Piecewise Functions

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Piecewise Functions · Grade 12 · Worksheet 3

1. Liam is designing a custom skateboard ramp that has a piecewise function for its cross-sectional height. The height function h(x) in meters is defined as: h(x) = { 0.5x + 1 for 0 ≤ x ≤ 2; -x² + 5x - 2 for 2 < x ≤ 4; 2 for 4 < x ≤ 6 }. Liam needs to determine if the ramp has any discontinuities in its height profile. At what x-value does a discontinuity occur, and what type of discontinuity is it? Answer: x = 4, jump discontinuity Solution: Piecewise functions can have different types of discontinuities where the function pieces meet. A jump discontinuity occurs when the left-hand limit and right-hand limit at a point both exist but are not equal.
   Full step-by-step solution

   Piecewise functions can have different types of discontinuities where the function pieces meet. A jump discontinuity occurs when the left-hand limit and right-hand limit at a point both exist but are not equal. To analyze continuity, we examine the function value and limits at the boundaries between different pieces of the function definition.

2. Liam is modeling the temperature in a chemical reaction chamber using a piecewise function. The temperature T(t) in degrees Celsius at time t minutes is defined as: T(t) = { 2t^2 - 8t + 10 for 0 ≤ t < 3; -t^2 + 10t - 11 for 3 ≤ t ≤ 7 }. Liam needs to determine the exact time when the temperature reaches its minimum value during the first 7 minutes of the reaction. At what time does this minimum temperature occur? Answer: 2 Solution: T(t) = 2t^2 - 8t + 10, for 0 ≤ t < 3 -t^2 + 10t - 11, for 3 ≤ t ≤ 7 For the first piece: T(t) = 2t^2 - 8t + 10 Derivative: T'(t) = 4t - 8 Set T'(t) = 0 → 4t - 8 = 0 → t = 2 Check: t = 2 is in [0, 3), so it's valid.
   Full step-by-step solution

   Let's solve step by step. We have the piecewise function: T(t) = 2t^2 - 8t + 10, for 0 ≤ t < 3 -t^2 + 10t - 11, for 3 ≤ t ≤ 7 --- **Step 1: Find critical points in each interval** For the first piece: T(t) = 2t^2 - 8t + 10 Derivative: T'(t) = 4t - 8 Set T'(t) = 0 → 4t - 8 = 0 → t = 2 Check: t = 2 is in [0, 3), so it's valid. For the second piece: T(t) = -t^2 + 10t - 11 Derivative: T'(t) = -2t + 10 Set T'(t) = 0 → -2t + 10 = 0 → t = 5 Check: t = 5 is in [3, 7], so it's valid. --- **Step 2: Evaluate T(t) at critical points and endpoints** First piece: At t = 0: T(0) = 2(0)^2 - 8(0) + 10 = 10 At t = 2: T(2) = 2(4) - 8(2) + 10 = 8 - 16 + 10 = 2 At t = 3 (use first piece since it's defined up to but not including 3, but we can check limit): T(3) from first piece = 2(9) - 8(3) + 10 = 18 - 24 + 10 = 4 Second piece: At t = 3: T(3) = -(9) + 10(3) - 11 = -9 + 30 - 11 = 10 At t = 5: T(5) = -(25) + 10(5) - 11 = -25 + 50 - 11 = 14 At t = 7: T(7) = -(49) + 10(7) - 11 = -49 + 70 - 11 = 10 --- **Step 3: Compare values** From first piece: t = 0 → 10, t = 2 → 2, t → 3⁻ → 4 From second piece: t = 3 → 10, t = 5 → 14, t = 7 → 10 The smallest value among these is 2 at t = 2. --- **Step 4: Check continuity at t = 3** At t = 3: First piece limit = 4 Second piece value = 10 So there's a jump, but the minimum is clearly at t = 2 with T = 2. --- **Conclusion:** The minimum temperature occurs at t = 2 minutes.

3. Charlotte is a financial analyst evaluating a special tax structure for a new investment fund. The fund's effective tax rate T(x) as a percentage of profit (in thousands of dollars) is defined by the piecewise function: T(x) = { 0.12x + 2 for 0 ≤ x < 7; 0.27x - 0.73 for 7 ≤ x ≤ 12 }, where x is the profit in thousands of dollars. Charlotte needs to determine the effective tax rate for profits of $2,000, $7,000, and $12,000. She also needs to identify any discontinuity in the tax rate at the boundary x = 7. What are T(2), T(7), and T(12), and is there a jump discontinuity at x = 7? Answer: T(2) = 2.24%, T(7) = 1.16% (from first piece) and 1.16% (from second piece), T(12) = 2.51%; no jump discontinuity at x = 7 because both pieces give the same value Solution: Evaluate T(2). Since 0 ≤ 2 < 7, use first piece: T(2) = 0.12(2) + 2 = 0.24 + 2 = 2.24%. So T(2) = 2.24%.
   Full step-by-step solution

   Step 1: Evaluate T(2). Since 0 ≤ 2 < 7, use first piece: T(2) = 0.12(2) + 2 = 0.24 + 2 = 2.24%. So T(2) = 2.24%. Step 2: Evaluate T(7). The first piece is defined for 0 ≤ x < 7, so x = 7 is NOT included in the first piece. The second piece is defined for 7 ≤ x ≤ 12, so x = 7 is included in the second piece. Using second piece: T(7) = 0.27(7) - 0.73 = 1.89 - 0.73 = 1.16%. Step 3: Evaluate T(12). Since 7 ≤ 12 ≤ 12, use second piece: T(12) = 0.27(12) - 0.73 = 3.24 - 0.73 = 2.51%. Step 4: Check for jump discontinuity at x = 7. Compute the left-hand limit as x approaches 7 from below (using first piece): lim_{x→7^-} T(x) = 0.12(7) + 2 = 0.84 + 2 = 2.84%. The actual value at x = 7 is T(7) = 1.16% from the second piece. Since the left-hand limit (2.84%) does not equal the function value (1.16%), there IS a jump discontinuity at x = 7. The jump size is |2.84 - 1.16| = 1.68%. Final answer: T(2) = 2.24%, T(7) = 1.16%, T(12) = 2.51%; yes, there is a jump discontinuity at x = 7.

4. Liam is designing a custom skateboard ramp that has a piecewise function for its cross-sectional profile. The ramp's height h(x) in meters at a horizontal distance x meters from the start is defined as: h(x) = { 0.2x² for 0 ≤ x ≤ 3; 1.8 + 0.4(x-3) for 3 < x ≤ 6; 3.0 - 0.1(x-6)² for 6 < x ≤ 9 }. Liam needs to determine the exact horizontal position where the ramp reaches its maximum height. At what x-value does the maximum height occur? Answer: 6 Solution: h(x) = 0.2x² for 0 ≤ x ≤ 3 1.8 + 0.4(x - 3) for 3 < x ≤ 6 3.0 - 0.1(x - 6)² for 6 < x ≤ 9 Examine the first piece (0 ≤ x ≤ 3) h(x) = 0.2x² This is a parabola opening upward, so it is increasing on [0, 3].
   Full step-by-step solution

   Let's go step-by-step. We have the piecewise function: h(x) = 0.2x² for 0 ≤ x ≤ 3 1.8 + 0.4(x - 3) for 3 < x ≤ 6 3.0 - 0.1(x - 6)² for 6 < x ≤ 9 --- **Step 1: Examine the first piece (0 ≤ x ≤ 3)** h(x) = 0.2x² This is a parabola opening upward, so it is increasing on [0, 3]. At x = 3: h(3) = 0.2 × 9 = 1.8. --- **Step 2: Examine the second piece (3 < x ≤ 6)** h(x) = 1.8 + 0.4(x - 3) This is a straight line with slope 0.4 > 0, so it is increasing on (3, 6]. At x = 6: h(6) = 1.8 + 0.4(3) = 1.8 + 1.2 = 3.0. --- **Step 3: Examine the third piece (6 < x ≤ 9)** h(x) = 3.0 - 0.1(x - 6)² This is a parabola opening downward (since -0.1 < 0), with vertex at x = 6. At x = 6: h(6) = 3.0 - 0.1(0)² = 3.0. For x > 6: h(x) = 3.0 - 0.1(positive) < 3.0. --- **Step 4: Compare values at transition points and within intervals** From piece 1: max at x=3 is 1.8. From piece 2: max at x=6 is 3.0. From piece 3: max at x=6 is 3.0, then decreases for x > 6. So the maximum height is 3.0 at x = 6. --- **Step 5: Check continuity and confirm maximum** At x = 6: From piece 2: h(6) = 3.0 From piece 3: h(6) = 3.0 So continuous at x = 6. For x < 6: h(x) < 3.0 (since pieces 1 and 2 are increasing up to 3.0). For x > 6: h(x) < 3.0 (since piece 3 is decreasing from 3.0). Thus, maximum height occurs at x = 6. --- **Final answer:** 6

5. Consider the piecewise function f(x) defined as: f(x) = 2x^3 - 3x^2 + 1 for x < 1, and f(x) = 4e^(x-1) - 2 for x ≥ 1. Determine the value of the derivative f'(1) if it exists, or state that it does not exist. Answer: 0 Solution: f(x) = 2x^3 - 3x^2 + 1, for x < 1 f(x) = 4e^(x-1) - 2, for x ≥ 1 We need f'(1). Since the rule changes at x = 1, we must check the left-hand derivative and right-hand derivative at x = 1.
   Full step-by-step solution

   Let's find f'(1) step by step. --- **Step 1: Understand the problem** We have a piecewise function: f(x) = 2x^3 - 3x^2 + 1, for x < 1 f(x) = 4e^(x-1) - 2, for x ≥ 1 We need f'(1). Since the rule changes at x = 1, we must check the left-hand derivative and right-hand derivative at x = 1. --- **Step 2: Compute the left-hand derivative at x = 1** For x < 1: f(x) = 2x^3 - 3x^2 + 1 Derivative: f'(x) = 6x^2 - 6x Left-hand derivative at x = 1 means the limit as x → 1⁻ of f'(x): f'(1⁻) = 6(1)^2 - 6(1) = 6 - 6 = 0 So left-hand derivative = 0. --- **Step 3: Compute the right-hand derivative at x = 1** For x ≥ 1: f(x) = 4e^(x-1) - 2 Derivative: f'(x) = 4e^(x-1) * 1 = 4e^(x-1) Right-hand derivative at x = 1 means the limit as x → 1⁺ of f'(x): f'(1⁺) = 4e^(1-1) = 4e^0 = 4 * 1 = 4 So right-hand derivative = 4. --- **Step 4: Compare left-hand and right-hand derivatives** Left-hand derivative = 0 Right-hand derivative = 4 Since 0 ≠ 4, the derivative f'(1) does not exist in the ordinary sense. --- **Step 5: Check for continuity at x = 1** Before concluding, let's check if the function is continuous at x = 1, because if it's not continuous, then it's definitely not differentiable. Left-hand limit as x → 1⁻: f(1⁻) = 2(1)^3 - 3(1)^2 + 1 = 2 - 3 + 1 = 0 Right-hand limit as x → 1⁺: f(1⁺) = 4e^(1-1) - 2 = 4(1) - 2 = 2 Since 0 ≠ 2, the function is not continuous at x = 1. --- **Step 6: Conclusion** Because the function is not continuous at x = 1, it cannot be differentiable there. Thus, f'(1) does not exist. --- **Final Answer:** f'(1) does not exist.

6. Mason is designing a custom suspension bridge. The height of the bridge's main cable above the deck, h(x) in meters, is modeled by the piecewise function: h(x) = { 0.1x² for 0 ≤ x < 10; 10 + 0.05(x-10)² for 10 ≤ x ≤ 30 }, where x is the horizontal distance in meters from the left tower. Graph this function and evaluate h(0), h(10), and h(20). Also, determine if the cable is continuous at x = 10. Answer: h(0)=0, h(10)=10, h(20)=15, continuous at x=10 Solution: Evaluate h(0). Since 0 is in the domain 0 ≤ x < 10, use the first piece: h(0) = 0.1(0)² = 0. Evaluate h(10).
   Full step-by-step solution

   Step 1: Evaluate h(0). Since 0 is in the domain 0 ≤ x < 10, use the first piece: h(0) = 0.1(0)² = 0. Step 2: Evaluate h(10). At x = 10, the second piece applies (10 ≤ x ≤ 30): h(10) = 10 + 0.05(10-10)² = 10 + 0 = 10. Step 3: Evaluate h(20). Since 20 is in the second piece: h(20) = 10 + 0.05(20-10)² = 10 + 0.05(100) = 10 + 5 = 15. Step 4: Check continuity at x = 10. Left-hand limit as x → 10⁻: from first piece, h(10) = 0.1(10)² = 10. Right-hand value at x = 10: h(10) = 10. Both equal 10, so the cable is continuous at x = 10. Final answer: h(0)=0, h(10)=10, h(20)=15, continuous at x=10.