Math 3.1-3.3 Review Solution Key

3.1-3.3 Review Name: Hour: Honors Pre-Calculus Graph the function. Identify the domain, range, asymptote, and y-intercept of the function. 1. f(x) = 2^x Domain: (-∞, ∞) Range: (0, ∞) Asymptote: y = 0 y-intercept: (0, 1) 2. g(x) = 2^(x+1) Domain: (-∞, ∞) Range: (0, ∞) Asymptote: y = 0 y-intercept: (0, 2) 3. h(x) = -2^(x-1) - 1 Domain: (-∞, ∞) Range: (-∞, -1) Asymptote: y = -1 y-intercept: (0, -9/8) 4. j(x) = (1/2)^x Domain: (-∞, ∞) Range: (0, ∞) Asymptote: y = 0 y-intercept: (0, 1) 5. k(x) = (1/3)^(x-1) Domain: (-∞, ∞) Range: (0, ∞) Asymptote: y = 0 y-intercept: (0, 3) 6. m(x) = (1/2)^(x+1) + 1 Domain: (-∞, ∞) Range: (1, ∞) Asymptote: y = 1 y-intercept: (0, 5/2) 7. When a certain medication is administered to a patient, the number of milligrams remaining in the patient's bloodstream after t hours is modeled by D(t) = 50e^(-0.2t). Round answers to the nearest tenth when necessary. a) What was the initial medication dose given to the patient? 50 mg b) How much medication remains in the bloodstream after 4 hours? 22.4 mg 8. Which of the given interest rates and compounding periods would provide the best investment? Option 1: 8.5% per year, compounded semiannually Option 2: 8.25% per year, compounded quarterly Option   3: 8% per year, compounded continuously   1(1 + 0.085/2)^(2*17) = 1.0868 1(1 + 0.0825/4)^(4*17) = 1.0851 e^(0.08*17) = 1.0833 Option 1 is the best investment. 9. An infectious disease begins to spread in a small city of population 10,000. After t days, the number of persons who have   contracted the disease is modeled by the function   v(t) = 10,000 / (5 + 12450e^(-0.5t)) a) How many infected people are there initially (at time t = 0)? 8 people b) Find the number of infected people after 1 day, 3 days, and 5 days. t = 1: 20.97 t   = 3: 157.32 t = 5: 678.13 10. Write the logarithmic function in exponential form. log₃81 = 4 --> 3^4 = 81 log₁₀(1/1000) = -3 --> 10^-3 = 1/1000 ln(1/e²) = -2 --> e^-2 = 1/e² ln(∛e²) = 2/3 --> e^(2/3) = ∛e² 11. Write the exponential function in logarithmic form. 9^(3/2) = 27 --> log₉27 = 3/2 10^(-4) = 0.0001 --> log₁₀0.0001 = -4 e^0 = 1 --> ln1 = 0 e^4 ≈ 54.6 --> ln54.6 ≈ 4 Evaluate the expression. 18. log₃8 3^x = 8 x = 3/5 19. log₈1 8^x = 1 x = 0 20. log√10000 10^x = (10^4)^(1/2) 10^x = 10² x = 2 21. ln e^-5 -5 22. 5 log₅26 log₅26⁵ 23. 1/2 log₄8 4^(1/2x) = 8^(1/2) 2^x = 2^(3/2) x = 3/4 24. ln √e ln(e^(1/2)) 1/2 25. -3 log₄4 -3 Graph the function. Identify the domain, range, asymptote, and x-intercept of the function. 26. f(x) = log₃x Domain: (0, ∞) Range: (-∞, ∞) Asymptote: x = 0 x-intercept: (1, 0) 27. g(x) = -log₃(x - 1) Domain: (1, ∞) Range: (-∞, ∞) Asymptote: x = 1 x-intercept: (2, 0) 28. h(x) = log₃(-x) - 1 Domain: (-∞, 0) Range: (-∞, ∞) Asymptote: x = 0 x-intercept: (-1, 0) 29. j(x) = ln x Domain: (0, ∞) Range: (-∞, ∞) Asymptote: x = 0 x-intercept: (1, 0) 30. k(x) = ln(x + 2) + 3 Domain: (-2, ∞) Range: (-∞, ∞) Asymptote: x = -2 x-intercept: (e^(-3) + 2, 0) 31. m(x) = ln(2 - x) Domain: (-∞, 2) Range: (-∞, ∞) Asymptote: x = 2 x-intercept: (1, 0) 32. College students in a psychology course took a final exam. As part of an experiment to see how much of the course content   they remembered over time, they took equivalent forms of the exam in monthly intervals thereafter. The average score for the group, f(t), after t months   is modeled by the function f(t) = 88   - 15ln(t + 1), 0 ≤ t ≤ 12. Round answers to the nearest tenth when necessary. a) What was the average score on the original exam? 88 b) What was the average score after 2 months? 4 months? 6 months?   10 months? One year? 2 months:   71.5 4 months: 63.9 6 months: 58.8 10 months: 51.0 12 months: 49.5 c) Use a graphing calculator to graph the function and describe what the graph indicates in terms of the material retained by the students.   As time increases, f(t) decreases, indicating a decline in retained knowledge.