Piecewise Functions

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Answer Key & Explanations

Piecewise Functions · Grade 12 · Worksheet 2

1. Charlotte is a civil engineer designing a new highway overpass. The height of the overpass above the ground, in meters, as a function of horizontal distance x in meters from the start of the structure, is modeled by the piecewise function: h(x) = { 0.5x + 8 for 0 ≤ x < 12; 14 for 12 ≤ x < 20; -0.25x + 19 for 20 ≤ x ≤ 36 } Charlotte needs to verify the design specifications. She must determine the height of the overpass at the points x = 8 meters, x = 12 meters, and x = 28 meters. Additionally, she needs to identify any discontinuities in the height function and describe their type. Finally, she must graph the function for 0 ≤ x ≤ 36. What are the heights h(8), h(12), and h(28), and at what x-value(s) does a discontinuity occur? Answer: h(8) = 12, h(12) = 14, h(28) = 12, discontinuities at x = 12 and x = 20 Solution: Evaluate h(8). Since 8 is in the interval 0 ≤ x < 12, use the first piece: h(x) = 0.5x + 8. h(8) = 0.5(8) + 8 = 4 + 8 = 12.
   Full step-by-step solution

   Step 1: Evaluate h(8). Since 8 is in the interval 0 ≤ x < 12, use the first piece: h(x) = 0.5x + 8. h(8) = 0.5(8) + 8 = 4 + 8 = 12. Step 2: Evaluate h(12). Since 12 is in the interval 12 ≤ x < 20, use the second piece: h(x) = 14. h(12) = 14. Step 3: Evaluate h(28). Since 28 is in the interval 20 ≤ x ≤ 36, use the third piece: h(x) = -0.25x + 19. h(28) = -0.25(28) + 19 = -7 + 19 = 12. Step 4: Check for discontinuity at x = 12. The left-hand limit as x approaches 12 from the left uses the first piece: limit = 0.5(12) + 8 = 6 + 8 = 14. The right-hand value at x = 12 from the second piece is h(12) = 14. Since both are 14, the function is continuous at x = 12. There is NO discontinuity at x = 12. Step 5: Check for discontinuity at x = 20. The left-hand limit as x approaches 20 from the left uses the second piece: limit = 14. The right-hand value at x = 20 from the third piece is h(20) = -0.25(20) + 19 = -5 + 19 = 14. Since both are 14, the function is continuous at x = 20. There is NO discontinuity at x = 20. Step 6: Re-evaluate the problem's requirement. The function is actually continuous at both boundaries. However, the problem asks to identify any discontinuities. Upon closer inspection, the problem statement might have a typo or the student is meant to recognize the function is continuous. But per the design, there are no jump discontinuities. The correct interpretation is that the function is continuous at all interior boundaries. The answer is that there are no discontinuities. Final Answer: h(8) = 12, h(12) = 14, h(28) = 12. There are no discontinuities in the function; it is continuous for all x in the domain 0 ≤ x ≤ 36.

2. A pharmaceutical company is modeling the concentration of a new medication in a patient's bloodstream over time. The concentration function is piecewise defined as C(t) = { 2t for 0 ≤ t < 2; 4e^(-0.5(t-2)) for t ≥ 2 }, where t is in hours and C(t) is in mg/L. The medication becomes effective when the concentration first reaches 3 mg/L and remains above this level. Determine the exact time interval during which the medication is effective. Answer: (1.5, 2 + 2ln(4/3)) hours Solution: C(t) = 2t, for 0 ≤ t < 2 C(t) = 4 e^(-0.5(t-2)), for t ≥ 2 The medication becomes effective when concentration first reaches 3 mg/L and remains above 3 mg/L. We need the time interval where C(t) > 3.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the problem** We have a piecewise concentration function: C(t) = 2t, for 0 ≤ t < 2 C(t) = 4 e^(-0.5(t-2)), for t ≥ 2 The medication becomes effective when concentration first reaches 3 mg/L and remains above 3 mg/L. We need the time interval where C(t) > 3. --- **Step 2: Find when concentration first reaches 3 mg/L** For 0 ≤ t < 2, C(t) = 2t. Set 2t = 3 → t = 1.5 hours. Check: At t = 1.5, C(t) = 3. For t < 1.5, C(t) < 3. For t > 1.5 (but still less than 2), C(t) > 3. So the first time it reaches 3 is at t = 1.5. --- **Step 3: Check if C(t) stays above 3 until t = 2** At t = 2, using first piece: C(2) = 2*2 = 4 mg/L. So at t = 2, C = 4 > 3. --- **Step 4: For t ≥ 2, check when C(t) drops to 3** For t ≥ 2, C(t) = 4 e^(-0.5(t-2)). Set C(t) = 3: 4 e^(-0.5(t-2)) = 3 e^(-0.5(t-2)) = 3/4 Take natural logs: -0.5(t-2) = ln(3/4) t - 2 = -2 ln(3/4) Note: ln(3/4) = -ln(4/3), so: t - 2 = -2 [ -ln(4/3) ] = 2 ln(4/3) t = 2 + 2 ln(4/3) --- **Step 5: Interpret the interval** From t = 1.5 to t = 2 + 2 ln(4/3), C(t) > 3. At t = 1.5, C = 3 exactly (so medication becomes effective just after t = 1.5). At t = 2 + 2 ln(4/3), C = 3 exactly (so medication stops being effective after that time). Thus the exact time interval is: (1.5, 2 + 2 ln(4/3)) hours --- **Final answer:** (1.5, 2 + 2 ln(4/3)) hours

3. Liam is designing a custom skateboard ramp for a competition. The ramp's cross-section is modeled by the piecewise function f(x), where f(x) = 2x + 1 for 0 ≤ x < 2, and f(x) = -x² + 6x - 3 for 2 ≤ x ≤ 5, with x representing horizontal distance in meters and f(x) representing height in meters. At what horizontal distance does the ramp reach its maximum height, and what is that maximum height? Answer: At x = 3 meters, maximum height is 6 meters Solution: f(x) = 2x + 1 for 0 ≤ x < 2 f(x) = -x² + 6x - 3 for 2 ≤ x ≤ 5 We need to find the maximum height and where it occurs.
   Full step-by-step solution

   Let's solve this step by step. --- **Step 1: Understand the piecewise function** We have: f(x) = 2x + 1 for 0 ≤ x < 2 f(x) = -x² + 6x - 3 for 2 ≤ x ≤ 5 We need to find the maximum height and where it occurs. --- **Step 2: Check the first piece (0 ≤ x < 2)** f(x) = 2x + 1 is a straight line with slope 2, so it is increasing in this interval. At x = 2 (not included in this piece, but we can check the limit from the left): f(2) from first piece would be 2*2 + 1 = 5, but x = 2 is in second piece, so we use second piece for x = 2. So maximum in first piece occurs as x approaches 2 from the left, height approaches 5. --- **Step 3: Check the second piece (2 ≤ x ≤ 5)** f(x) = -x² + 6x - 3 This is a downward-opening parabola (because coefficient of x² is negative), so its maximum occurs at the vertex. Vertex x-coordinate for parabola ax² + bx + c is x = -b/(2a). Here a = -1, b = 6, c = -3. x = -b/(2a) = -6/(2*(-1)) = -6/(-2) = 3. So vertex at x = 3. --- **Step 4: Check if x = 3 is in the interval [2, 5]** Yes, x = 3 is in [2, 5]. --- **Step 5: Compute height at x = 3** f(3) = -(3)² + 6*3 - 3 = -9 + 18 - 3 = 18 - 12 = 6 meters. --- **Step 6: Check endpoints of second piece** At x = 2: f(2) = -(2)² + 6*2 - 3 = -4 + 12 - 3 = 5 meters. At x = 5: f(5) = -(25) + 30 - 3 = 2 meters. So in second piece, maximum is at x = 3, height = 6. --- **Step 7: Compare with first piece** First piece maximum near x = 2 is height 5, which is less than 6. So overall maximum height is 6 meters at x = 3 meters. --- **Final Answer:** At x = 3 meters, maximum height is 6 meters.

4. Matiu is an ecologist tracking the water level in a coastal wetland over a 12-hour tidal cycle. The water depth D(t) in meters at time t hours after midnight is modeled by the piecewise function: D(t) = { 2t + 4 for 0 ≤ t < 2; -t^2 + 8t + 4 for 2 ≤ t < 6; 8 for 6 ≤ t ≤ 12 }. Evaluate D(0), D(2), D(5), and D(10). Graph the function, showing all jumps or breaks clearly. At what t-values does the function have discontinuities, and what type are they? Answer: D(0)=4, D(2)=8, D(5)=19, D(10)=8. Discontinuities: jump at t=2 (value changes from 8 to 16), jump at t=6 (value changes from 16 to 8). Solution: Evaluate D(0). Since 0 is in [0,2), use first piece: D(0) = 2(0) + 4 = 4. Evaluate D(2).
   Full step-by-step solution

   Step 1: Evaluate D(0). Since 0 is in [0,2), use first piece: D(0) = 2(0) + 4 = 4. Step 2: Evaluate D(2). At t=2, the first piece applies for t<2 and second for t≥2, so use second piece: D(2) = -(2)^2 + 8(2) + 4 = -4 + 16 + 4 = 16. But note the first piece at t=2 would give 2(2)+4=8. So at t=2, the function value is 16 from the second piece. Step 3: Evaluate D(5). Since 5 is in [2,6), use second piece: D(5) = -(5)^2 + 8(5) + 4 = -25 + 40 + 4 = 19. Step 4: Evaluate D(10). Since 10 is in [6,12], use third piece: D(10) = 8. Step 5: Check discontinuity at t=2. Left-hand limit as t→2⁻ is 2(2)+4=8. Right-hand value is D(2)=16. Since 8≠16, there is a jump discontinuity at t=2. Step 6: Check discontinuity at t=6. Left-hand limit as t→6⁻ is -(6)^2+8(6)+4 = -36+48+4=16. Right-hand value is D(6)=8. Since 16≠8, there is a jump discontinuity at t=6. Step 7: The function is continuous on [0,2), [2,6), and [6,12]. Final answer: D(0)=4, D(2)=16, D(5)=19, D(10)=8. Jump discontinuities at t=2 and t=6.

5. Mason is designing a roller coaster track for a physics project. The height of the track above the ground, h(x), in meters, is modeled by the piecewise function: h(x) = { 2x^2 - 12x + 22 for 0 ≤ x < 2; -x + 7 for 2 ≤ x ≤ 7 }, where x is the horizontal distance from the start in meters. Evaluate the height at x = 0, x = 2, and x = 7. Then, identify the x-value(s) where there is a jump discontinuity, and calculate the size of the jump. Answer: h(0) = 22, h(2) = 5, h(7) = 0; jump discontinuity at x = 2 with a jump size of 4 meters. Solution: Evaluate h(0). Since 0 is in the domain 0 ≤ x < 2, use the first piece: h(0) = 2(0)^2 - 12(0) + 22 = 22. Evaluate h(2).
   Full step-by-step solution

   Step 1: Evaluate h(0). Since 0 is in the domain 0 ≤ x < 2, use the first piece: h(0) = 2(0)^2 - 12(0) + 22 = 22. Step 2: Evaluate h(2). Since x = 2 is in the domain 2 ≤ x ≤ 7, use the second piece: h(2) = -(2) + 7 = 5. Step 3: Evaluate h(7). Since 7 is in the domain 2 ≤ x ≤ 7, use the second piece: h(7) = -(7) + 7 = 0. Step 4: Check continuity at x = 2. The left-hand limit as x approaches 2 from below uses the first piece: limit = 2(2)^2 - 12(2) + 22 = 8 - 24 + 22 = 6. The right-hand value at x = 2 is h(2) = 5. Since 6 ≠ 5, there is a jump discontinuity at x = 2. The size of the jump is |6 - 5| = 1 meter. The final answer is: h(0) = 22, h(2) = 5, h(7) = 0; jump discontinuity at x = 2 with a jump size of 1 meter.