$ at the point $\left(4,1\right)$

Answer to Problem 50E

The equation of the tangent line at the point $\left(4,1\right)$ is $y=-x+5$

Explanation of Solution

Given:

The function $f\left(x\right)=\frac{1}{x-3}$ and the point $\left(4,1\right)$ .

And from part a.) slope of the graph of $f\left(x\right)=\frac{1}{x-3}$ at the point $\left(4,1\right)$ is $m=-1$

Concept Used:

The equation of the tangent line to the graph of a function $f$ at the point $\left(a,b\right)$ with slope $m$ is given by

  $y-b=m\left(x-a\right)$

Calculation:

From part a.) the slope of the graph of $f\left(x\right)=\frac{1}{x-3}$ at the point $\left(4,1\right)$ is $m=-1$

The equation of the tangent line at the point $\left(a,b\right)$ is given by

  $y-b=m\left(x-a\right)$

  $\begin{array}{c}y-1=-1\left(x-4\right)\\ y-1=-x+4\\ y=-x+4+1\\ y=-x+5\end{array}$

Therefore, the equation of the tangent line at the point $\left(4,1\right)$ is $y=-x+5$

Conclusion:

The equation of the tangent line at the point $\left(4,1\right)$ is $y=-x+5$

c.

To determine

ToGraph: The function $f\left(x\right)=\frac{1}{x-3}$ and the tangent line $y=-x+5$

Answer to Problem 50E

The graph of the function $f\left(x\right)=\frac{1}{x-3}$ and the tangent line $y=-x+5$ is

  

Explanation of Solution

Given:

The function $f\left(x\right)=\frac{1}{x-3}$ and the point $\left(4,1\right)$ .

And from part b.) the equation of tangent line is $y=-x+5$

Concept Used:

The graph of a function is the smooth curve, which is the collection of all the points that satisfy the particular function.

Calculation:

For the function $f\left(x\right)=\frac{1}{x-3}$

And the tangent line $y=-x+5$

Plotting both of them on the same graph using a graphing utility, we have

  

Conclusion:

The graph of the function $f\left(x\right)=\frac{1}{x-3}$ and the tangent line $y=-x+5$ is