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 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].

3. A civil engineer is designing a parabolic arch bridge that follows the equation f(x) = -x² + 8x - 12, where x represents the horizontal distance from the left support in meters. To ensure the bridge's structural analysis can be properly modeled, she needs to restrict the domain to make the function invertible while maintaining the portion where the arch is increasing. Determine the largest possible interval of the form [a, b] where the function is strictly increasing and therefore invertible. Answer: [2, 4] Solution: Find the derivative of f(x) = -x² + 8x - 12 f'(x) = -2x + 8 Set the derivative equal to zero to find critical points -2x + 8 = 0 -2x = -8 x = 4 For x < 4: f'(x) = -2x + 8 > 0 (since -2x + 8 > 0 when x < 4) For x > 4: f'(x) = -2x + 8 < 0 (since -2x + 8 < 0 when x > 4) The function is increasing…
   Full step-by-step solution

   Step 1: Find the derivative of f(x) = -x² + 8x - 12 f'(x) = -2x + 8 Step 2: Set the derivative equal to zero to find critical points -2x + 8 = 0 -2x = -8 x = 4 Step 3: Analyze the sign of the derivative For x < 4: f'(x) = -2x + 8 > 0 (since -2x + 8 > 0 when x < 4) For x > 4: f'(x) = -2x + 8 < 0 (since -2x + 8 < 0 when x > 4) Step 4: The function is increasing on (-∞, 4) and decreasing on (4, ∞) Step 5: Find the left endpoint of the bridge by solving f(x) = 0 -x² + 8x - 12 = 0 Multiply by -1: x² - 8x + 12 = 0 Factor: (x - 2)(x - 6) = 0 x = 2 or x = 6 Step 6: The bridge spans from x = 2 to x = 6 Step 7: The largest interval where the function is strictly increasing is from the left support to the vertex This is [2, 4] The answer is [2, 4].

4. f(x) = (x - 6)² + 1. Restrict the domain to make it invertible. Answer: x ≥ 6 Solution: The function f(x) = (x - 6)² + 1 is a parabola opening upward with vertex at (6, 1). This parabola fails the horizontal line test over its entire domain because it is symmetric about x = 6.
   Full step-by-step solution

   Step 1: The function f(x) = (x - 6)² + 1 is a parabola opening upward with vertex at (6, 1). Step 2: This parabola fails the horizontal line test over its entire domain because it is symmetric about x = 6. Step 3: To make it invertible, we restrict the domain 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, making the function one-to-one and invertible. Step 5: Therefore, the domain restriction is x ≥ 6.

5. A solid is formed by rotating the region bounded by the curve y = x³, the x-axis, and the vertical line x = 1 about the y-axis. Using the method of cylindrical shells, set up the integral expression for the volume of this solid. Describe the visual geometric elements: the cubic curve, the bounded region under the curve from x=0 to x=1, and the rotation around the y-axis creating a three-dimensional volume. Answer: 2π∫₀¹ x⁴ dx Solution: Identify the region being rotated: bounded by y = x³, y = 0 (x-axis), and x = 1 from x = 0 to x = 1. When rotating around the y-axis using cylindrical shells, the radius of a shell at position x is x.
   Full step-by-step solution

   Step 1: Identify the region being rotated: bounded by y = x³, y = 0 (x-axis), and x = 1 from x = 0 to x = 1. Step 2: When rotating around the y-axis using cylindrical shells, the radius of a shell at position x is x. Step 3: The height of a shell at position x is given by the function value y = x³. Step 4: The thickness of each shell is dx. Step 5: The volume of each shell is 2π × radius × height × thickness = 2π × x × x³ × dx = 2πx⁴ dx. Step 6: Integrate from x = 0 to x = 1: Volume = ∫₀¹ 2πx⁴ dx. Step 7: The integral expression is 2π∫₀¹ x⁴ dx.

6. 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.