Invertible Functions

lessonbunny.com

Answer Key & Explanations

Invertible Functions · Grade 12 · Worksheet 2

1. Consider the function f(x) = (x - 6)^2 - 1, which represents a parabola opening upward with its vertex at (6, -1). Visualize this U-shaped curve on a coordinate plane. The function fails the horizontal line test over its entire domain because a horizontal line can intersect the parabola at two points. Determine the largest possible restriction of the domain to the right of the vertex (including the vertex) so that the resulting function is invertible. Express your answer using interval notation. Answer: [6, infinity) Solution: The function f(x) = (x - 6)^2 - 1 is a parabola with vertex at (6, -1). It opens upward, so it decreases on (-infinity, 6] and increases on [6, infinity).
   Full step-by-step solution

   Step 1: The function f(x) = (x - 6)^2 - 1 is a parabola with vertex at (6, -1). It opens upward, so it decreases on (-infinity, 6] and increases on [6, infinity). Step 2: The function is not invertible over its entire domain because it fails the horizontal line test (most horizontal lines above y = -1 intersect the graph at two points). Step 3: To make the function invertible, we restrict the domain to one side of the vertex. The problem specifies the right side (including the vertex). Step 4: On the interval [6, infinity), the function is strictly increasing. A strictly increasing function passes the horizontal line test because each horizontal line intersects the graph at most once. Step 5: Therefore, the largest possible restriction to the right of the vertex is [6, infinity). The answer is [6, infinity).

2. f(x) = (x - 6)² + 2; restrict domain to make invertible Answer: x ≥ 6 Solution: The function f(x) = (x - 6)² + 2 is a parabola opening upward with vertex at (6, 2) Since it's a parabola, it fails the horizontal line test over its entire domain To make it invertible, we restrict to either x ≥ 6 (right side of vertex) or x ≤ 6 (left side of vertex) The standard convention is…
   Full step-by-step solution

   Step 1: The function f(x) = (x - 6)² + 2 is a parabola opening upward with vertex at (6, 2) Step 2: Since it's a parabola, it fails the horizontal line test over its entire domain Step 3: To make it invertible, we restrict to either x ≥ 6 (right side of vertex) or x ≤ 6 (left side of vertex) Step 4: The standard convention is to restrict to x ≥ 6, which gives the increasing portion of the parabola Step 5: With domain x ≥ 6, each y-value corresponds to exactly one x-value, making the function invertible Step 6: The restricted domain is x ≥ 6

3. Noah is a pharmaceutical researcher modeling the concentration of a new antibiotic in a patient's bloodstream. The concentration, in mg/L, is given by the function C(t) = 12t^2 - t^3, where t is the time in hours after administration, for 0 ≤ t ≤ 12. To use the model for predicting the exact time when a specific concentration occurs, Noah needs to restrict the domain to a smaller interval where the function is one-to-one (invertible). Determine the largest possible interval of the form [a, b] within [0, 12] where C(t) is strictly decreasing, making it invertible. Answer: [8, 12] Solution: Find the derivative of C(t) = 12t^2 - t^3. C'(t) = 24t - 3t^2. Set the derivative equal to zero to find critical points: 24t - 3t^2 = 0.
   Full step-by-step solution

   Step 1: Find the derivative of C(t) = 12t^2 - t^3. C'(t) = 24t - 3t^2. Step 2: Set the derivative equal to zero to find critical points: 24t - 3t^2 = 0. Factor: 3t(8 - t) = 0. So t = 0 or t = 8. Step 3: Test the sign of C'(t) on intervals within [0, 12]: - For 0 < t < 8, choose t = 1: C'(1) = 24(1) - 3(1)^2 = 24 - 3 = 21 > 0, so C is increasing on (0, 8). - For 8 < t < 12, choose t = 10: C'(10) = 24(10) - 3(100) = 240 - 300 = -60 < 0, so C is decreasing on (8, 12). Step 4: The function is strictly decreasing on the interval [8, 12]. Since it is strictly decreasing there, it passes the horizontal line test and is invertible on that interval. Step 5: This is the largest possible interval within [0, 12] where C(t) is strictly decreasing. The answer is [8, 12].

4. f(x) = (x - 12)² + 3 is not invertible. Find the largest domain restriction of the form x ≥ k that makes it invertible. Answer: x ≥ 12 Solution: The function f(x) = (x - 12)² + 3 is a parabola opening upward. The vertex occurs where x - 12 = 0, so at x = 12. To the left of x = 12, the function is decreasing.
   Full step-by-step solution

   Step 1: The function f(x) = (x - 12)² + 3 is a parabola opening upward. Step 2: The vertex occurs where x - 12 = 0, so at x = 12. Step 3: To the left of x = 12, the function is decreasing. To the right of x = 12, the function is increasing. Step 4: For the function to be one-to-one and invertible, we must restrict to either x ≤ 12 or x ≥ 12. Step 5: The problem asks for the largest domain of the form x ≥ k, so we choose x ≥ 12. Step 6: With this restriction, the function passes the horizontal line test and is invertible. The answer is x ≥ 12.

5. f(x) = x² - 6x + 5. Find the largest possible domain restriction of the form x ≤ a or x ≥ a that makes f invertible. Answer: x ≤ 3 or x ≥ 3 Solution: Find the vertex of the parabola f(x) = x² - 6x + 5 The x-coordinate of the vertex is given by x = -b/(2a) = -(-6)/(2×1) = 6/2 = 3 The parabola opens upward (since a = 1 > 0), so it decreases on (-∞, 3] and increases on [3, ∞) To make f invertible, we need to restrict the domain to either (-∞, 3]…
   Full step-by-step solution

   Step 1: Find the vertex of the parabola f(x) = x² - 6x + 5 Step 2: The x-coordinate of the vertex is given by x = -b/(2a) = -(-6)/(2×1) = 6/2 = 3 Step 3: The parabola opens upward (since a = 1 > 0), so it decreases on (-∞, 3] and increases on [3, ∞) Step 4: To make f invertible, we need to restrict the domain to either (-∞, 3] or [3, ∞) Step 5: Both x ≤ 3 and x ≥ 3 are valid domain restrictions that make f invertible Step 6: The largest possible domain restriction of the requested form is x ≤ 3 or x ≥ 3

6. A civil engineer is designing a suspension bridge where the cable shape follows the parabolic function f(x) = -x² + 6x - 5, where x represents the horizontal distance from one tower in hundreds of meters. To ensure the bridge's stability calculations are mathematically sound, she needs to restrict the domain to an interval where the function is invertible. Determine the largest possible domain containing x = 4 where f(x) is one-to-one. Answer: [3, ∞) Solution: Quadratic functions are generally not one-to-one over their entire domain because they are symmetric about their vertex.
   Full step-by-step solution

   Quadratic functions are generally not one-to-one over their entire domain because they are symmetric about their vertex. To make them invertible, we restrict the domain to either side of the vertex where the function is strictly increasing or decreasing. For a downward-opening parabola, the function decreases after the vertex, making the right side of the vertex a suitable domain for invertibility.

7. f(x) = (x - 3)² - 4 is not invertible. Find the largest domain restriction x ≤ a that makes it invertible. Answer: 3 Solution: The function f(x) = (x - 3)² - 4 is a parabola opening upward with vertex at (3, -4). Since it opens upward, the function is decreasing on (-∞, 3] and increasing on [3, ∞).
   Full step-by-step solution

   Step 1: The function f(x) = (x - 3)² - 4 is a parabola opening upward with vertex at (3, -4). Step 2: Since it opens upward, the function is decreasing on (-∞, 3] and increasing on [3, ∞). Step 3: To make it invertible, we need to restrict to either the left or right side of the vertex. Step 4: The largest domain restriction x ≤ a would be the decreasing portion, so a = 3. Step 5: On the domain (-∞, 3], the function passes the horizontal line test and is invertible. The answer is 3.