Chapter 2.3, Problem 53E

(a)

To determine

To find: The formula that gives the charge for $x$ hours with $0\le x\le 24$.

Answer to Problem 53E

The formula that gives the charge for $x$ hours with $0\le x\le 24$ is $\{\begin{array}{l}-1.10\left(\mathrm{int}\left(-x\right)\right),0\le x\le 6\\ 7.25,6<x\le 24\end{array}$.

Explanation of Solution

Given information: The charge of a value park is $\$1.10$ per hour or fraction of an hour for airport parking. The maximum charge per day is $\$7.25$.

Calculation:

The charges are taken for the fraction of an hour. So, for the time period of $0$ to $1$ hour the charge is $\$1.10$ and the charges for $1$ to $2$ hour is $\$2.20$ and so on.

The charges for $5$ to $6$ hour is $\$6.60$ and for more than $6$ hours the charges becomes more than $\$7.25$ . So, for the time period of more than $6$ hours the charges is $\$7.25$.

So, the function is constant and equal to $7.25$ for between $6$ to $24$ hours and the function for $0$ to $6$ hour is $-1.10\left(\mathrm{int}\left(-x\right)\right)$.

Therefore, the formula that gives the charge for $x$ hours with $0\le x\le 24$ is $\{\begin{array}{l}-1.10\left(\mathrm{int}\left(-x\right)\right),0\le x\le 6\\ 7.25,6<x\le 24\end{array}$.

(b)

To determine

To find: The graph of the function in part (a) and the points of discontinuity of function.

Answer to Problem 53E

The graph of the function is shown in Figure (1) and the points of discontinuity of function are $x=0$ , $1$ , $2$ , $3$ , $4$ , $5$ and $6$.

Explanation of Solution

Given information: The charge of a value park is $\$1.10$ per hour or fraction of an hour for airport parking. The maximum charge per day is $\$7.25$.

Calculation:

To graph a function $\{\begin{array}{l}-1.10\left(\mathrm{int}\left(-x\right)\right),0\le x\le 6\\ 7.25,6<x\le 24\end{array}$ , 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 $f\left(x\right)=-1.10\left(\mathrm{int}\left(-x\right)\right)$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\begin{array}{l}\boxed{(-)}\boxed{1.10}\boxed{(}\boxed{\text{MATH}}\boxed{\to }\boxed{5}\boxed{(-)}\boxed{\text{X}}\boxed{)}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{4}\boxed{0}\boxed{2\text{nd}}\\ \boxed{\text{MATH}}\boxed{\to }\boxed{1}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{6}\boxed{6}\boxed{)}\end{array}$.

Now, press the $\boxed{\text{ENTER}}$ button and enter right hand side of the equation $f\left(x\right)=7.25$ after the symbol ${\text{Y}}_2$ . Enter the keystrokes $\boxed{(}\boxed{7.25}\boxed{)}\boxed{÷}\boxed{(}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{3}\boxed{6}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{\to }\boxed{1}\boxed{\text{X}}\boxed{2\text{nd}}\boxed{\text{MATH}}\boxed{6}\boxed{24}\boxed{)}$.

The display will show the equations,

  $\begin{array}{l}{\text{Y}}_{\text{1}}=-1.10\left(\mathrm{int}\left(-\text{X}\right)\right)/\left(\text{X}\ge 0\text{and}\text{X}\le 6\right)\\ {\text{Y}}_2=\left(7.25\right)/\left(\text{X}>6\text{and}\text{X}\le 24\right)\end{array}$

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)

As observed from the graph, the function is constant and continuous for all the values of $x$ between $6$ and $24$.

Also it is observed from the graph that the function is step function for values of $x$ between $0$ and $6$ . Step function is always discontinuous at integer points.

So, all the integer points between $0$ and $6$ are the points of discontinuity for $f\left(x\right)$.

Therefore, the points of discontinuity of function are $x=0$ , $1$ , $2$ , $3$ , $4$ , $5$ and $6$.