Chapter 1.6, Problem 44E

a.

To determine

The graph of the model. The domain of the piecewise function for each part.

Answer to Problem 44E

Domain of .505x2−1.47x+6.3:[1,6]

Domain of −1.97x+26.3:(6,12]

Explanation of Solution

Given information:

The table shows the monthly revenue $y$ (in thousands of dollars) of a landscaping business for each month of the year $2013$ ,with $x=1$ representing January:

  

A mathematical model that represents these data is$f(x)=\{\begin{array}{l}-1.97x+2.63\\ 0.505x^2-1.47x+6.3\end{array}$

  $(a)$ Use a graphing utility to graph the model.

Calculation:

A mathematical model represents the data:

  $f(x)=\{\begin{array}{l}-1.97x+2.63\\ 0.505x^2-1.47x+6.3\end{array}$

From the given information, the model of the graph are:

  

Now calculate the domain by checking where each part of the graph begins and ends.

Domain of .505x2−1.47x+6.3:[1,6]

Domain of −1.97x+26.3:(6,12]

Thus, the domain of the graph are $.505x^2-1.47x+6.3:[1,6]$ and $-1.97x+26.3:(6,12]$ .

b.

To determine

The functions of the numbers.

Answer to Problem 44E

  $\$11.575$ and $\$4.63$ .

Explanation of Solution

Given information:

The table shows the monthly revenue $y$ (in thousands of dollars) of a landscaping business for each month of the year $2013$ , with $x=1$ representing January:

  

A mathematical model that represents these data is$f(x)=\{\begin{array}{l}-1.97x+2.63\\ 0.505x^2-1.47x+6.3\end{array}$

  $(b)$ Find

  $f(5)$ and $f(11)$ .

Calculation:

A mathematical model represents the data:

  $f(x)=\{\begin{array}{l}-1.97x+2.63\\ 0.505x^2-1.47x+6.3\end{array}$

To find the function of f(5) and $f(11)$ , plug the value $5$ and $11$ into the appropriate equations are as shown below:

  $\begin{array}{l}$ f(5)=.505{(5)}^2-1.47(5)+6.3=11.575$f(11)=-1.97(11)+26.3=4.63\end{array}$

Thus,

  $\$11.575$ thousand in revenue for May, and

  $\$4.63$ thousand in revenue for November.

c.

To determine

The values obtained from the model and compare the actual data values.

Answer to Problem 44E

Both pairs of values are very close.

Explanation of Solution

Given information:

The table shows the monthly revenue $y$ (in thousands of dollars) of a landscaping business for each month of the year $2013$ , with $x=1$ representing January:

  

A mathematical model that represents these data is$f(x)=\{\begin{array}{l}-1.97x+2.63\\ 0.505x^2-1.47x+6.3\end{array}$

  $(c)$ How the values do obtained from the model in part (a) compare with the actual data values.

Calculation:

A mathematical model represents the data:

  $f(x)=\{\begin{array}{l}-1.97x+2.63\\ 0.505x^2-1.47x+6.3\end{array}$

From the given information (a),

  $\$11.575$ thousand in revenue for May, and

  $\$4.63$ thousand in revenue for November.

For actual data,

  $\$11.5$ thousand in revenue for May, and

  $\$4.5$ thousand in revenue for November.

Both pairs of values are very close.