Piecewise Functions - IB Mathematics Applications and Interpretation

1. A customer made a call that cost $8.34. How long was the call in minutes?
   Answer: 22 The cost $8.34 is greater than $0.99, so we must use the second piece of the function to find the duration $t$. Set $0.99 + 0.35(t-1) = 8.34$. Subtract 0.99 from both sides: $0.35(t-1) = 7.35$. Divide by 0.35: $t-1 = \frac{7.35}{0.35} = 21$. Solve for $t$: $t = 21 + 1 = 22$. So the call was 22 minutes long. A common mistake is to solve $0.99 + 0.35t = 8.34$, which gives $t \approx 20.9$, or to forget to add 1 at the end, giving 21 minutes (Distractor A).
2. Last week, the company made a profit of $185,000. How many units did they sell?
   Answer: 150 The profit is given in thousands of dollars, so a profit of $185,000 corresponds to $P=185$. We need to find $x$ such that $P(x) = 185$. Check the first piece: at $x=100$, $P(100) = -5 + 1.5(100) = -5 + 150 = 145$. Since $185 > 145$, the solution must be in the second interval. Set $145 + 0.8(x-100) = 185$. Subtract 145 from both sides: $0.8(x-100) = 40$. Divide by 0.8: $x-100 = \frac{40}{0.8} = 50$. Solve for $x$: $x = 150$. This value is in the domain $x > 100$, so it is the correct solution. The company sold 150 units.
3. Calculate the value of the stamp in the year 2025, rounded to the nearest dollar.
   Answer: $1553 The year 2025 is 15 years after 2010, so $t=15$. Since $t=15$ is greater than 10, we use the second piece of the function. First, calculate the value at $t=10$: $V(10) = 800(1.05)^{10} \approx 800 \times 1.62889 \approx 1303.12$. Now, use this in the second formula for $t=15$: $V(15) = 1303.12 + 50(15-10)$. This simplifies to $V(15) = 1303.12 + 50(5) = 1303.12 + 250 = 1553.12$. Rounding to the nearest dollar gives $1553. Distractor A is the value at t=10. Distractor C uses the wrong formula (continuing exponential growth): $800(1.05)^{15} \approx 1663$. Distractor D is an unrealistic overestimate.
4. Let $S$ be the total monthly sales in dollars. Which function correctly models the salesperson's total monthly earnings, $E(S)$?
   Answer: $E(S) = \begin{cases} 1500 + 0.02S & 0 \le S \le 20000 \\ 1900 + 0.05(S-20000) & S > 20000 \end{cases}$ When you encounter piecewise commission problems, you need to carefully track how each piece of the salary structure contributes to the total earnings. The key is understanding that higher commission rates typically apply only to sales above certain thresholds, not to all sales once you reach that level. Let's build this function step by step. For sales up to $20,000, the salesperson earns the base salary ($1,500) plus 2% commission on all sales: $E(S) = 1500 + 0.02S$. This part is consistent across all options. For sales above $20,000, the commission structure changes. The salesperson still gets 2% on the first $20,000 (which equals $400), plus 5% on only the amount exceeding $20,000. So the total becomes: base salary + commission on first $20,000 + commission on excess = $1500 + 400 + 0.05(S - 20000) = 1900 + 0.05(S - 20000)$ This matches option A exactly. Option B incorrectly applies the 5% rate to all sales when $S > 20000$, ignoring the tiered structure. Option C forgets to include the $400 commission earned on the first $20,000 when calculating earnings above that threshold. Option D omits the fixed salary entirely from the first piece and uses an incorrect base value of 400 in the second piece. Study tip: In tiered commission problems, always remember that higher rates apply only to the excess above each threshold, and you must carry forward all previously earned amounts. Test your piecewise function at the boundary point ($20,000 here) to ensure continuity.
5. What is the best interpretation of this model at $t=3$?
   Answer: An external factor caused the population growth rate to change from exponential to linear. When you encounter piecewise functions modeling real-world phenomena, focus on analyzing what happens at the transition points and what causes the mathematical behavior to change. Let's examine what occurs at $t = 3$. For $0 \le t \le 3$, the bacteria follow exponential growth: $P(t) = 500(2^t)$. At $t = 3$: $P(3) = 500(2^3) = 4000$ bacteria. For $t > 3$, growth becomes linear: $P(t) = 4000 + 1000(t-3)$. At $t = 3$: $P(3) = 4000 + 1000(0) = 4000$ bacteria. The function is continuous at $t = 3$ since both pieces give the same value, but the growth pattern fundamentally shifts from exponential to linear. This suggests an external intervention or environmental change affected the bacterial culture's growth mechanism. A is correct because it recognizes that some external factor (like nutrient depletion, space limitations, or temperature change) caused the fundamental growth pattern to change from exponential to linear. B is incomplete - while growth does become linear after 3 hours, this doesn't explain why the change occurred, missing the key insight about external factors. C misinterprets the mathematics. No bacteria are "added" at $t = 3$; the 1000 represents the linear growth rate (bacteria per hour), not an instantaneous addition. D is mathematically incorrect. The function is continuous at $t = 3$ since both pieces equal 4000, even though the growth rates differ. Study tip: In piecewise function problems, always check continuity at transition points and ask yourself what real-world factors would cause such mathematical changes in growth patterns.
6. Let $t$ be the time in hours a car is parked. Which of the following piecewise functions correctly models the parking cost, $C(t)$, for $t > 0$? (Assume $t$ is not necessarily an integer).
   Answer: $C(t) = \begin{cases} 8 & 0 < t \le 2 \\ 8 + 3.50(t-2) & t > 2 \end{cases}$ For the first 2 hours ($0 < t \le 2$), the cost is a flat $8. For any time longer than 2 hours ($t > 2$), the cost is the initial $8 plus $3.50 for each additional hour. The number of additional hours is $t-2$. Therefore, the cost for $t > 2$ is $8 + 3.50(t-2)$. This matches option C. Option A incorrectly charges for the total time $t$ in the second piece. Option B forgets to include the initial $8 fee in the second piece. Option D incorrectly models the first 2 hours as a linear function from 0.
7. A household used 920 kWh of electricity in one month. Calculate their bill.
   Answer: $150.60 Since the usage of 920 kWh is greater than 500 kWh, the second part of the piecewise function must be used. The cost is calculated as $C(920) = 75 + 0.18(920-500)$. This simplifies to $C(920) = 75 + 0.18(420)$. $0.18 \times 420 = 75.6$. So, the total cost is $75 + 75.6 = \$150.60$. Distractor A comes from incorrectly using the first formula: $15 + 0.12(920) = 15 + 110.40 = 125.40$ (not in options). Distractor C comes from miscalculating the base charge: $15 + 0.12(500) + 0.18(920-500) = 15+60+75.6 = 150.6$. Distractor C comes from applying the higher rate to all usage: $15 + 0.18(920) = 15 + 165.6 = 180.60$. Distractor D is a general calculation error.
8. What is the height of the rocket after 10 seconds?
   Answer: 176 m When you encounter a piecewise function, the critical first step is determining which piece to use based on the given input value. Since we need the height at $t = 10$ seconds, and $10 > 8$, we use the second piece of the function: $h(t) = 128 + 32(t-8) - 4(t-8)^2$. Substituting $t = 10$: $h(10) = 128 + 32(10-8) - 4(10-8)^2$ $h(10) = 128 + 32(2) - 4(2)^2$ $h(10) = 128 + 64 - 4(4)$ $h(10) = 128 + 64 - 16 = 176$ metres This confirms answer A is correct. Looking at the wrong answers: Answer B (128 m) represents the height exactly at $t = 8$ seconds, which you might get if you mistakenly used the boundary value instead of calculating for $t = 10$. Answer C (200 m) could result from calculation errors, such as forgetting the negative sign in the quadratic term or arithmetic mistakes. Answer D (256 m) might come from incorrectly using the first piece of the function ($2t^2 = 2(10)^2 = 200$) or from other computational errors involving the wrong formula. Study tip: With piecewise functions, always check which interval your input falls into before substituting. Write out each step of your calculation clearly, especially with nested operations like $(t-8)^2$. Double-check that you're using the correct piece of the function—this is where most errors occur on IB exams.
9. A customer paid $39.00 to ship a package. What was the weight of the package in kilograms?
   Answer: 8.00 We need to find which piece of the function yields a cost of $39.00. 1. The first piece gives a constant cost of $12.00, so this is not correct. 2. For the second piece, we set $12.00 + 4.50(w-1) = 39.00$. This gives $4.50(w-1) = 27.00$, so $w-1 = 6$, and $w=7$. However, this value $w=7$ is outside the domain for this piece ($1 < w \le 5$), so this is not the solution. 3. For the third piece, we set $30.00 + 3.00(w-5) = 39.00$. This gives $3.00(w-5) = 9.00$, so $w-5 = 3$, and $w=8$. This value $w=8$ is within the domain for this piece ($w > 5$). Therefore, the weight of the package was 8 kg.
10. What is the maximum depth of the water in the harbour between midnight ($t=0$) and noon ($t=12$)?
    Answer: 11.0 m We need to find the maximum value of the function $D(t)$ on the interval $[0, 12]$. For the first piece, $D(t) = 8 + 0.5t$ is a linear function with a positive slope, so its maximum value on $[0, 6]$ will occur at $t=6$. $D(6) = 8 + 0.5(6) = 8 + 3 = 11$. For the second piece, $D(t) = 11 - 0.2(t-6)$ is a linear function with a negative slope, so its maximum value on $(6, 12]$ will occur at the leftmost point, as $t$ approaches 6. The limit as $t \to 6^+$ is $11 - 0.2(6-6) = 11$. The value at the right endpoint is $D(12) = 11 - 0.2(12-6) = 11 - 0.2(6) = 11 - 1.2 = 9.8$. Comparing the maximum values from each part (11 m) and the endpoints ($D(0)=8$, $D(12)=9.8$), the overall maximum depth is 11.0 m.
11. At what time $t$, in seconds, is the particle's velocity equal to 9 m/s?
    Answer: 3 and 14 We need to solve $v(t) = 9$ for each piece of the function. 1. For $0 \le t \le 5$: $3t = 9 \implies t = 3$. This value is within the domain for this piece, so it is a valid solution. 2. For $5 < t \le 12$: $v(t) = 15$. There is no solution here as the velocity is constant at 15 m/s. 3. For $12 < t \le 17$: $51 - 3t = 9 \implies 42 = 3t \implies t = 14$. This value is within the domain for this piece, so it is also a valid solution. Therefore, the velocity is 9 m/s at $t=3$ and $t=14$ seconds.
12. A customer realizes that buying 10 cookies is cheaper than buying 9 cookies. What is the price difference between buying 10 cookies and buying 9 cookies?
    Answer: $2.50 First, calculate the cost of 9 cookies using the first piece of the function: $C(9) = 2.50 \times 9 = \$22.50$. Next, calculate the cost of 10 cookies using the second piece of the function: $C(10) = 2.00 \times 10 = \$20.00$. The price difference is $\$22.50 - \$20.00 = \$2.50$. This type of pricing model creates a 'price drop' discontinuity at the boundary, making a larger quantity cheaper.
13. The drone is considered to be at 'cruising altitude' when its height is exactly 80 metres. For how long, in minutes, is the drone at cruising altitude?
    Answer: 6 We need to find the duration for which $h(t) = 80$. 1. In the first interval, $20t = 80 \implies t=4$. So it reaches 80m at $t=4$. 2. In the second interval, $4 < t \le 10$, the height is constant at $h(t) = 80$. The duration of this interval is $10 - 4 = 6$ minutes. 3. In the third interval, $80 - 8(t-10) = 80 \implies -8(t-10)=0 \implies t=10$. This is the moment it starts descending from 80m. So the drone is at a constant height of 80 metres from $t=4$ to $t=10$. The total time is $10-4=6$ minutes.
14. A customer buys a ticket 3 weeks before the concert. What is the price of the ticket?
    Answer: $88 Three weeks before the concert is $3 \times 7 = 21$ days. So we need to calculate $P(21)$. The value $d=21$ falls into the second interval ($7 < d \le 30$). We use the formula for this interval: $P(21) = 60 + 2(21-7)$. This simplifies to $P(21) = 60 + 2(14) = 60 + 28 = \$88$. Distractor A uses the first interval. Distractor D uses the third interval. Distractor C is a calculation error, perhaps $60 + 2 \times 21 = 102$.
15. According to the model, when does the population first drop to 1400 fish?
    Answer: 15 years We need to solve $N(t) = 1400$. First, check the first piece: $1200 + 100t = 1400 \implies 100t = 200 \implies t=2$. At $t=2$ years, the population reaches 1400. However, the question asks when it drops to 1400. This implies we are looking for a solution on a downward trend. The first piece shows population increasing. Let's check the second piece, which shows a decreasing population: $2000 - 60(t-5) = 1400$. Subtract 2000 from both sides: $-60(t-5) = -600$. Divide by -60: $t-5 = 10$. Solve for $t$: $t = 15$. This value is in the domain $t \ge 5$, so it is a valid solution. The population drops to 1400 after 15 years.

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