Invertible Functions

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Invertible Functions · Grade 12 · Worksheet 3

1. Sophia is an astrophysicist modeling the gravitational potential energy of a satellite orbiting a planet. The energy (in gigajoules) as a function of orbital radius r (in thousands of kilometers) is given by E(r) = 2r^3 - 30r^2 + 126r - 10. For her analysis of orbital stability, she needs to restrict the domain to an interval where the energy function is strictly decreasing, ensuring the function is invertible. Determine the largest possible interval of the form [a, b] where E(r) is strictly decreasing and therefore invertible. Answer: [3, 7] Solution: Find the derivative of E(r) = 2r^3 - 30r^2 + 126r - 10 E'(r) = 6r^2 - 60r + 126 Set the derivative equal to zero to find critical points 6r^2 - 60r + 126 = 0 Divide by 6: r^2 - 10r + 21 = 0 Factor: (r - 3)(r - 7) = 0 Critical points: r = 3 and r = 7 Analyze the sign of the derivative on the…
   Full step-by-step solution

   Step 1: Find the derivative of E(r) = 2r^3 - 30r^2 + 126r - 10 E'(r) = 6r^2 - 60r + 126 Step 2: Set the derivative equal to zero to find critical points 6r^2 - 60r + 126 = 0 Divide by 6: r^2 - 10r + 21 = 0 Factor: (r - 3)(r - 7) = 0 Critical points: r = 3 and r = 7 Step 3: Analyze the sign of the derivative on the intervals determined by the critical points Interval 1: r < 3. Test r = 0: E'(0) = 6(0)^2 - 60(0) + 126 = 126 > 0 (increasing) Interval 2: 3 < r < 7. Test r = 5: E'(5) = 6(25) - 60(5) + 126 = 150 - 300 + 126 = -24 < 0 (decreasing) Interval 3: r > 7. Test r = 10: E'(10) = 6(100) - 60(10) + 126 = 600 - 600 + 126 = 126 > 0 (increasing) Step 4: The function is strictly decreasing on the interval (3, 7). Since the derivative is zero at the endpoints but negative in between, the function is strictly decreasing on the closed interval [3, 7]. Step 5: The largest possible interval where E(r) is strictly decreasing and thus invertible is [3, 7]. The answer is [3, 7].

2. Sophia is an electrical engineer analyzing the voltage output of a prototype circuit over time. The voltage is modeled by the function V(t) = t^3 - 6t^2 + 9t + 1, where t is time in seconds and V(t) is in volts. To design a feedback control system that requires a one-to-one relationship between time and voltage, Sophia must restrict the domain to an interval where the function is strictly decreasing. Determine the largest possible interval of the form [a, b] on which V(t) is strictly decreasing and therefore invertible. Answer: [1, 3] Solution: Find the derivative of V(t) = t^3 - 6t^2 + 9t + 1. V'(t) = 3t^2 - 12t + 9. Set the derivative equal to zero to find critical points.
   Full step-by-step solution

   Step 1: Find the derivative of V(t) = t^3 - 6t^2 + 9t + 1. V'(t) = 3t^2 - 12t + 9. Step 2: Set the derivative equal to zero to find critical points. 3t^2 - 12t + 9 = 0 Divide both sides by 3: t^2 - 4t + 3 = 0 Factor: (t - 1)(t - 3) = 0 So t = 1 and t = 3 are critical points. Step 3: Test the sign of V'(t) in each interval. - For t < 1 (e.g., t = 0): V'(0) = 3(0)^2 - 12(0) + 9 = 9 > 0, so V is increasing. - For 1 < t < 3 (e.g., t = 2): V'(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 < 0, so V is decreasing. - For t > 3 (e.g., t = 4): V'(4) = 3(16) - 12(4) + 9 = 48 - 48 + 9 = 9 > 0, so V is increasing. Step 4: The function is strictly decreasing on the interval (1, 3). Since the derivative is zero at the endpoints, the function is not strictly increasing or decreasing exactly at t = 1 and t = 3, but to include them in a closed interval where it is strictly decreasing, we consider the interval [1, 3] where the function is strictly decreasing for all interior points. Step 5: The largest possible interval where V(t) is strictly decreasing and therefore invertible is [1, 3]. The answer is [1, 3].

3. Sophia is a pharmaceutical researcher modeling the rate at which a new antibiotic is absorbed into bacterial cells. The absorption rate (in micrograms per minute) is given by the function f(x) = x^3 - 15x^2 + 63x - 49, where x represents the time in minutes after the antibiotic is introduced. To analyze the period when the absorption rate is strictly decreasing, Sophia needs to restrict the domain of f(x) to make it invertible. Determine the largest possible interval of the form [a, b] where f(x) is strictly decreasing and therefore invertible. Answer: [3, 7] Solution: Find the derivative of f(x) = x^3 - 15x^2 + 63x - 49 f'(x) = 3x^2 - 30x + 63 Set the derivative equal to zero to find critical points 3x^2 - 30x + 63 = 0 Divide both sides by 3: x^2 - 10x + 21 = 0 Factor: (x - 3)(x - 7) = 0 So x = 3 or x = 7 Analyze the sign of the derivative in the intervals…
   Full step-by-step solution

   Step 1: Find the derivative of f(x) = x^3 - 15x^2 + 63x - 49 f'(x) = 3x^2 - 30x + 63 Step 2: Set the derivative equal to zero to find critical points 3x^2 - 30x + 63 = 0 Divide both sides by 3: x^2 - 10x + 21 = 0 Factor: (x - 3)(x - 7) = 0 So x = 3 or x = 7 Step 3: Analyze the sign of the derivative in the intervals determined by the critical points For x < 3, choose x = 0: f'(0) = 3(0)^2 - 30(0) + 63 = 63 > 0, so f is increasing on (-infinity, 3) For 3 < x < 7, choose x = 5: f'(5) = 3(25) - 30(5) + 63 = 75 - 150 + 63 = -12 < 0, so f is decreasing on (3, 7) For x > 7, choose x = 8: f'(8) = 3(64) - 30(8) + 63 = 192 - 240 + 63 = 15 > 0, so f is increasing on (7, infinity) Step 4: The function is strictly decreasing on the open interval (3, 7). The largest closed interval where it is strictly decreasing includes the endpoints, since at the endpoints the derivative is zero but the function is still one-to-one on the closed interval. Therefore, the largest possible interval where f(x) is strictly decreasing and invertible is [3, 7]. The answer is [3, 7].

4. Mason is an engineer analyzing the water flow in a pipe system. The flow rate (in liters per second) is modeled by the function f(x) = 2x^2 - 28x + 97, where x represents the time in seconds after a valve is opened. To create a mathematical model that can invert time from flow rate for a specific control algorithm, Mason needs to restrict the domain to the interval where the function is strictly decreasing. Determine the largest possible interval of the form (-∞, b] that makes f(x) invertible. Answer: (-∞, 7] Solution: Identify the function f(x) = 2x^2 - 28x + 97. This is a quadratic function with a = 2, b = -28, c = 97. Since a > 0, the parabola opens upward.
   Full step-by-step solution

   Step 1: Identify the function f(x) = 2x^2 - 28x + 97. This is a quadratic function with a = 2, b = -28, c = 97. Since a > 0, the parabola opens upward. Step 2: Find the vertex (turning point) of the parabola using the derivative. Compute f'(x) = 4x - 28. Step 3: Set the derivative equal to zero to find the critical point: 4x - 28 = 0 => 4x = 28 => x = 7. Step 4: Analyze the sign of the derivative on each side of x = 7. For x < 7, choose x = 0: f'(0) = 4(0) - 28 = -28 < 0, so f is strictly decreasing on (-∞, 7). For x > 7, choose x = 8: f'(8) = 4(8) - 28 = 32 - 28 = 4 > 0, so f is strictly increasing on (7, ∞). Step 5: The function is strictly decreasing on the interval (-∞, 7]. At x = 7, the function attains its minimum and is still one-to-one on this interval because the horizontal line test passes (no two distinct x-values give the same y-value). Step 6: The largest possible interval of the form (-∞, b] that makes f invertible is (-∞, 7]. The answer is (-∞, 7].

5. f(x) = (x - 15)² + 10. Find the largest domain restriction x ≤ a or x ≥ a that makes f invertible. Answer: 15 Solution: The function f(x) = (x - 15)² + 10 is a parabola opening upward with vertex at (15, 10). Since it's a parabola opening upward, it fails the horizontal line test over its entire domain.
   Full step-by-step solution

   Step 1: The function f(x) = (x - 15)² + 10 is a parabola opening upward with vertex at (15, 10). Step 2: Since it's a parabola opening upward, it fails the horizontal line test over its entire domain. Step 3: To make it invertible, we restrict to either x ≤ 15 or x ≥ 15. Step 4: The problem asks for the largest domain restriction of the form x ≤ a or x ≥ a. Step 5: The vertex x-coordinate is 15, so a = 15. Step 6: Therefore, the domain restriction is either x ≤ 15 or x ≥ 15. The answer is 15.

6. f(x) = (x - 8)² + 3 is not invertible. Find the domain restriction x ≥ a that makes it invertible. Answer: 8 Solution: The function f(x) = (x - 8)² + 3 is a parabola opening upward. The vertex is at x = 8, where the function changes from decreasing to increasing.
   Full step-by-step solution

   Step 1: The function f(x) = (x - 8)² + 3 is a parabola opening upward. Step 2: The vertex is at x = 8, where the function changes from decreasing to increasing. Step 3: To make the function one-to-one (pass the horizontal line test), we need to restrict to either x ≥ 8 or x ≤ 8. Step 4: The standard domain restriction for upward-opening parabolas is x ≥ h, where h is the x-coordinate of the vertex. Step 5: Since the vertex is at x = 8, the domain restriction is x ≥ 8. Step 6: Therefore, a = 8.