Practice 1 Solutions

Match each graph below (numbered) with its correct piecewise function and intervals (letters). Problems 1–3 have two matches.

$f (x) = \{\begin{cases} - 2 \\ x^{2} - 4 \\ - \frac{3}{2} x + 6 \end{cases}$

$g (x) = \{\begin{cases} 2 \\ | x + 3 | \\ {(x - 2)}^{3} \end{cases}$

$if - 6 < x \leq - 4 if - 4 < x < 1 if 1 \leq x \leq 3$

$if - 4 \leq x < - 3 if - 3 < x < - 1 if - 1 \leq x \leq 2$

$if - 5 < x \leq - 2 if - 2 < x < 2 if 2 \leq x \leq 6$

$h (x) = \{\begin{cases} - x - 3 \\ x + 3 \\ x^{2} \end{cases}$

$if - 6 < x \leq - 4 if - 4 < x < 1 if 1 \leq x \leq 3 if 3 < x \leq 6$

B, G 1)

A, E 2)

Note

The intervals are partially correct for C, but it is missing the interval $(3, 6]$ .

F, D 3)

Graph the piecewise function.

4 $h (x) = \{\begin{cases} - \frac{1}{3} x & if x < - 3 \\ x & if - 3 \leq x \leq 3 \\ - 2 x + 9 & if x > 4 \end{cases}$

$b (x) = \{\begin{cases} {(x + 4)}^{2} + 1 & if x \leq - 2 \\ - | x + 1 | + 6 & if x > - 2 \end{cases}$

Find the domain and range of problem 5.

$Domain : \{x | x \in ℝ\} Range : \{y | y \in ℝ\}$

$f (x) = ⌊x + 3⌋ + 1$

$g (x) = ⌈- x⌉$

$h (x) = \{\begin{cases} \sqrt{- x} - 2 \\ x - 2 & if x > 0 \end{cases}$

Note

The equation $h (x) = \sqrt{- x} - 2$ has the domain $x \leq 0 .$ Therefore it is unnecessary to write the domain as a part of the piecewise function.

Find the domain and range of problem 9.

$Domain : \{x | x \in ℝ\} Range : \{y | y \in ℝ, y \geq - 2\}$

Write the absolute value function as a piecewise function.

$f (x) = \{\begin{cases} - 2 x + 1 & if x < 2 \\ - 3 & if x = 2 \\ 2 x - 7 & if x > 2 \end{cases}$

Note

The equation is $f (x) = 2 | x - 2 | - 3$

You may choose to use the definition of absolute value to find the two parts. $y = - 2 (x - 2) - 3 y = 2 (x - 2) - 3$

Use the information below to complete problems 12–15. 
The STEM club will host a catered dinner to celebrate their accomplishments. Tickets will be sold over a period of 45 days and are more expensive the closer it is to the event. Use the table to complete the following problems.

Dates tickets will be sold	Price of tickets
Day 0–14	$30
Day 15–30	$35
Day 31–45	$40

Write a piecewise function and draw a graph to represent the scenario.

$y = \{\begin{cases} 30 & if 0 \leq x < 15 \\ 35 & if 15 \leq x < 31 \\ 40 & if 31 \leq x \leq 45 \end{cases}$

Note

Q: Why does $y = 30$ not include 15?

A: On day 15, the price goes up to $35. The student has all day (up until the last second) of day 14 to pay $30.

Determine the cost for 5 students on the 16th day.

$15 < 16 < 30 5 ($ 35) = $ 175$

$175

Calculate the cost when 3 tickets were purchased on day 6, and then 2 additional tickets were purchased on the 32nd day.

$0 \leq 6 \leq 14 31 < 32 < 45 6 ($ 30) = $ 180 2 ($ 40) = $ 80 180 + 80 = $ 260$

$260

15 What is the cost for a student on the 46th day?

$46 > 45$

This is not possible because it is past the ticket sales time.

Use the information below to complete problems 16–21. 
Natalee left her house for a drive at 8:06 a.m. At 8:12 a.m., she was three miles from the house. At 8:20 a.m. she was 10 miles from the house. She then turned around. It took 10 minutes to drive home.

Graph the scenario.

This image has an empty alt attribute; its file name is image-21-1024x414.png

Note

Q: What is the dependent variable and unit?

A: Distance from home in miles 

Q: What is the independent variable and unit?

A: Time in minutes

Write the rate of change as a ratio in words.

$\frac{Miles}{Minutes}$

What is the rate of change from the start of the trip at 8:06 until 8:12 a.m.?

$(6, 0), (12, 3) m = \frac{3 - 0}{12 - 6} = \frac{1}{2}$

$\frac{1}{2}$

Calculate the rate of change between 8:20 and 8:30 a.m.. Explain your reasoning for the sign of the slope in your solution.

$m = \frac{10 - 0}{20 - 30} = \frac{10}{- 10} = - 1$

Natalee is returning home from 10 miles away. The distance from home at the end of her trip is 0. The distance decreases while time continues to increase.

How far did Natalee travel round trip?

20 miles

Note

The farthest distance from home was 10 miles. Then Natalee returned home.

$(10 + 10 = 20)$

What would Natalee’s speed be in miles per hour from 8:12 to 8:20 a.m.? (Hint: You need to convert miles per minute to miles per hour.)

$(12, 3), (20, 10) m = \frac{10 - 3}{20 - 12} = \frac{7 miles}{8 \min} \frac{7 miles}{8 \min} \cdot \frac{60 \min}{1 hour} = \frac{7 (60)}{(8)} = 52.5 mph$

52.5 mph