Invertible Functions

lessonbunny.com

Answer Key & Explanations

Invertible Functions · Grade 12 · Worksheet 1

1. Mere is a marine biologist studying the population of a certain fish species in a lake. She models the fish population (in thousands) over time using the function f(t) = t^3 - 15t^2 + 63t, where t represents time in years since the study began. The population initially increases, then decreases, and then increases again. To create an inverse function that can predict the time when the population first reaches a given value during the initial increasing phase, Mere needs to restrict the domain to the interval where the function is strictly increasing and includes t = 0. Determine the largest possible domain of the form [0, a] that makes f(t) invertible. Answer: [0, 3] Solution: Find the derivative of f(t) = t^3 - 15t^2 + 63t. f'(t) = 3t^2 - 30t + 63. Set the derivative equal to zero to find critical points: 3t^2 - 30t + 63 = 0.
   Full step-by-step solution

   Step 1: Find the derivative of f(t) = t^3 - 15t^2 + 63t. f'(t) = 3t^2 - 30t + 63. Step 2: Set the derivative equal to zero to find critical points: 3t^2 - 30t + 63 = 0. Divide by 3: t^2 - 10t + 21 = 0. Factor: (t - 3)(t - 7) = 0. So t = 3 and t = 7. Step 3: Analyze the sign of the derivative on intervals defined by these critical points. For t < 3, test t = 0: f'(0) = 63 > 0, so f is increasing on [0, 3). For 3 < t < 7, test t = 5: f'(5) = 3(25) - 30(5) + 63 = 75 - 150 + 63 = -12 < 0, so f is decreasing on (3, 7). For t > 7, test t = 8: f'(8) = 3(64) - 30(8) + 63 = 192 - 240 + 63 = 15 > 0, so f is increasing on (7, ∞). Step 4: The function is strictly increasing on [0, 3] (including t = 3 where f'(3) = 0 but the function continues to increase after a momentary plateau at a single point, which still allows invertibility on a closed interval). Step 5: The largest domain of the form [0, a] where f(t) is strictly increasing and therefore invertible is [0, 3]. The answer is [0, 3].

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. f(x) = (x - 4)² is not invertible. Find the domain restriction x ≥ k that makes it invertible. Answer: 4 Solution: The function f(x) = (x - 4)² is a parabola opening upward with vertex at x = 4. To make the function one-to-one (and therefore invertible), we need to restrict the domain to either x ≥ 4 or x ≤ 4.
   Full step-by-step solution

   Step 1: The function f(x) = (x - 4)² is a parabola opening upward with vertex at x = 4. Step 2: To make the function one-to-one (and therefore invertible), we need to restrict the domain to either x ≥ 4 or x ≤ 4. Step 3: The problem asks for the domain restriction x ≥ k, so we need the right side of the parabola. Step 4: The vertex is at x = 4, so the restriction should be x ≥ 4. Step 5: Therefore, k = 4. The answer is 4.

4. A biomedical engineer is modeling the concentration of a medication in a patient's bloodstream using the function f(x) = x² - 4x + 3, where x represents time in hours. To ensure the model accurately represents the medication's behavior during peak effectiveness, she needs to restrict the domain to make the function invertible while maintaining the interval where concentration increases. Over what interval should she restrict the domain to achieve this? Answer: [2, ∞) Solution: f(x) = x² - 4x + 3 This is a quadratic function, which is a parabola opening upward (since the coefficient of x² is positive).
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the problem** We have f(x) = x² - 4x + 3 This is a quadratic function, which is a parabola opening upward (since the coefficient of x² is positive). A parabola is not one-to-one over its entire domain, so to make it invertible we must restrict the domain to either the left side or the right side of the vertex. The problem says: restrict the domain to make the function invertible **while maintaining the interval where concentration increases**. Since the parabola opens upward, it decreases on the left side of the vertex and increases on the right side of the vertex. So we want the right side of the vertex. --- **Step 2: Find the vertex** For a quadratic in the form ax² + bx + c, the vertex occurs at x = -b/(2a) Here a = 1, b = -4 So x = -(-4)/(2*1) = 4/2 = 2. --- **Step 3: Determine increasing/decreasing intervals** Since the parabola opens upward: - For x < 2, the function is decreasing. - For x > 2, the function is increasing. We want the concentration to be increasing, so we choose x ≥ 2. --- **Step 4: Ensure invertibility** On the interval [2, ∞), the function is strictly increasing (one-to-one), so it is invertible there. --- **Step 5: Final answer** The restricted domain is [2, ∞). --- **Final answer:** [2, ∞)

5. f(x) = (x - 6)² + 1 is not invertible. Find the domain restriction x ≥ a that makes it invertible. Answer: 6 Solution: The function f(x) = (x - 6)² + 1 is a parabola opening upward. The vertex of this parabola is at x = 6. To make the function one-to-one (pass the horizontal line test), we restrict the domain to either x ≥ 6 or x ≤ 6.
   Full step-by-step solution

   Step 1: The function f(x) = (x - 6)² + 1 is a parabola opening upward. Step 2: The vertex of this parabola is at x = 6. Step 3: To make the function one-to-one (pass the horizontal line test), we restrict the domain to either x ≥ 6 or x ≤ 6. Step 4: The problem asks for the domain restriction x ≥ a, so we use x ≥ 6. Step 5: Therefore, a = 6. The answer is 6.

6. f(x) = (x - 7)² + 2 is not invertible. Find the domain restriction x ≥ a that makes it invertible. Answer: 7 Solution: The function f(x) = (x - 7)² + 2 is a parabola opening upward. The vertex occurs at x = 7, where (x - 7) = 0. To the left of x = 7, the function is decreasing.
   Full step-by-step solution

   Step 1: The function f(x) = (x - 7)² + 2 is a parabola opening upward. Step 2: The vertex occurs at x = 7, where (x - 7) = 0. Step 3: To the left of x = 7, the function is decreasing. To the right of x = 7, the function is increasing. Step 4: To make the function one-to-one (pass the horizontal line test), we restrict to either x ≤ 7 or x ≥ 7. Step 5: Since the problem asks for x ≥ a, we use the right side of the parabola where the function is increasing. Step 6: Therefore, the domain restriction is x ≥ 7. The answer is 7.