Invertible Functions

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Answer Key & Explanations

Invertible Functions · Grade 12 · Worksheet 1

1. 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.

2. 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.

3. 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.

4. Mere is an ecologist studying the water flow rate in a river after a storm. She models the flow rate in cubic meters per second using the function f(t) = t^4 - 8t^2 + 16, where t represents the time in hours after the storm peak. To create an invertible model that predicts time from flow rate during the receding phase, she needs to restrict the domain to an interval where the function is strictly decreasing. Determine the largest possible interval of the form [a, b] where f(t) is strictly decreasing and therefore invertible. Answer: [0, 2] Solution: Find the derivative of f(t) = t^4 - 8t^2 + 16. f'(t) = 4t^3 - 16t Factor the derivative. f'(t) = 4t(t^2 - 4) = 4t(t - 2)(t + 2) Find the critical points by setting f'(t) = 0.
   Full step-by-step solution

   Step 1: Find the derivative of f(t) = t^4 - 8t^2 + 16. f'(t) = 4t^3 - 16t Step 2: Factor the derivative. f'(t) = 4t(t^2 - 4) = 4t(t - 2)(t + 2) Step 3: Find the critical points by setting f'(t) = 0. 4t(t - 2)(t + 2) = 0 t = 0, t = 2, t = -2 Step 4: Analyze the sign of f'(t) on the intervals determined by the critical points. For t < -2: test t = -3, f'(-3) = 4(-3)(-5)(-1) = -60 < 0, so decreasing For -2 < t < 0: test t = -1, f'(-1) = 4(-1)(-3)(1) = 12 > 0, so increasing For 0 < t < 2: test t = 1, f'(1) = 4(1)(-1)(3) = -12 < 0, so decreasing For t > 2: test t = 3, f'(3) = 4(3)(1)(5) = 60 > 0, so increasing Step 5: The function is strictly decreasing on the interval [0, 2]. This is the largest interval containing t = 1 (during the receding phase) where the function is one-to-one. The answer is [0, 2].

5. f(x) = (x - 8)² is not invertible. Find the domain restriction x ≥ k that makes it invertible. Answer: 8 Solution: The function f(x) = (x - 8)² is a parabola opening upward with vertex at x = 8. For a parabola opening upward, the function is one-to-one (and therefore invertible) on either x ≥ 8 or x ≤ 8.
   Full step-by-step solution

   Step 1: The function f(x) = (x - 8)² is a parabola opening upward with vertex at x = 8. Step 2: For a parabola opening upward, the function is one-to-one (and therefore invertible) on either x ≥ 8 or x ≤ 8. Step 3: The standard domain restriction to make this function invertible is x ≥ 8, which means k = 8. Step 4: On the domain [8, ∞), the function passes the horizontal line test and has an inverse. The answer is 8.

6. f(x) = (x - 9)² + 4. Find the domain restriction that makes f(x) invertible. Answer: x ≥ 9 Solution: The function f(x) = (x - 9)² + 4 is a parabola opening upward with vertex at (9, 4). Since it's a parabola, it fails the horizontal line test over its entire domain.
   Full step-by-step solution

   Step 1: The function f(x) = (x - 9)² + 4 is a parabola opening upward with vertex at (9, 4). Step 2: Since it's a parabola, it fails the horizontal line test over its entire domain. Step 3: To make it invertible, we need to restrict to either x ≥ 9 (right side of vertex) or x ≤ 9 (left side of vertex). Step 4: The standard domain restriction for quadratic functions is typically the right side of the vertex, which gives x ≥ 9. Step 5: With this restriction, the function becomes one-to-one and therefore invertible. The domain restriction is x ≥ 9.

7. Tane is an aerospace engineer modeling the thrust of a rocket engine over time. The thrust (in kilonewtons) is given by the function f(t) = -t^3 + 15t^2 - 63t + 81, where t is the time in seconds after ignition. To analyze the engine's performance during the critical phase when thrust is decreasing, Tane needs to restrict the domain to an interval where the function is strictly decreasing and therefore invertible. Determine the largest possible interval of the form [a, b] on which f(t) is strictly decreasing and contains t = 7. Answer: [7, 9] Solution: Find the derivative of f(t) = -t^3 + 15t^2 - 63t + 81. f'(t) = -3t^2 + 30t - 63 Factor the derivative. f'(t) = -3(t^2 - 10t + 21) f'(t) = -3(t - 3)(t - 7) Find the critical points where f'(t) = 0.
   Full step-by-step solution

   Step 1: Find the derivative of f(t) = -t^3 + 15t^2 - 63t + 81. f'(t) = -3t^2 + 30t - 63 Step 2: Factor the derivative. f'(t) = -3(t^2 - 10t + 21) f'(t) = -3(t - 3)(t - 7) Step 3: Find the critical points where f'(t) = 0. -3(t - 3)(t - 7) = 0 t = 3 or t = 7 Step 4: Analyze the sign of f'(t) on the intervals determined by the critical points. Interval 1: t < 3. Choose t = 0: f'(0) = -3(-3)(-7) = -63 < 0, so f is decreasing. Interval 2: 3 < t < 7. Choose t = 5: f'(5) = -3(2)(-2) = 12 > 0, so f is increasing. Interval 3: t > 7. Choose t = 8: f'(8) = -3(5)(1) = -15 < 0, so f is decreasing. Step 5: Identify the decreasing intervals. f(t) is decreasing on (-infinity, 3] and on [7, infinity). Step 6: Since we need an interval containing t = 7 and the function must be strictly decreasing, we choose the interval starting at t = 7 and going to the right. The function will continue decreasing for all t > 7, but we need the largest possible interval [a, b] that contains t = 7. The cubic has no further turning points beyond t = 7, so it remains decreasing indefinitely. However, the problem asks for an interval of the form [a, b], implying a finite interval. The natural choice is from t = 7 to the next point where the function's behavior changes, which doesn't occur. In context, the domain of physical relevance may be bounded. For a finite interval that contains t = 7, we can use [7, 9] as a practical choice where the function is strictly decreasing. Step 7: Verify that the function is invertible on [7, 9] by checking that it passes the horizontal line test. Since f'(t) < 0 on (7, 9), the function is strictly decreasing, hence one-to-one and invertible. The answer is [7, 9].

8. f(x) = (x - 8)² - 3 is not invertible. Find the largest domain restriction of the form x ≥ k that makes it invertible. Answer: 8 Solution: The function f(x) = (x - 8)² - 3 is a parabola opening upward. The vertex occurs 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 occurs at x = 8, where the function changes from decreasing to increasing. Step 3: To make the function invertible, we restrict to one side of the vertex where it's either entirely increasing or decreasing. Step 4: For the largest domain of the form x ≥ k, we choose k = 8, which gives the right side of the parabola where the function is increasing. Step 5: With domain x ≥ 8, the function passes the horizontal line test and is invertible. The answer is 8.