Invertible Functions

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Answer Key & Explanations

Invertible Functions · Grade 12 · Worksheet 2

1. Noah is an environmental engineer modeling the temperature of a chemical reaction over time. The temperature in degrees Celsius is given by the function f(x) = x^3 - 12x^2 + 36x + 1, where x represents time in minutes. To analyze the reaction's behavior during a specific phase, Noah needs to restrict the domain to an interval where the function is strictly decreasing, making it invertible. Determine the largest possible interval of the form [a, b] where f(x) is strictly decreasing and therefore invertible. Answer: [2, 6] Solution: Find the derivative of f(x) = x^3 - 12x^2 + 36x + 1. f'(x) = 3x^2 - 24x + 36. Set the derivative equal to zero to find critical points: 3x^2 - 24x + 36 = 0.
   Full step-by-step solution

   Step 1: Find the derivative of f(x) = x^3 - 12x^2 + 36x + 1. f'(x) = 3x^2 - 24x + 36. Step 2: Set the derivative equal to zero to find critical points: 3x^2 - 24x + 36 = 0. Divide by 3: x^2 - 8x + 12 = 0. Factor: (x - 2)(x - 6) = 0. So x = 2 and x = 6 are critical points. Step 3: Analyze the sign of the derivative on the intervals determined by the critical points: For x < 2, test x = 0: f'(0) = 3(0)^2 - 24(0) + 36 = 36 > 0, so f is increasing. For 2 < x < 6, test x = 4: f'(4) = 3(16) - 24(4) + 36 = 48 - 96 + 36 = -12 < 0, so f is decreasing. For x > 6, test x = 7: f'(7) = 3(49) - 24(7) + 36 = 147 - 168 + 36 = 15 > 0, so f is increasing. Step 4: The function is strictly decreasing on the interval [2, 6] (including the endpoints where the derivative is zero, as the function is still one-to-one on that closed interval). This is the largest interval where f is strictly decreasing. The answer is [2, 6].

2. 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.

3. A pharmaceutical company is modeling the concentration of a new drug in the bloodstream over time using the function C(t) = (t² - 9)/(t - 3), where t represents hours after administration. The function is undefined at t = 3 due to division by zero. To make the function continuous and invertible for their pharmacokinetic analysis, they need to restrict the domain by removing the problematic point. What is the maximum domain on which this function becomes both continuous and invertible? Answer: (-∞, 3) ∪ (3, ∞) Solution: C(t) = (t² - 9)/(t - 3) t² - 9 = (t - 3)(t + 3) C(t) = (t - 3)(t + 3) / (t - 3) For t ≠ 3, we can cancel (t - 3): C(t) = t + 3 The original function is undefined at t = 3 because of division by zero.
   Full step-by-step solution

   Let's go step by step. --- **Step 1: Simplify the function** We are given: C(t) = (t² - 9)/(t - 3) Factor the numerator: t² - 9 = (t - 3)(t + 3) So: C(t) = (t - 3)(t + 3) / (t - 3) For t ≠ 3, we can cancel (t - 3): C(t) = t + 3 --- **Step 2: Identify the discontinuity** The original function is undefined at t = 3 because of division by zero. After simplification, we get C(t) = t + 3, but this simplification is only valid for t ≠ 3. At t = 3, the original function is undefined, so there is a hole in the graph at (3, 6). --- **Step 3: Continuity requirement** A function is continuous on an interval if it has no holes, jumps, or vertical asymptotes in that interval. Here, the only discontinuity is at t = 3. So, if we remove t = 3 from the domain, the function becomes continuous on each of the intervals (-∞, 3) and (3, ∞). --- **Step 4: Invertibility requirement** A function is invertible if it is one-to-one (passes the horizontal line test). C(t) = t + 3 is a straight line with slope 1, so it is one-to-one over its entire domain. But if we include t = 3, the function is not defined there, so we cannot include it. Also, if we take the entire real line except t = 3, the function is still one-to-one because for any two different t₁ and t₂ in (-∞, 3) ∪ (3, ∞), we have C(t₁) ≠ C(t₂) unless t₁ and t₂ are symmetric around 3? Let's check: Suppose C(a) = C(b) ⇒ a + 3 = b + 3 ⇒ a = b. So it's one-to-one on the whole domain excluding t = 3. --- **Step 5: Maximum domain for both continuity and invertibility** We need the largest set of real numbers where C(t) is both continuous and invertible. If we take (-∞, 3) ∪ (3, ∞): - Continuous? Yes, because the only discontinuity (t = 3) is removed. - Invertible? Yes, because it's one-to-one on this domain. If we take only (-∞, 3) or only (3, ∞), it would still be continuous and invertible, but that's not the *maximum* domain — the problem asks for the maximum domain. Maximum domain = all real numbers except t = 3. --- **Step 6: Final answer** In interval notation: (-∞, 3) ∪ (3, ∞) --- **Final Answer:** (-∞, 3) ∪ (3, ∞)

4. Consider the function f(x) = sqrt(x^2 - 6x + 8). Find the smallest integer value in the domain of f(x) that would make f invertible on the restricted domain [a,∞). Answer: 4 Solution: Find the domain by solving x^2 - 6x + 8 ≥ 0 Factor the quadratic: (x - 2)(x - 4) ≥ 0 The roots are x = 2 and x = 4, and the parabola opens upward The inequality holds when x ≤ 2 or x ≥ 4 To make f invertible, we need to restrict to either (-∞, 2] or [4, ∞) The vertex of x^2 - 6x + 8 is at x = 3…
   Full step-by-step solution

   Step 1: Find the domain by solving x^2 - 6x + 8 ≥ 0 Step 2: Factor the quadratic: (x - 2)(x - 4) ≥ 0 Step 3: The roots are x = 2 and x = 4, and the parabola opens upward Step 4: The inequality holds when x ≤ 2 or x ≥ 4 Step 5: To make f invertible, we need to restrict to either (-∞, 2] or [4, ∞) Step 6: The vertex of x^2 - 6x + 8 is at x = 3 Step 7: On [4, ∞), the function is strictly increasing, making it one-to-one Step 8: The smallest integer in [4, ∞) is 4 Step 9: Therefore, the smallest integer value in the restricted domain that makes f invertible is 4

5. A biologist is modeling the growth of a bacterial culture using the function f(x) = x² - 4x + 3. To make this function invertible for her population study, she needs to restrict its domain. Determine the largest possible domain containing x = 5 where f(x) becomes one-to-one. Answer: [4, ∞) Solution: For a quadratic function to be invertible, we must restrict its domain to either side of the vertex where it is strictly increasing or decreasing.
   Full step-by-step solution

   For a quadratic function to be invertible, we must restrict its domain to either side of the vertex where it is strictly increasing or decreasing. The vertex represents the turning point where the function changes from increasing to decreasing or vice versa. By choosing the appropriate side of the vertex, we ensure the function passes the horizontal line test and becomes one-to-one.

6. A solid is formed by rotating the region bounded by the curves y = x² and y = 4 about the x-axis. This creates a three-dimensional volume. Using the method of cylindrical shells, set up the integral that represents the volume of this solid. Describe the geometric configuration and the reasoning behind choosing the shell method for this particular bounded region. Answer: 2π∫₀² x(4 - x²) dx Solution: The method of cylindrical shells is particularly useful when integrating with respect to the variable perpendicular to the axis of rotation.
   Full step-by-step solution

   The method of cylindrical shells is particularly useful when integrating with respect to the variable perpendicular to the axis of rotation. For volumes of revolution, shells are cylindrical elements whose volume is the circumference times height times thickness. This approach often simplifies integration when the region is bounded by functions that are easier to work with in a particular orientation relative to the axis of rotation.