Piecewise Functions

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Answer Key & Explanations

Piecewise Functions · Grade 12 · Worksheet 1

1. f(x) = {5x + 10 if x < -5; -x^2 + 20 if -5 ≤ x < 0; 15 if x ≥ 0}. Evaluate f(-6), f(-5), f(-2), f(0), f(5) Answer: f(-6) = -20, f(-5) = -5, f(-2) = 16, f(0) = 15, f(5) = 15 Solution: Evaluate f(-6). Since -6 < -5, use the first piece: f(x) = 5x + 10. f(-6) = 5(-6) + 10 = -30 + 10 = -20.
   Full step-by-step solution

   Step 1: Evaluate f(-6). Since -6 < -5, use the first piece: f(x) = 5x + 10. f(-6) = 5(-6) + 10 = -30 + 10 = -20. Step 2: Evaluate f(-5). Since -5 ≤ -5 < 0, use the second piece: f(x) = -x^2 + 20. f(-5) = -(-5)^2 + 20 = -25 + 20 = -5. Step 3: Evaluate f(-2). Since -5 ≤ -2 < 0, use the second piece: f(x) = -x^2 + 20. f(-2) = -(-2)^2 + 20 = -4 + 20 = 16. Step 4: Evaluate f(0). Since 0 ≥ 0, use the third piece: f(x) = 15. f(0) = 15. Step 5: Evaluate f(5). Since 5 ≥ 0, use the third piece: f(x) = 15. f(5) = 15. Final answer: f(-6) = -20, f(-5) = -5, f(-2) = 16, f(0) = 15, f(5) = 15.

2. f(x) = {4x + 8 if x < -4; 2x^2 - 6 if -4 ≤ x < 2; 10 if x ≥ 2}. Evaluate f(-6), f(-4), f(0), f(2), f(6) Answer: f(-6) = -16, f(-4) = 26, f(0) = -6, f(2) = 10, f(6) = 10 Solution: Evaluate f(-6). Since -6 < -4, use the first piece: f(x) = 4x + 8. f(-6) = 4(-6) + 8 = -24 + 8 = -16.
   Full step-by-step solution

   Step 1: Evaluate f(-6). Since -6 < -4, use the first piece: f(x) = 4x + 8. f(-6) = 4(-6) + 8 = -24 + 8 = -16. Step 2: Evaluate f(-4). Since -4 ≤ -4 < 2, use the second piece: f(x) = 2x^2 - 6. f(-4) = 2(-4)^2 - 6 = 2(16) - 6 = 32 - 6 = 26. Step 3: Evaluate f(0). Since -4 ≤ 0 < 2, use the second piece: f(x) = 2x^2 - 6. f(0) = 2(0)^2 - 6 = 0 - 6 = -6. Step 4: Evaluate f(2). Since 2 ≥ 2, use the third piece: f(x) = 10. f(2) = 10. Step 5: Evaluate f(6). Since 6 ≥ 2, use the third piece: f(x) = 10. f(6) = 10. The final answers are: f(-6) = -16, f(-4) = 26, f(0) = -6, f(2) = 10, f(6) = 10.

3. f(x) = {2x + 7 if x < -2; 3x^2 - 2 if -2 ≤ x < 2; 7x - 2 if x ≥ 2}. Evaluate f(-3), f(-2), f(1), f(2), f(5). Answer: f(-3) = -1, f(-2) = 10, f(1) = 1, f(2) = 12, f(5) = 33 Solution: Evaluate f(-3). Since -3 < -2, use the first piece: f(x) = 2x + 7. f(-3) = 2(-3) + 7 = -6 + 7 = 1.
   Full step-by-step solution

   Step 1: Evaluate f(-3). Since -3 < -2, use the first piece: f(x) = 2x + 7. f(-3) = 2(-3) + 7 = -6 + 7 = 1. Step 2: Evaluate f(-2). Since -2 ≤ -2 < 2, use the second piece: f(x) = 3x^2 - 2. f(-2) = 3(-2)^2 - 2 = 3(4) - 2 = 12 - 2 = 10. Step 3: Evaluate f(1). Since -2 ≤ 1 < 2, use the second piece: f(x) = 3x^2 - 2. f(1) = 3(1)^2 - 2 = 3 - 2 = 1. Step 4: Evaluate f(2). Since 2 ≥ 2, use the third piece: f(x) = 7x - 2. f(2) = 7(2) - 2 = 14 - 2 = 12. Step 5: Evaluate f(5). Since 5 ≥ 2, use the third piece: f(x) = 7x - 2. f(5) = 7(5) - 2 = 35 - 2 = 33. The final answers are: f(-3) = 1, f(-2) = 10, f(1) = 1, f(2) = 12, f(5) = 33.

4. A piecewise function is defined as f(x) = { 3x - 2 for x < 1; x² + 1 for x ≥ 1 }. The graph consists of a line segment and a parabolic curve. At what x-coordinate do these two pieces intersect on the coordinate plane? Answer: 1 Solution: To find where the two pieces intersect, set the expressions equal to each other: 3x - 2 = x² + 1 Rearrange the equation: x² - 3x + 3 = 0 Solve using the quadratic formula: x = [3 ± sqrt(9 - 12)]/2 = [3 ± sqrt(-3)]/2 The discriminant is negative (-3), so there are no real solutions from the…
   Full step-by-step solution

   Step 1: To find where the two pieces intersect, set the expressions equal to each other: 3x - 2 = x² + 1 Step 2: Rearrange the equation: x² - 3x + 3 = 0 Step 3: Solve using the quadratic formula: x = [3 ± sqrt(9 - 12)]/2 = [3 ± sqrt(-3)]/2 Step 4: The discriminant is negative (-3), so there are no real solutions from the quadratic equation Step 5: Check the boundary point where the function definition changes at x = 1 Step 6: For x < 1: f(1) is not defined by 3x - 2, but we can find the limit as x approaches 1 from the left: 3(1) - 2 = 1 Step 7: For x ≥ 1: f(1) = (1)² + 1 = 2 Step 8: The y-values don't match (1 ≠ 2), so the pieces don't meet at x = 1 Step 9: Since there are no real solutions to the equation 3x - 2 = x² + 1, and the pieces don't connect at the boundary, these two curve segments do not intersect at any x-coordinate in the real plane. The correct answer is that there is no intersection point.

5. f(x) = {2x + 11 if x < -6; x^2 - 1 if -6 ≤ x < 1; 6 if x ≥ 1}. Evaluate f(-7), f(-6), f(0), f(1), f(6) Answer: f(-7) = -3, f(-6) = 35, f(0) = -1, f(1) = 6, f(6) = 6 Solution: Evaluate f(-7). Since -7 < -6, use the first piece: f(x) = 2x + 11. f(-7) = 2(-7) + 11 = -14 + 11 = -3.
   Full step-by-step solution

   Step 1: Evaluate f(-7). Since -7 < -6, use the first piece: f(x) = 2x + 11. f(-7) = 2(-7) + 11 = -14 + 11 = -3. Step 2: Evaluate f(-6). Since -6 ≤ -6 < 1, use the second piece: f(x) = x^2 - 1. f(-6) = (-6)^2 - 1 = 36 - 1 = 35. Step 3: Evaluate f(0). Since -6 ≤ 0 < 1, use the second piece: f(x) = x^2 - 1. f(0) = 0^2 - 1 = 0 - 1 = -1. Step 4: Evaluate f(1). Since 1 ≥ 1, use the third piece: f(x) = 6. f(1) = 6. Step 5: Evaluate f(6). Since 6 ≥ 1, use the third piece: f(x) = 6. f(6) = 6. Final answer: f(-7) = -3, f(-6) = 35, f(0) = -1, f(1) = 6, f(6) = 6.

6. f(x) = {x^2 - 7 if x < 3; 2x + 1 if 3 ≤ x < 8; 17 if x ≥ 8}. Evaluate f(2), f(3), f(7), f(9) Answer: f(2) = -3, f(3) = 7, f(7) = 15, f(9) = 17 Solution: Evaluate f(2) Since 2 < 3, use the first piece: f(x) = x^2 - 7 f(2) = (2)^2 - 7 = 4 - 7 = -3 Evaluate f(3) Since 3 ≤ 3 < 8, use the second piece: f(x) = 2x + 1 f(3) = 2(3) + 1 = 6 + 1 = 7 Evaluate f(7) Since 3 ≤ 7 < 8, use the second piece: f(x) = 2x + 1 f(7) = 2(7) + 1 = 14 + 1 = 15 Evaluate…
   Full step-by-step solution

   Step 1: Evaluate f(2) Since 2 < 3, use the first piece: f(x) = x^2 - 7 f(2) = (2)^2 - 7 = 4 - 7 = -3 Step 2: Evaluate f(3) Since 3 ≤ 3 < 8, use the second piece: f(x) = 2x + 1 f(3) = 2(3) + 1 = 6 + 1 = 7 Step 3: Evaluate f(7) Since 3 ≤ 7 < 8, use the second piece: f(x) = 2x + 1 f(7) = 2(7) + 1 = 14 + 1 = 15 Step 4: Evaluate f(9) Since 9 ≥ 8, use the third piece: f(x) = 17 f(9) = 17 Final answer: f(2) = -3, f(3) = 7, f(7) = 15, f(9) = 17

7. 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.

8. 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.

9. f(x) = {4x - 6 if x < -2; 2x^2 - 8 if -2 ≤ x < 4; 10 if x ≥ 4}. Evaluate f(-3), f(0), f(4), f(6) Answer: f(-3) = -18, f(0) = -8, f(4) = 10, f(6) = 10 Solution: Evaluate f(-3) Since -3 < -2, use the first piece: f(x) = 4x - 6 f(-3) = 4(-3) - 6 = -12 - 6 = -18 Evaluate f(0) Since -2 ≤ 0 < 4, use the second piece: f(x) = 2x^2 - 8 f(0) = 2(0)^2 - 8 = 0 - 8 = -8 Evaluate f(4) Since 4 ≥ 4, use the third piece: f(x) = 10 f(4) = 10 Evaluate f(6) Since 6 ≥ 4,…
   Full step-by-step solution

   Step 1: Evaluate f(-3) Since -3 < -2, use the first piece: f(x) = 4x - 6 f(-3) = 4(-3) - 6 = -12 - 6 = -18 Step 2: Evaluate f(0) Since -2 ≤ 0 < 4, use the second piece: f(x) = 2x^2 - 8 f(0) = 2(0)^2 - 8 = 0 - 8 = -8 Step 3: Evaluate f(4) Since 4 ≥ 4, use the third piece: f(x) = 10 f(4) = 10 Step 4: Evaluate f(6) Since 6 ≥ 4, use the third piece: f(x) = 10 f(6) = 10 The final answers are: f(-3) = -18, f(0) = -8, f(4) = 10, f(6) = 10