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 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

2. f(x) = {5x + 3 if x < -1; -x^2 + 9 if -1 ≤ x < 5; 11 if x ≥ 5}. Evaluate f(-3), f(-1), f(3), f(5), f(7). Answer: f(-3) = -12, f(-1) = 8, f(3) = 0, f(5) = 11, f(7) = 11 Solution: Evaluate f(-3). Since -3 < -1, use the first piece: f(x) = 5x + 3. f(-3) = 5(-3) + 3 = -15 + 3 = -12.
   Full step-by-step solution

   Step 1: Evaluate f(-3). Since -3 < -1, use the first piece: f(x) = 5x + 3. f(-3) = 5(-3) + 3 = -15 + 3 = -12. Step 2: Evaluate f(-1). Since -1 ≥ -1 and -1 < 5, use the second piece: f(x) = -x^2 + 9. f(-1) = -(-1)^2 + 9 = -1 + 9 = 8. Step 3: Evaluate f(3). Since -1 ≤ 3 < 5, use the second piece: f(x) = -x^2 + 9. f(3) = -(3)^2 + 9 = -9 + 9 = 0. Step 4: Evaluate f(5). Since 5 ≥ 5, use the third piece: f(x) = 11. f(5) = 11. Step 5: Evaluate f(7). Since 7 ≥ 5, use the third piece: f(x) = 11. f(7) = 11. The final answers are: f(-3) = -12, f(-1) = 8, f(3) = 0, f(5) = 11, f(7) = 11.

3. A pharmaceutical company is modeling the concentration of a new drug in a patient's bloodstream using the piecewise function C(t) = { 2t for 0 ≤ t < 3; 6 + 4e^(-0.5(t-3)) for t ≥ 3 }, where C(t) is the concentration in mg/L and t is time in hours. The drug becomes effective when concentration reaches 5 mg/L and remains effective while concentration stays above 3 mg/L. Determine the total time interval during which the drug is effective. Answer: 4.5 hours Solution: In pharmacokinetics, drugs often have different absorption and elimination phases.
   Full step-by-step solution

   Piecewise functions model real-world scenarios where behavior changes at specific points. In pharmacokinetics, drugs often have different absorption and elimination phases. To find effective duration, we solve for when concentration equals the threshold values in each relevant piece of the function, ensuring we check continuity at the transition point between pieces.

4. 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.

5. f(x) = {3x - 5 if x < -3; 7 - x^2 if -3 ≤ x < 1; 2x + 9 if x ≥ 1}. Evaluate f(-5), f(-3), f(0), f(1), f(3). Answer: f(-5) = -20, f(-3) = -2, f(0) = 7, f(1) = 11, f(3) = 15 Solution: Evaluate f(-5). Since -5 < -3, use the first piece: f(x) = 3x - 5. f(-5) = 3(-5) - 5 = -15 - 5 = -20.
   Full step-by-step solution

   Step 1: Evaluate f(-5). Since -5 < -3, use the first piece: f(x) = 3x - 5. f(-5) = 3(-5) - 5 = -15 - 5 = -20. Step 2: Evaluate f(-3). Since -3 ≤ -3 < 1, use the second piece: f(x) = 7 - x^2. f(-3) = 7 - (-3)^2 = 7 - 9 = -2. Step 3: Evaluate f(0). Since -3 ≤ 0 < 1, use the second piece: f(x) = 7 - x^2. f(0) = 7 - 0^2 = 7. Step 4: Evaluate f(1). Since 1 ≥ 1, use the third piece: f(x) = 2x + 9. f(1) = 2(1) + 9 = 2 + 9 = 11. Step 5: Evaluate f(3). Since 3 ≥ 1, use the third piece: f(x) = 2x + 9. f(3) = 2(3) + 9 = 6 + 9 = 15. Final answer: f(-5) = -20, f(-3) = -2, f(0) = 7, f(1) = 11, f(3) = 15.

6. f(x) = {x² - 6 if x < 1; 4x - 1 if 1 ≤ x < 6; 16 if x ≥ 6}. Evaluate f(-1), f(1), f(6), f(8) Answer: f(-1) = -5, f(1) = 3, f(6) = 16, f(8) = 16 Solution: Evaluate f(-1). Since -1 < 1, use the first piece: f(x) = x² - 6. f(-1) = (-1)² - 6 = 1 - 6 = -5.
   Full step-by-step solution

   Step 1: Evaluate f(-1). Since -1 < 1, use the first piece: f(x) = x² - 6. f(-1) = (-1)² - 6 = 1 - 6 = -5. Step 2: Evaluate f(1). Since 1 ≤ 1 < 6, use the second piece: f(x) = 4x - 1. f(1) = 4(1) - 1 = 4 - 1 = 3. Step 3: Evaluate f(6). Since 6 ≥ 6, use the third piece: f(x) = 16. f(6) = 16. Step 4: Evaluate f(8). Since 8 ≥ 6, use the third piece: f(x) = 16. f(8) = 16. Final answer: f(-1) = -5, f(1) = 3, f(6) = 16, f(8) = 16.

7. f(x) = {3x - 11 if x < -4; 2x^2 + 5 if -4 ≤ x < 2; 9x - 13 if x ≥ 2}. Evaluate f(-5), f(-4), f(1), f(2), f(6). Answer: f(-5) = -26, f(-4) = 37, f(1) = 7, f(2) = 5, f(6) = 41 Solution: Evaluate f(-5). Since -5 < -4, use the first piece: f(x) = 3x - 11. f(-5) = 3(-5) - 11 = -15 - 11 = -26.
   Full step-by-step solution

   Step 1: Evaluate f(-5). Since -5 < -4, use the first piece: f(x) = 3x - 11. f(-5) = 3(-5) - 11 = -15 - 11 = -26. Step 2: Evaluate f(-4). Since -4 ≤ -4 < 2, use the second piece: f(x) = 2x^2 + 5. f(-4) = 2(-4)^2 + 5 = 2(16) + 5 = 32 + 5 = 37. Step 3: Evaluate f(1). Since -4 ≤ 1 < 2, use the second piece: f(x) = 2x^2 + 5. f(1) = 2(1)^2 + 5 = 2(1) + 5 = 2 + 5 = 7. Step 4: Evaluate f(2). Since 2 ≥ 2, use the third piece: f(x) = 9x - 13. f(2) = 9(2) - 13 = 18 - 13 = 5. Step 5: Evaluate f(6). Since 6 ≥ 2, use the third piece: f(x) = 9x - 13. f(6) = 9(6) - 13 = 54 - 13 = 41. The final answers are: f(-5) = -26, f(-4) = 37, f(1) = 7, f(2) = 5, f(6) = 41.