[tex]\begin{gathered} a)\mathbf{C(x)=5.50x+12}\text{ (}\forall x\in\lbrack0,30\rbrack) \\ b)(8,56), \\ c)(16,100) \\ d)0\leq x\leq30 \\ e)\mathbf{12\leq C(x)\leq177} \\ f)\: Check\: the\: graph\: below,\: please \end{gathered}[/tex]
3)
a) Since the local towing company charges $5.50 per mile, and add to that a reservation fee of $12 for a maximum of 30 miles we can write out the following equation:
[tex]\begin{gathered} \\ C(x)=5.50x+12,x\leq30 \end{gathered}[/tex]
b) To determine C(8) is to evaluate the function
[tex]\begin{gathered} C(8)=5.50(8)+12 \\ C(8)=56 \end{gathered}[/tex]
So for 8 miles, there's a cost of $56. We can write it out as (8,56)
c) Let's find out the number of miles that the company charges for $100:
[tex]\begin{gathered} 100=5.50x+12 \\ 100-12=5.50x+12-12 \\ 88=5.50x \\ \frac{88}{5.50}=\frac{5.50x}{5.50} \\ x=16 \end{gathered}[/tex]
So $100 dollars is the price of towing a car for 16 miles. We have (16,100)
As the ordered pair.
d) The practical Domain, is the Domain of the function that in this case will only allow values for x in this interval: [0, 30] Because that towing company has restricted its mileage up to 30 miles:
[tex]0\leq x\leq30[/tex]
e) the range, will take into account the restriction of the Domain
[tex]\begin{gathered} \mathbf{12\leq C(x)\leq177} \\ C(x)=5.50(0)+12=12 \\ C(x)=5.50(30)+12=177 \end{gathered}[/tex]
So if one person makes a reservation but in fact does not close the deal, at least the company makes $12. And, if a customer requests a 30 mile towing then the maximum profit will be $177
f) To construct a graph we need to write out a table for at least three values:
x | C(x) =5.5x +12
1 | 17.5
2 | 23
3 | 28.5
So the points we're going to locate are (1, 17.5), (2,23), and (3,28.5)
Since the slope is greater than 1, we can trace an increasing line passing through those points:
Note that in this graph there is not the practical Domain and Range, but rather the Domain and Range defined for the Real Set.
Hence, the answers are:
