Chapter 11.2, Problem 66E

a.

To determine

To graph: The function.

Explanation of Solution

Given: The cost of hooking up and towing a car is $80 for the first mile and $3 for each additional mile or portion of a mile. A model for the cost C (in dollars) is $\mathrm{C}\left(\mathrm{x}\right)=80-33\left[|-\left(\mathrm{x}-1\right)|\right]$ , where x is the distance in miles.

Using graphing utility, graph of function $\mathrm{C}\left(\mathrm{x}\right)=80-33\left[|-\left(\mathrm{x}-1\right)|\right]$ .

  

b.

To determine

To complete: The table and observe the behavior of C as x approaches 5.5.

Answer to Problem 66E

  $\begin{array}{cccccccc}\mathrm{x}& 5& 5.3& 5.4& 5.5& 5.6& 5.7& 6\\ \mathrm{C}\left(\mathrm{x}\right)& 92& 95& 95& 95& 95& 95& 95\end{array}$

Explanation of Solution

Given: The cost of hooking up and towing a car is $80 for the first mile and $3 for each additional mile or portion of a mile. A model for the cost C (in dollars) is $\mathrm{C}\left(\mathrm{x}\right)=80-33\left[|-\left(\mathrm{x}-1\right)|\right]$ , where x is the distance in miles.

  $\mathrm{C}\left(\mathrm{x}\right)=80-33\left[|-\left(\mathrm{x}-1\right)|\right]$

  $\begin{array}{cccccccc}\mathrm{x}& 5& 5.3& 5.4& 5.5& 5.6& 5.7& 6\\ \mathrm{C}\left(\mathrm{x}\right)& 92& 95& 95& 95& 95& 95& 95\end{array}$

From graph, $\mathrm{C}\left(5.5\right)=95$

From table, $\mathrm{C}\left(5.5\right)=95$

  $\begin{array}{l}\text{LHL, }\mathrm{L}=\underset{\mathrm{x}\to {5.5}^{+}}{\lim }\mathrm{C}\left(\mathrm{x}\right)=95\\ \text{RHL, }\mathrm{L}=\underset{\mathrm{x}\to {5.5}^{-}}{\lim }\mathrm{C}\left(\mathrm{x}\right)=95\end{array}$

So, limit exist at $\mathrm{x}=5.5$

  $\mathrm{L}=\underset{\mathrm{x}\to 5.5}{\lim }\mathrm{C}\left(\mathrm{x}\right)=95$

c.

To determine

To complete: The table and observe the behavior of C as x approaches 5.

Answer to Problem 66E

  $\begin{array}{cccccccc}\mathrm{x}& 4& 4.5& 4.9& 5& 5.1& 5.5& 6\\ \mathrm{C}\left(\mathrm{x}\right)& 89& 92& 92& 92& 95& 95& 95\end{array}$

Explanation of Solution

Given: The cost of hooking up and towing a car is $80 for the first mile and $3 for each additional mile or portion of a mile. A model for the cost C (in dollars) is $\mathrm{C}\left(\mathrm{x}\right)=80-33\left[|-\left(\mathrm{x}-1\right)|\right]$ , where x is the distance in miles.

  $\mathrm{C}\left(\mathrm{x}\right)=80-33\left[|-\left(\mathrm{x}-1\right)|\right]$

  $\begin{array}{cccccccc}\mathrm{x}& 4& 4.5& 4.9& 5& 5.1& 5.5& 6\\ \mathrm{C}\left(\mathrm{x}\right)& 89& 92& 92& 92& 95& 95& 95\end{array}$

  $\begin{array}{l}\text{LHL, }\mathrm{L}=\underset{\mathrm{x}\to 5^{-}}{\lim }\mathrm{C}\left(\mathrm{x}\right)=92\\ \text{RHL, }\mathrm{L}=\underset{\mathrm{x}\to 5^{+}}{\lim }\mathrm{C}\left(\mathrm{x}\right)=95\end{array}$

  $\mathrm{L}\mathrm{H}\mathrm{L}\ne \mathrm{R}\mathrm{H}\mathrm{L}$

So, limit does not exist at $\mathrm{x}=5$