Polynomial Technology

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Polynomial Technology · Grade 12 · Worksheet 3

1. Liam is designing a roller coaster and needs to analyze the polynomial function f(x) = x³ - 6x² + 9x + 2 that models the track's elevation. He wants to determine the intervals where the track is increasing in elevation. Find all x-values where f'(x) > 0. Answer: (1,3) Solution: The derivative of a function represents its instantaneous rate of change. When the derivative is positive, the original function is increasing.
   Full step-by-step solution

   The derivative of a function represents its instantaneous rate of change. When the derivative is positive, the original function is increasing. For polynomial functions, we can find increasing intervals by solving the inequality where the derivative is greater than zero. This involves finding critical points where the derivative equals zero, then testing values in each interval to determine where the derivative is positive.

2. Use Desmos to graph f(x)=x³-9x²+24x-16 and find the sum of all x-coordinates where the local maximum and minimum occur. Answer: 6 Solution: Find the derivative of f(x)=x³-9x²+24x-16. f'(x)=3x²-18x+24. Set the derivative equal to zero: 3x²-18x+24=0.
   Full step-by-step solution

   Step 1: Find the derivative of f(x)=x³-9x²+24x-16. f'(x)=3x²-18x+24. Step 2: Set the derivative equal to zero: 3x²-18x+24=0. Step 3: Divide by 3: x²-6x+8=0. Step 4: Factor: (x-2)(x-4)=0. Step 5: Critical points are x=2 and x=4. Step 6: Use the second derivative test: f''(x)=6x-18. f''(2)=6(2)-18=-6 (negative, local maximum). f''(4)=6(4)-18=6 (positive, local minimum). Step 7: Sum the x-coordinates: 2+4=6. The answer is 6.

3. Use Desmos to graph f(x)=x⁴-8x³+22x²-24x+8 and find the sum of all x-coordinates where the graph has local minima. Answer: 4 Solution: Find the derivative of f(x)=x⁴-8x³+22x²-24x+8. f'(x)=4x³-24x²+44x-24 Set the derivative equal to zero to find critical points. 4x³-24x²+44x-24=0 Divide by 4: x³-6x²+11x-6=0 Factor the cubic.
   Full step-by-step solution

   Step 1: Find the derivative of f(x)=x⁴-8x³+22x²-24x+8. f'(x)=4x³-24x²+44x-24 Step 2: Set the derivative equal to zero to find critical points. 4x³-24x²+44x-24=0 Divide by 4: x³-6x²+11x-6=0 Step 3: Factor the cubic. Test x=1: 1-6+11-6=0, so (x-1) is a factor. Divide: (x³-6x²+11x-6) ÷ (x-1) = x²-5x+6 Factor: x²-5x+6 = (x-2)(x-3) So critical points are x=1, x=2, x=3. Step 4: Use the second derivative test. f''(x)=12x²-48x+44 f''(1)=12-48+44=8 > 0, so local minimum at x=1 f''(2)=48-96+44=-4 < 0, so local maximum at x=2 f''(3)=108-144+44=8 > 0, so local minimum at x=3 Step 5: Sum the x-coordinates of local minima: 1+3=4. The answer is 4.

4. A polynomial function f(x) is graphed on a coordinate plane. The graph shows the function crossing the x-axis at (-4, 0) and (2, 0), and touching but not crossing the x-axis at (-1, 0). The graph passes through the point (1, -12). The end behavior shows that as x → -∞, f(x) → -∞ and as x → ∞, f(x) → ∞. Determine the equation of this polynomial in standard form. Answer: f(x) = x^4 + 4x^3 - 3x^2 - 14x + 8 Solution: Identify the roots and their multiplicities from the graph behavior. The equation in standard form is f(x) = (3/5)x^4 + (12/5)x^3 - (9/5)x^2 - (42/5)x - (24/5).
   Full step-by-step solution

   Step 1: Identify the roots and their multiplicities from the graph behavior. - Crossing at x = -4: odd multiplicity (use 1) - Crossing at x = 2: odd multiplicity (use 1) - Touching at x = -1: even multiplicity (use 2) Step 2: Write the factored form with unknown leading coefficient 'a': f(x) = a(x + 4)(x - 2)(x + 1)^2 Step 3: Use the point (1, -12) to solve for 'a': -12 = a(1 + 4)(1 - 2)(1 + 1)^2 -12 = a(5)(-1)(2)^2 -12 = a(5)(-1)(4) -12 = a(-20) -12 = -20a a = -12/-20 a = 3/5 Step 4: Write the function with the found coefficient: f(x) = (3/5)(x + 4)(x - 2)(x + 1)^2 Step 5: Expand the factors step by step: First expand (x + 4)(x - 2) = x^2 + 4x - 2x - 8 = x^2 + 2x - 8 Then expand (x + 1)^2 = x^2 + 2x + 1 Now multiply: (x^2 + 2x - 8)(x^2 + 2x + 1) = x^2(x^2 + 2x + 1) + 2x(x^2 + 2x + 1) - 8(x^2 + 2x + 1) = x^4 + 2x^3 + x^2 + 2x^3 + 4x^2 + 2x - 8x^2 - 16x - 8 = x^4 + 4x^3 - 3x^2 - 14x - 8 Step 6: Multiply by the leading coefficient: f(x) = (3/5)(x^4 + 4x^3 - 3x^2 - 14x - 8) = (3/5)x^4 + (12/5)x^3 - (9/5)x^2 - (42/5)x - (24/5) Step 7: Verify end behavior: With positive leading coefficient (3/5 > 0) and even degree (4), as x → ±∞, f(x) → ∞, which matches the given end behavior. The equation in standard form is f(x) = (3/5)x^4 + (12/5)x^3 - (9/5)x^2 - (42/5)x - (24/5).

5. An environmental engineer is modeling the concentration of a pollutant in a river system using the polynomial function P(t) = -0.01t⁴ + 0.12t³ + 0.2t² - 1.2t + 2, where P(t) represents the pollutant concentration in parts per million and t is the time in days after a chemical spill. Environmental regulations require the pollutant concentration to remain below 1.5 ppm. Determine the time intervals during which the pollutant concentration exceeds the regulatory limit. Answer: (0, 2) and (5, 10) Solution: Set up the inequality: -0.01t⁴ + 0.12t³ + 0.2t² - 1.2t + 2 > 1.5 Rearrange: -0.01t⁴ + 0.12t³ + 0.2t² - 1.2t + 0.5 > 0 Multiply by -100 to simplify: t⁴ - 12t³ - 20t² + 120t - 50 < 0 Find the roots by testing values: P(0) = 2 > 1.5, P(1) = 1.11 < 1.5, P(2) = 1.52 > 1.5, P(3) = 1.37 < 1.5, P(4) =…
   Full step-by-step solution

   Step 1: Set up the inequality: -0.01t⁴ + 0.12t³ + 0.2t² - 1.2t + 2 > 1.5 Step 2: Rearrange: -0.01t⁴ + 0.12t³ + 0.2t² - 1.2t + 0.5 > 0 Step 3: Multiply by -100 to simplify: t⁴ - 12t³ - 20t² + 120t - 50 < 0 Step 4: Find the roots by testing values: P(0) = 2 > 1.5, P(1) = 1.11 < 1.5, P(2) = 1.52 > 1.5, P(3) = 1.37 < 1.5, P(4) = 0.68 < 1.5, P(5) = 1.5 = 1.5, P(6) = 2.96 > 1.5, P(7) = 3.67 > 1.5, P(8) = 3.08 > 1.5, P(9) = 1.49 < 1.5, P(10) = 1.0 < 1.5 Step 5: The roots are approximately t = 2, t = 5, and t = 9 Step 6: Test intervals: (0,2): P(1) = 1.11 < 1.5, so below limit; (2,5): P(3) = 1.37 < 1.5, so below limit; (5,9): P(6) = 2.96 > 1.5, so above limit; (9,∞): P(10) = 1.0 < 1.5, so below limit Step 7: The pollutant exceeds 1.5 ppm during the intervals (0,2) and (5,9)

6. A biomedical engineer is modeling the concentration of a new medication in a patient's bloodstream using the polynomial function C(t) = -0.02t³ + 0.36t² - 1.8t + 2.4, where C(t) represents the concentration in milligrams per liter and t is the time in hours after administration. The medication becomes effective when the concentration reaches its maximum value. At what time does the medication become effective, and what is the maximum concentration? Answer: t = 6 hours, C = 2.4 mg/L Solution: In optimization problems involving polynomial functions, we use calculus to find extreme values.
   Full step-by-step solution

   In optimization problems involving polynomial functions, we use calculus to find extreme values. The derivative of a function gives us its rate of change, and setting this derivative equal to zero helps identify critical points where the function may reach maximum or minimum values. For real-world applications like drug concentration modeling, we need to verify which critical point represents the maximum by checking the behavior of the function or using the second derivative test.

7. Use Desmos to graph f(x) = x³ - 12x² + 47x - 62 and find the sum of all x-coordinates where the local maximum and minimum occur. Answer: 8 Solution: Find the derivative of f(x) = x³ - 12x² + 47x - 62. f'(x) = 3x² - 24x + 47 Set the derivative equal to zero to find critical points. 3x² - 24x + 47 = 0 Use the quadratic formula to solve.
   Full step-by-step solution

   Step 1: Find the derivative of f(x) = x³ - 12x² + 47x - 62. f'(x) = 3x² - 24x + 47 Step 2: Set the derivative equal to zero to find critical points. 3x² - 24x + 47 = 0 Step 3: Use the quadratic formula to solve. x = [24 ± sqrt(576 - 564)] / 6 x = [24 ± sqrt(12)] / 6 x = [24 ± 2√3] / 6 x = 4 ± √3/3 Step 4: The two critical points are x = 4 + √3/3 and x = 4 - √3/3. Step 5: Sum the x-coordinates. (4 + √3/3) + (4 - √3/3) = 8 The answer is 8.