Inverse Functions

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Answer Key & Explanations

Inverse Functions · Grade 12 · Worksheet 2

1. If f(x) = 12x³ + 19 and g(x) is its inverse, then g(175) = ? Answer: ∛13 Solution: Since g(x) is the inverse of f(x), g(175) means find x such that f(x) = 175.
   Full step-by-step solution

   Step 1: Since g(x) is the inverse of f(x), g(175) means find x such that f(x) = 175. Step 2: Set up the equation: 12x³ + 19 = 175 Step 3: Subtract 19 from both sides: 12x³ = 156 Step 4: Divide both sides by 12: x³ = 13 Step 5: Take the cube root of both sides: x = ∛13 Step 6: Therefore, g(175) = ∛13 The answer is ∛13.

2. Aroha is a volcanologist studying the cooling of lava after an eruption. The temperature T(t) in degrees Celsius of a lava sample t hours after collection is modeled by the function T(t) = 35 + 45e^(-0.15t) for t ≥ 0. Aroha wants to create a graph that shows both the original function and its inverse to better understand how temperature and time are related. She knows the inverse function will be a reflection of T(t) across the line y = x. Find the inverse function T⁻¹(x) that gives the time in hours required for the lava to cool to any given temperature x degrees Celsius. Answer: T⁻¹(x) = -20/3 * ln((x - 35)/45) Solution: Start with the original function: T(t) = 35 + 45e^(-0.15t) Replace T(t) with y: y = 35 + 45e^(-0.15t) Swap the variables x and y: x = 35 + 45e^(-0.15y) Subtract 35 from both sides: x - 35 = 45e^(-0.15y) Divide both sides by 45: (x - 35)/45 = e^(-0.15y) Take the natural logarithm of both sides:…
   Full step-by-step solution

   Step 1: Start with the original function: T(t) = 35 + 45e^(-0.15t) Step 2: Replace T(t) with y: y = 35 + 45e^(-0.15t) Step 3: Swap the variables x and y: x = 35 + 45e^(-0.15y) Step 4: Subtract 35 from both sides: x - 35 = 45e^(-0.15y) Step 5: Divide both sides by 45: (x - 35)/45 = e^(-0.15y) Step 6: Take the natural logarithm of both sides: ln((x - 35)/45) = -0.15y Step 7: Divide both sides by -0.15: y = ln((x - 35)/45) / (-0.15) Step 8: Simplify the fraction: 1/(-0.15) = -20/3, so y = (-20/3) * ln((x - 35)/45) Step 9: Write in inverse notation: T⁻¹(x) = -20/3 * ln((x - 35)/45) The inverse function is T⁻¹(x) = -20/3 * ln((x - 35)/45).

3. Liam is analyzing the temperature fluctuations in a chemical reaction. The temperature T(t) in degrees Celsius at time t hours is modeled by the function T(t) = 3t² - 12t + 15 for t ≥ 0. Liam wants to determine at what time the temperature reaches a specific value. If the inverse function exists for the domain t ≥ 2, find T⁻¹(15). Answer: 4 Solution: T(t) = 3t² - 12t + 15, for t ≥ 0. We are told the inverse function exists for the restricted domain t ≥ 2. We need to find T⁻¹(15).
   Full step-by-step solution

   We are given the temperature function: T(t) = 3t² - 12t + 15, for t ≥ 0. We are told the inverse function exists for the restricted domain t ≥ 2. We need to find T⁻¹(15). --- **Step 1: Interpret T⁻¹(15)** T⁻¹(15) means the time t such that T(t) = 15, with t ≥ 2. --- **Step 2: Set up the equation** T(t) = 15 3t² - 12t + 15 = 15 --- **Step 3: Simplify the equation** Subtract 15 from both sides: 3t² - 12t + 15 - 15 = 0 3t² - 12t = 0 --- **Step 4: Factor** Factor out 3t: 3t(t - 4) = 0 --- **Step 5: Solve for t** So t = 0 or t = 4. --- **Step 6: Apply the domain restriction** The inverse exists for t ≥ 2. Thus t = 0 is not in the restricted domain. Therefore, t = 4 is the only valid solution. --- **Step 7: Conclusion** T⁻¹(15) = 4. --- **Final answer:** 4

4. The function f(x) = x^3 + 2x - 4 has an inverse function f⁻¹(x). Given that f(1) = -1, what is the value of the derivative of the inverse function at x = -1, that is, find (f⁻¹)'(-1)? Answer: 0.2 Solution: We know that for an inverse function, (f⁻¹)'(a) = 1/f'(f⁻¹(a)) We're given that f(1) = -1, which means f⁻¹(-1) = 1 Find the derivative of f(x): f'(x) = 3x^2 + 2 Evaluate f'(1) = 3(1)^2 + 2 = 3 + 2 = 5 Apply the inverse derivative formula: (f⁻¹)'(-1) = 1/f'(f⁻¹(-1)) = 1/f'(1) = 1/5 1/5 = 0.2 The…
   Full step-by-step solution

   Step 1: We know that for an inverse function, (f⁻¹)'(a) = 1/f'(f⁻¹(a)) Step 2: We're given that f(1) = -1, which means f⁻¹(-1) = 1 Step 3: Find the derivative of f(x): f'(x) = 3x^2 + 2 Step 4: Evaluate f'(1) = 3(1)^2 + 2 = 3 + 2 = 5 Step 5: Apply the inverse derivative formula: (f⁻¹)'(-1) = 1/f'(f⁻¹(-1)) = 1/f'(1) = 1/5 Step 6: 1/5 = 0.2 The answer is 0.2.

5. Aroha is an environmental scientist studying the infiltration rate of a pollutant into a groundwater aquifer. She models the concentration C(t) in parts per billion (ppb) at a monitoring well t days after a spill using the function C(t) = 12 + 60 * (2)^(-t/9) for t ≥ 0. To create a quick-reference chart that tells her how many days have passed since the spill for any measured concentration level, she needs to find the inverse function. On the same set of axes, she graphs C(t) and its inverse function t(C). She then verifies that these two graphs are reflections of each other across the line y = x. Determine the equation of the inverse function t(C) that Aroha will use to find the time t (in days) corresponding to a given concentration C (in ppb). Answer: t(C) = -9 * log_2((C - 12)/60) Solution: Write the function using y notation: y = 12 + 60 * (2)^(-t/9). Swap the variables t and y to begin finding the inverse: t = 12 + 60 * (2)^(-y/9).
   Full step-by-step solution

   Step 1: Write the function using y notation: y = 12 + 60 * (2)^(-t/9). Step 2: Swap the variables t and y to begin finding the inverse: t = 12 + 60 * (2)^(-y/9). Step 3: Isolate the exponential term by subtracting 12 from both sides: t - 12 = 60 * (2)^(-y/9). Step 4: Divide both sides by 60: (t - 12)/60 = (2)^(-y/9). Step 5: Apply the base-2 logarithm to both sides: log_2((t - 12)/60) = -y/9. Step 6: Multiply both sides by -9 to solve for y: -9 * log_2((t - 12)/60) = y. Step 7: Rewrite using inverse function notation with the appropriate variables. Since the original function was C(t) = 12 + 60 * (2)^(-t/9), the inverse function, which gives time t from concentration C, is: t(C) = -9 * log_2((C - 12)/60). Step 8: (Graphical verification) The original function has a horizontal asymptote at C = 12, so the inverse function has a vertical asymptote at t = 12. The point (0, 72) is on C(t) because C(0) = 12 + 60*1 = 72. Therefore, the point (72, 0) must be on the graph of t(C). Substituting C = 72 into the inverse: t(72) = -9 * log_2((72 - 12)/60) = -9 * log_2(60/60) = -9 * log_2(1) = -9 * 0 = 0. This confirms the reflection property. The final answer is t(C) = -9 * log_2((C - 12)/60).

6. Given f(x) = 5x + 10, sketch the graph of f and its inverse f⁻¹ on the same coordinate plane. Then, verify that the point (0, 10) on f reflects to (10, 0) on f⁻¹ across the line y = x. Answer: f⁻¹(x) = (x - 10)/5 Solution: Write f(x) = 5x + 10. This is a line with slope 5 and y-intercept 10. Find the inverse algebraically.
   Full step-by-step solution

   Step 1: Write f(x) = 5x + 10. This is a line with slope 5 and y-intercept 10. Step 2: Find the inverse algebraically. Replace f(x) with y: y = 5x + 10. Swap x and y: x = 5y + 10. Solve for y: x - 10 = 5y, so y = (x - 10)/5. Thus f⁻¹(x) = (x - 10)/5. Step 3: Graph f by plotting points. For x = 0, f(0) = 10 gives point (0, 10). For x = 5, f(5) = 35 gives (5, 35). For x = -2, f(-2) = 0 gives (-2, 0). Step 4: Graph f⁻¹ by swapping coordinates of points on f. (0, 10) becomes (10, 0). (5, 35) becomes (35, 5). (-2, 0) becomes (0, -2). Plot these and draw the line through them. Step 5: Draw the line y = x (dashed) as the mirror. Verify that (0, 10) on f reflects to (10, 0) on f⁻¹: the midpoint of (0, 10) and (10, 0) is (5, 5), which lies on y = x, and the segment connecting them is perpendicular to y = x (slope of segment is -1, slope of y = x is 1). The answer is f⁻¹(x) = (x - 10)/5.