MATH 107 QUIZ 5 NAME: _______________________________ 1. Which of these graphs represent a one-to-one function? (A) (B) (C) (D) 2. The students in history class took a final exam and then took equivalent forms of the exam at monthly intervals thereafter. The average score S(t), as percent, after t months was found to be given by the function: S(t) = 78-15 log(t+1), t0 (a) What was the average score when the sudents initially took the test, t=0? (b) What was the average score after 4 months? 3. Convert to a logarithmic equation: 5x = 251 A. log 5 251 = B. log 251 = 5 C. log 5 = 251 D. log 5 = 251 4. Solve the equation. Check all proposed solutions. + 4 + 2 = 5. Simplify: (a) log 8 1 = (b) log 5 6. Let f (x) = e 1 625 x+1 = - 3. (a) Which describes how the graph of f can be obtained from the graph of y = ex ? A. Shift the graph of y = ex to the left by 1 unit and down by 3 units. B. Shift the graph of y = ex to the right by 1 unit and down by 3 units. C. Reflect the graph of y = ex across the x-axis and shift up by 3 units. D. Reflect the graph of y = ex across the y-axis and shift up by 3 units. (b) What is the domain of f ? (c) What is the range of f ? (d) Sketch the graph of f ? 7. QUADRATIC REGRESSION Data: On a particular summer day, the outdoor temperature was recorded at 8 times of the day, and the following table was compiled. A scatterplot was produced and the parabola of best fit was determined. 90 Temperature on a Summer Day 80 Temperature (degrees) t= Time y = Outdoor of day Temperature (hour) (degrees F.) 7 52 9 67 11 73 13 76 14 78 17 79 20 76 23 61 70 60 50 y = -0.3476t2 + 10.948t - 6.0778 R = 0.9699 40 30 20 10 0 0 4 8 12 16 20 24 Time of Day (hour) Quadratic Polynomial of Best Fit: y = 0.3476t2 + 10.948t 6.0778 where t = Time of day (hour) and y = Temperature (in degrees) REMARKS: The times are the hours since midnight. For instance, 7 means 7 am, and 13 means 1 pm. (a) Using algebraic techniques we have learned, find the maximum temperature predicted by the quadratic model and find the time when it occurred. Report the time to the nearest quarter hour (i.e., __:00 or __:15 or __:30 or __:45). (For instance, a time of 18.25 hours is reported as 6:15 pm.) Report the maximum temperature to the nearest tenth of a degree. (b) Use the quadratic polynomial to estimate the outdoor temperature at 9:30 am, to the nearest tenth of a degree. (c) Use the quadratic polynomial y = 0.3476t + 10.948t 6.0778 together with algebra to estimate the time(s) of day when the outdoor temperature y was 70 degrees. State your results clearly; report the time(s) to the nearest quarter hour. 2 8. Let f(x) = 3x + 1 and g (x) = x2 - 2x - 6 (a) (b) (c) (d) Find the composite function ( )() and simplify the results Find ( )(2) Find the composite function ( )() and simplify the results Find ( )(2) 9. Let ( ) = +6 3 4 (a) Find f 1 , the inverse function of f. (b) What is the domain of f ? (c) What is the domain of the inverse function? (d) What is f (2) ? (e) What is f 1 ( ____ ), where the number in the blank is your answer from part (d)? 10. EXPONENTIAL REGRESSION Data: A cup of hot coffee was placed in a room maintained at a constant temperature of 69 degrees, and the coffee temperature was recorded periodically, in Table 1. TABLE 1 t = Time Elapsed (minutes) 166.0 140.5 125.2 110.3 104.5 98.4 93.9 Temperature Difference (degrees) 0 10 20 30 40 50 60 C = Coffee Temperature (degrees F.) REMARKS: Common sense tells us that the coffee will be cooling off and its temperature will decrease and approach the ambient temperature of the room, 69 degrees. So, the temperature difference between the coffee temperature and the room temperature will decrease to 0. TABLE 2 t = Time Elapsed (minutes) y = C 69 Temperature Difference (degrees F.) 0 10 20 30 40 50 60 97.0 71.5 56.2 41.3 35.5 29.4 24.9 We will fit the temperature difference data (Table 2) to an bt exponential curve of the form y = A e . Notice that as t gets large, y will get closer and closer to 0, which is what the temperature difference will do. So, we want to analyze the data where t = time elapsed and y = C 69, the temperature difference between the coffee temperature and the room temperature. 120 Temperature Difference between Coffee and Room 100 80 y = 89.976e-0.023t R = 0.9848 60 40 20 0 0 10 20 30 40 50 60 70 Time Elapsed (minutes) Exponential Function of Best Fit (using the data in Table 2): y = 89.976 e 0.023 t where t = Time Elapsed (minutes) and y = Temperature Difference (in degrees) Use the exponential function to estimate the coffee temperature C when 35 minutes have elapsed. Report your estimated temperature difference to the nearest tenth of a degree