AP Precalculus SL Fall 2024 Exam Review PDF
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Uploaded by PatientEuphemism2210
Ball State University
2024
AP
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Summary
This is an AP Precalculus SL exam review for Fall 2024, covering topics including exponential functions, average rate of change calculations, and solving equations. Sample questions are included.
Full Transcript
# AP Precalculus – SL (Fall 2024) ## Exam Review 1. A wind turbine uses the power of wind to generate electricity. The blades of the turbine make a noise that can be heard at a distance from the turbine. At a distance of d = 0 meters from the turbine, the noise level is 105 decibels. At a distance...
# AP Precalculus – SL (Fall 2024) ## Exam Review 1. A wind turbine uses the power of wind to generate electricity. The blades of the turbine make a noise that can be heard at a distance from the turbine. At a distance of d = 0 meters from the turbine, the noise level is 105 decibels. At a distance of d = 100 meters from the turbine, the noise level is 49 decibels. The noise level can be modeled by the function S given by S(d) = *ab<sup>d</sup>*, where S(d) is the noise level, in decibels, at a distance of d meters from the turbine. a. Use the given data to write two equations that can be used to find the values for constants *a* and *b* in the expression for S(d). * S(0) = 105 * S(100) = 49 * ab<sup>0</sup> = 105 * ab<sup>100</sup> = 49 b. Find the values for *a* and *b*. * a = 105 * ab<sup>100</sup> = 49 * 105b<sup>100</sup> = 49 * b<sup>100</sup> = 49/ 105 * b = (49/105)<sup>1/100</sup> * b ≈ 0.992408 c. Use the given data to find the average rate of change of the noise level, in decibels per meter, from d = 0 to d = 100 meters. Express your answer as a decimal approximation. Show the computations that lead to your answer. * Average Rate of Change = (S(100) - S(0)) / (100 - 0) * = (49 - 105) / 100 * = -0.56 decibels per meter d. Interpret the meaning of your answer from (c) in the context of the problem. On average, as the distance from the turbine increases from 0 to 100 meters, the noise level decreases 0.56 decibels per meter. e. Use the average rate of change found in (c) to estimate the noise level, in decibels, at a distance of d = 120 meters. Show the work that leads to your answer. * y - y<sub>1</sub> = m(x - x<sub>1</sub>) * y - 105 = -0.56(120 - 0) * y - 105 =- 67.2 * y = 37.8 The noise level at a distance of d = 120 meters is 37.8 decibels. f. Use the model S to find the value of m such that the noise level is 20 decibels at a distance of m meters from the turbine. * S(d) = *ab<sup>d</sup>* * S(d) = 105(0.992408)<sup>d</sup> * 20 = 105(0.992408)<sup>d</sup> * d = log(20/105) / log(0.992408) * d ≈ 217.585 2. a. Use the following functions to solve the following problems. * g(x) = 4x - 1 * h(x) = 5 - 2x * j(x) = 1 - x<sup>2</sup> i. (g o h)(2) * (g o h)(x) = g(h(x)) * g(h(2)) = g(5 - 2 * 2) = g(1) * g(1) = 4(1) - 1 = 3 * (g o h)(2) = 3 ii. (j o h)(x) * (j o h)(x) = j(h(x)) * j(h(x)) = j(5 - 2x) = 1 - (5 - 2x)<sup>2</sup> * = 1 - (25 - 20x + 4x<sup>2</sup>) * = 1 - 25 + 20x - 4x<sup>2</sup> * = -4x<sup>2</sup> + 20x - 24 * = x<sup>2</sup> - 5x + 6 * = (x - 2)(x - 3) b. Find f<sup>-1</sup>(x) when f(x) = (x - 4)<sup>2</sup> + 3. For what values of x is f<sup>-1</sup>(f(x)) = x? For what values of x is f(f<sup>-1</sup>(x)) = x? * y = (x - 4)<sup>2</sup> + 3 * x = (y - 4)<sup>2</sup> + 3 * x - 3 = (y - 4)<sup>2</sup> * √(x - 3) = y - 4 * y = √(x - 3) + 4 * f<sup>-1</sup>(x) = √(x - 3) + 4; x ≥ 3 For values of x ≥ 3, f<sup>-1</sup>(f(x)) = x and f(f<sup>-1</sup>(x)) = x. 3. What exponential function goes through the following points a. (1, 8) and (2, 4) * f(x) = Ca<sup>x</sup> * a = 4 / 8 = 1/2 * f(1) = Ca<sup>1</sup> = 8 * C(1/2) = 8 * C = 16 * f(x) = 16(1/2)<sup>x</sup> b. (0, 4) and (1, 12) * f(x) = Ca<sup>x</sup> * a = 12 / 4 = 3 * f(0) = Ca<sup>0</sup> = 4 * C = 4 * f(x) = 4 * 3<sup>x</sup> 4. a. Solve the exponential equation 9<sup>-x+15</sup> = 27<sup>x</sup> for x. Express your answer as a decimal approximation. * 9<sup>-x+15</sup> = 27<sup>x</sup> * 3<sup>2(-x+15)</sup> = 3<sup>3x</sup> * 2(-x + 15) = 3x * -2x + 30 = 3x * 5x = 30 * x = 6 b. If f(x) = 2x<sup>2</sup> + 5 and g(x) = 3x + a, find a so that the graph of f o g crosses the y-axis at 23. * (f o g)(x) = 2(3x + a)<sup>2</sup> + 5 * = 2(9x<sup>2</sup> + 6xa + a<sup>2</sup>) + 5 * = 18x<sup>2</sup> + 12xa + 2a<sup>2</sup> + 5 The graph crosses the y-axis when x = 0. * 18(0)<sup>2</sup> + 12(0)a + 2a<sup>2</sup> + 5 = 23 * 2a<sup>2</sup> + 5 = 23 * 2a<sup>2</sup> = 18 * a<sup>2</sup> = 9 * a = ±3