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# Solved Examples on Piecewise Functions For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on the questions you solve in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions
There is no negative penalty for a wrong answer.

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(1.) ACT The function $f(x)$ is defined below. What is the value of $f(-1)$?

$f(x) = \begin{cases} |x| + 1, & \quad \text{if }\: x \: \le -1 \\[3ex] |x| - 2, & \quad \text{if }\: x \gt -1 \end{cases} \\[10ex] F.\;\; -3 \\[3ex] G.\;\; -1 \\[3ex] H.\;\; 0 \\[3ex] J.\;\; 1 \\[3ex] K.\;\; 2 \\[3ex]$

$x = -1 \\[3ex] First\:\: piece \\[3ex] f(x) = |x| + 1 \\[3ex] f(-1) = |-1| + 1 \\[3ex] f(-1) = 1 + 1 \\[3ex] f(-1) = 2$
(2.) ACT In the standard $(x, y)$ coordinate plane, what is the $y-intercept$ of the graph of the function $y = f(x)$ defined below?

$f(x) = \begin{cases} x^2 - 1 & \quad \text{for}\: x \: \lt -3 \\[3ex] 2x - 5 & \quad \text{for}\: -3 \le x \le 2 \\[3ex] |x - 3| & \quad \text{for}\; x \gt 2 \end{cases} \\[10ex] A.\;\; -5 \\[3ex] B.\;\; -3 \\[3ex] C.\;\; -1 \\[3ex] D.\;\; 2.5 \\[3ex] E.\;\; 3 \\[3ex]$

$At\;\;y-intercept;\;\; x = 0 \\[3ex] This\;\;falls\;\;in\;\;the\;\;second\;\;piece \\[3ex] f(x) = 2x - 5 \\[3ex] f(0) = 2(0) - 5 \\[3ex] f(0) = 0 - 5 \\[3ex] f(0) = -5$
(3.) ACT The DigiPhone Company advertises the following calling plans:

DigiPhone Calling Plans
600 minutes* for $89.99** per month 1,000 minutes* for$119.99** per month
1,400 minutes * for $149.99** per month *The charge for each additional minute is$0.20**
**Taxes are NOT included.

Luis chose the $149.99 plan. There are certain months when he uses more than 1,400 minutes of calling time. For a month in which he uses x minutes of calling time, where x is an integer greater than 1,400, which of the following expressions gives the amount, in dollars, of his bill before any taxes are added?$ F.\;\; 0.20x \\[3ex] G.\;\; 149.99 + 0.20x \\[3ex] H.\;\; 149.99 + 0.20(1,400 - x) \\[3ex] J.\;\; 149.99 + 0.20(1,400 + x) \\[3ex] K.\;\; 149.99 + 0.20(x - 1,400) \\[3ex]  Total\;\;time = x\;minutes \\[3ex] \underline{3rd\;\;Basic\;\;Plan} \\[3ex] charge = \$149.99 \\[3ex] time = 1400\;minutes \\[3ex] \underline{Extra} \\[3ex] charge\;\;per\;\;minute = \$0.20 \\[3ex] time = x - 1400\;minutes \\[3ex] charge = 0.20(x - 1400) \\[3ex] \underline{Total\;\;charge\;\;before\;\;taxes} \\[3ex] Total\;\;charge = Basic\;\;charge + Extra\;\;charge \\[3ex] = 149.99 + 0.20(x - 1400) $(4.) ACT Given the function below, what is$f(4)$?$ f(x) = \begin{cases} 2x + 1; & \quad \;x\: \lt 4 \\[3ex] -\dfrac{1}{2}x - 3; & \quad \;x\: \ge 4 \end{cases} \\[10ex] F.\;\; -5 \\[3ex] G.\;\; \dfrac{1}{2} \\[5ex] H.\;\; 4 \\[3ex] J.\;\; 9 \\[3ex] K.\;\; 14 \\[3ex]  4\;\;falls\;\;in\;\;the\;\;second\;\;piece \\[3ex] f(x) = -\dfrac{1}{2}x - 3 \\[5ex] f(4) = -\dfrac{1}{2}(4) - 3 \\[5ex] f(4) = -1(2) - 3 \\[3ex] f(4) = -2 - 3 \\[3ex] f(4) = -5 $(5.) ACT The table below shows an electric utility company's old and new rates. In the table,$kWh$stands for kilowatt-hour, a standard unit for measuring electrical energy.  Reunion Electric Association Rates Old rate New rate Monthly service charge$\$7.00$ $\$11.00$Energy-use charge first$1,500$kWh$6.6¢/kWh6.7¢/kWhkWh$over$1,5006.2¢/kWh6.2¢/kWh$LaTasha is a computer programmer for the electric company. She updated the program for calculating customers' monthly bills. Which of the following is an expression that uses the new rates to calculate the bill, in dollars, of a customer who uses$x\: kWh$of electricity during the month, where$x \gt 1,500$?$ F.\:\: 7 + 0.062(1,500) + 0.067(x - 1,500) \\[3ex] G.\:\: 7 + 0.066(1,500) + 0.062(x - 1,500) \\[3ex] H.\:\: 11 + 0.062x \\[3ex] J.\:\: 11 + 0.062(1,500) + 0.067(x - 1,500) \\[3ex] K.\:\: 11 + 0.067(1,500) + 0.062(x - 1,500) \\[3ex]  x \gt 1500 \\[3ex] \underline{New\:\: rate} \\[3ex] Monthly\:\: service\:\: charge = 11.00 \\[3ex] First\:\: 1500 \\[3ex] 6.7¢/kWh = \$0.067/kWh \\[3ex] Energy-use\:\: charge = 1500 @ 6.7¢/kWh = 1500(0.067) \\[3ex] Over\:\: 1,500 \\[3ex] Remaining:\:\: (x - 1500) \\[3ex] 6.2¢/kWh = \$0.062/kWh \\[3ex] Energy-use\:\: charge = (x - 1500) @ 6.2¢/kWh = (x - 1500)(0.062) \\[3ex] Total\:\: charge = 11 + 1500(0.067) + (x - 1500)(0.062) \\[3ex] Total\:\: charge = 11 + 0.067(1500) + 0.062(x - 1500) \\[3ex] $Please stop here because the ACT is a timed exam. The answer is Option$K$But, I want to simplify completely.$ Total\:\: charge = 11 + 100.5 + 0.062x - 93 \\[3ex] Total\:\: charge = 0.062x + 18.5 $(6.) ACT The cheerleading squad wants to purchase new uniforms to wear at the regional championship competition. They decide to sell candy bars for the$1.00 each
The squad will receive $0.40 for each of the first 200 candy bars sold. For each of the next 300 sold, the squad will receive$0.50
For each additional candy bar sold, the squad will receive $0.60 How many candy bars must the squad sell to reach their goal of rasing$350.00?

$F.\;\; 350 \\[3ex] G.\;\; 584 \\[3ex] H.\;\; 667 \\[3ex] J.\;\; 700 \\[3ex] K.\;\; 875 \\[3ex]$

$\underline{First\;\;Piece} \\[3ex] Maximum\;\;amount = 0.4(200) = 80 \\[3ex] \underline{Second\;\;Piece} \\[3ex] Maximum\;\;amount = 0.5(300) = 150 \\[3ex] \$80 + \$150 = \$230 \\[3ex] 230 \lt 350 \\[3ex] So,;\;move\;\;to\;\;the\;\;third\;\;piece \\[3ex] \underline{Third\;\;Piece} \\[3ex] Let\;\;x = number\;\;of\;\;candy\;\;bars \\[3ex] \implies \\[3ex] 230 + 0.6(x - 500) = 350 \\[3ex] 230 + 0.6x - 300 = 350 \\[3ex] 0.6x - 70 = 350 \\[3ex] 0.6x = 350 + 70 \\[3ex] 0.6x = 420 \\[3ex] x = \dfrac{420}{0.6} \\[5ex] x = 700 \\[3ex] $The squad must sell 700 candy bars to reach their goal of raising$350.00

$\underline{Check} \\[3ex] \underline{Third\;\;Piece} \\[3ex] Maximum\;\;amount = 0.6(700 - 500) = 0.6(200) = 120 \\[3ex] \underline{All\;\;Pieces} \\[3ex] Maximum\;\;amount = 230 + 120 = \$350 $(7.) (8.) ACT A function P is defined as follows: for$x \gt 0$,$P(x) = x^5 + x^4 - 36x - 36$for$x \lt 0$,$P(x) = -x^5 + x^4 + 36x - 36$What is the value of$P(-1)$?$ A.\;\; -70 \\[3ex] B.\;\; -36 \\[3ex] C.\;\; 0 \\[3ex] D.\;\; 36 \\[3ex] E.\;\; 70 \\[3ex]  -1 \lt 0 ...2nd\;\;Piece \\[3ex] P(x) = -x^5 + x^4 + 36x - 36 \\[3ex] P(-1) = -(-1)^5 + (-1)^4 + 36(-1) - 36 \\[3ex] P(-1) = -(-1) + 1 - 36 - 36 \\[3ex] P(-1) = 1 + 1 - 36 - 36 \\[3ex] P(-1) = -70 \$
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