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As a summer activity, Tearrik participates in charity events for his community. He is volunteering for a food drive event this weekend.

He went to a shopping center multiple times over the week to collect boxes of food for the food drive. On each day, he collected the following number of boxes.

• $5$ boxes on Monday
• $3$ boxes on Tuesday
• $2$ boxes on Wednesday
• $3$ boxes on Thursday
• $4$ boxes on Friday

Zain is working as a server in a restaurant for a week before their summer vacation ends.

Most of their payment comes from the tips they receive. Zain made note of how much they received in tips, starting from Monday and through their last day working on Saturday.

• $\$49.50$ on Monday
• $\$38.17$ on Tuesday
• $\$41.45$ on Wednesday
• $\$58.33$ on Thursday
• $\$60.55$ on Friday
• $\$57.00$ on Saturday

Answer

a $\begin{array}{c}f(x)=\{\begin{array}{ll}0,& \text{if }0\le x<1\\ 49.50,& \text{if }1\le x<2\\ 87.67,& \text{if }2\le x<3\\ 129.12,& \text{if }3\le x<4\\ 187.45,& \text{if }4\le x<5\\ 248.00,& \text{if }5\le x<6\\ 305.00,& \text{if }6\le x<7\end{array}\end{array}$

b

Solution

a The first thing to consider when writing a function is the domain. The domain of this function is defined as the days of the week, starting from the Sunday before Zain's week of work. This Sunday will be marked as $x=0.$ Each day of the week will correspond to the next consecutive integer.

$$\begin{array}{c}f(x)=0,\text{if }0\le x<1\end{array}$$

This indicates that Zain received $\$0$ in tips on Sunday before they start working. On Monday $x=1,$ Zain received $\$49.50$ in tips. This defines the next piece of the function.

$$\begin{array}{c}f(x)=49.50,\text{if }1\le x<2\end{array}$$

Since the function reflects the total amount of money Zain receives in tips, the value of each of the pieces can be determined by adding the new tip value to the total tip amount from the day before. Using this information, the pieces can be written in a table.

Day	Tips Received	Total Tips	$x$	$f(x)$
Sunday	-	$0$	$0\le x<1$	$0$
Monday	$49.50$	$0+49.50=49.50$	$1\le x<2$	$49.50$
Tuesday	$38.17$	$49.50+38.17=87.67$	$2\le x<3$	$87.67$
Wednesday	$41.45$	$87.67+41.45=129.12$	$3\le x<4$	$129.12$
Thursday	$58.33$	$129.12+58.33=187.45$	$4\le x<5$	$187.45$
Friday	$60.55$	$187.45+60.55=248.00$	$5\le x<6$	$248.00$
Saturday	$57.00$	$248.00+57.00=305.00$	$6\le x<7$	$305.00$

Finally, the step function can be written based on the table.

$$\begin{array}{c}f(x)=\{\begin{array}{ll}0,& \text{if }0\le x<1\\ 49.50,& \text{if }1\le x<2\\ 87.67,& \text{if }2\le x<3\\ 129.12,& \text{if }3\le x<4\\ 187.45,& \text{if }4\le x<5\\ 248.00,& \text{if }5\le x<6\\ 305.00,& \text{if }6\le x<7\end{array}\end{array}$$

b To graph a step function, each piece should be graphed separately on the same coordinate plane. The first section is a horizontal line at $0,$ limited to $x=0$ and $x=1.$ The first circle will be closed and the second will be open.

The next piece is the horizontal segment at $49.50$ with a closed endpoint at $x=1$ and an open endpoint at $x=2.$

The remaining pieces can be added to the graph by following the same process.

Dominika is going to a movie at a local theater on her last day of vacation.

The cost to park in the theater lot is $\$10$ for less than an hour. An additional $\$2.50$ is charged for each hour of parking.

Answer

a

$f(x)=2.5\lfloor x\rfloor +10$

b

Solution

a The first thing to notice to write the function is that $\$2.50$ is charged for each hour of parking. This means that as soon as one hour is reached, $\$2.50$ is added to the total cost of parking. At the two hour mark, the next $\$2.50$ is charged. This can be written by multiplying $\lfloor x\rfloor$ by $2.5,$ where $x$ is the parking time in hours.

$$\begin{array}{c}{f}_1(x)=2.5\lfloor x\rfloor \end{array}$$

Additionally, the lot requires an initial payment of $\$10$ for the first $59$ minutes of parking. This information can be used to add $10$ to the value of the obtained function $f_1.$

$$\begin{array}{c}f(x)=2.5\lfloor x\rfloor +10\end{array}$$

This completes the required function, as the price to enter is considered and $\$2.50$ is added as each hour is reached.

b This function can be graphed by modifying the graph of the greatest integer function. Since only the positive values of $x$ are relevant, the graph will be considered only in the first quadrant. This graph is shown as follows.

Next, since the greatest integer function is multiplied by $2.5,$ each value of $y$ is multiplied by $2.5,$ vertically stretching the spaces between each horizontal segment.

Then, since $10$ is added to the product, each segment is translated vertically $10$ units up.

Finally, the scope of the coordinate plane will be adjusted so that more steps of the graph can be seen.

During his summer vacation, Ignacio went to private math lessons. After learning about the greatest integer function, Ignacio was asked by his math tutor to graph the numbers $y$ greater than or equal to $\lfloor x\rfloor .$

Answer

Solution

The first step to graph an inequality is to graph the border function. The border function of the given inequality is given by the greatest integer function. This function is a step function whose output is the greatest integer less than or equal to the input $x.$ Note that the inequality is non-strict, so the horizontal lines are drawn as solid lines.

Now a random point can be used to test which region satisfies the inequality. Any point that does not lie on the border function can be chosen. For this example, $(2,0)$ will be used.

$y\ge \lfloor x\rfloor$

SubstituteII

$x=2$, $y=0$

$0\stackrel{?}{\ge }\lfloor 2\rfloor$

$\lfloor 2\rfloor =2$

$0\ngeqq 2$

Since the point does not satisfy the inequality, the region on the other side of the border function is solution set of the inequality.

Finally, it is important to verify if the borders of the vertical segments between each horizontal segment are solid or dashed, to avoid confusion whether the points that lie on this line satisfy the inequality or not. To verify this, as with the initial solution region, a test point will be selected and substituted into the inequality. In this case, $(1,0.5)$ will be used.

$y\ge \lfloor x\rfloor$

SubstituteII

$x=1$, $y=0.5$

$0.5\stackrel{?}{\ge }\lfloor 1\rfloor$

$\lfloor 1\rfloor =1$

$0.5\ngeqq 1$

Since the point does not satisfy the inequality, the vertical segments should be dashed.

Finally, the following is the complete graph of the inequality that the teacher asked for.

At the beginning of this lesson, it was asked that a function for the number of soda cans in Mark's refrigerator be written. Mark starts the week with $11$ cans in the fridge and then does the following.

• On Monday, Mark drank $1$ can.
• On Wednesday, Mark bought $2$ cans at a convenience store and put them into the fridge.
• On Friday, Mark had a party and a total of $10$ cans were consumed.
• On Saturday, Mark bought a $12\text{-}$pack of soda from a supermarket and put them away in the refrigerator.

Answer

a $\begin{array}{c}f(x)=\{\begin{array}{ll}11,& \text{if }0\le x<1\\ 10,& \text{if }1\le x<3\\ 12,& \text{if }3\le x<5\\ 2,& \text{if }5\le x<6\\ 14,& \text{if }6\le x<7\end{array}\end{array}$

b

Solution

a The first thing to do when writing a function is to determine its domain. In this case, the domain can be written as the number of days passed from the beginning of the week, starting from Sunday as $x=0.$ With the given information, it is possible to write the first part of the piecewise function.

$$\begin{array}{c}f(x)=11,\text{if }0\le x<1\end{array}$$

Mark drank one can of soda on Monday. Therefore, he has now $11-1=10$ cans left in the fridge that day. Also, since he neither drank nor bought any soda on Tuesday, the function's output is still $10$ until Wednesday, when $x=3.$

$$\begin{array}{c}f(x)=10,\text{if }1\le x<3\end{array}$$

Mark bought $2$ cans on Wednesday, meaning that there are $10+2=12$ cans of soda in the fridge that day. Since he neither bought nor drank any of the soda in the refrigerator on Thursday, he has $12$ cans until Friday.

$$\begin{array}{c}f(x)=12,\text{if }3\le x<5\end{array}$$

At Friday's party, a total of $10$ cans were drunk. Therefore, he has $12-10=2$ cans left in the refrigerator for Saturday. Then, on Saturday, he bought a $12\text{-}$pack of soda cans, making it for $2+12=14$ cans in the refrigerator for that day. Because no more information was given, the whole function rule can now be written.

$$\begin{array}{c}f(x)=\{\begin{array}{ll}11,& \text{if }0\le x<1\\ 10,& \text{if }1\le x<3\\ 12,& \text{if }3\le x<5\\ 2,& \text{if }5\le x<6\\ 14,& \text{if }6\le x<7\end{array}\end{array}$$

b Since every part of this function is a constant value, this is a step function. The graph of a step function consists of horizontal lines limited to specific inputs. It is important to remember the values for which the inequality for $x$ is strict or non-strict.