8.2 Exercises. Section 8.2 Exponential Functions 783

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1 Section 8.2 Eponential Functions Eercises 1. The current population of Fortuna is 10,000 heart souls. It is known that the population is growing at a rate of 4% per ear. Assuming this rate remains constant, perform each of the following tasks. a. Set up an equation that models the population P (t) as a function of time t. b. Use the model in the previous part to predict the population 40 ears from now. c. Use our calculator to sketch the graph of the population over the net 40 ears. 2. The population of the town of Imagination currentl numbers 12,000 people. It is known that the population is growing at a rate of 6% per ear. Assuming this rate remains constant, perform each of the following tasks. a. Set up an equation that models the population P (t) as a function of time t. b. Use the model in the previous part to predict the population 30 ears from now. c. Use our calculator to sketch the graph of the population over the net 30 ears. 3. The population of the town of Despairia currentl numbers 1,000 individuals. It is known that the population is decaing at a rate of % per ear. Assuming this rate remains constant, perform each of the following tasks. a. Set up an equation that models the population P (t) as a function of time t. b. Use the model in the previous part to predict the population 0 ears from now. c. Use our calculator to sketch the graph of the population over the net 0 ears. 4. The population of the town of Hopeless currentl numbers 2,000 individuals. It is known that the population is decaing at a rate of 6% per ear. Assuming this rate remains constant, perform each of the following tasks. a. Set up an equation that models the population P (t) as a function of time t. b. Use the model in the previous part to predict the population 40 ears from now. c. Use our calculator to sketch the graph of the population over the net 40 ears. In Eercises -12, perform each of the following tasks for the given function. a. Find the -intercept of the graph of the function. Also, use our calculator to find two points on the graph to the right of the -ais, and two points to the left. b. Using our five points from (a) as a guide, set up a coordinate sstem on graph paper. Choose and label appropriate scales for each ais. Plot the five points, and an additional points ou feel are necessar to dis- 1 Coprighted material. See:

2 784 Chapter 8 Eponential and Logarithmic Functions cern the shape of the graph. c. Draw the horizontal asmptote with a dashed line, and label it with its equation. d. Sketch the graph of the function. e. Use interval notation to describe both the domain and range of the function f() = (2.) 6. f() = (0.1) 7. f() = (0.7) 8. f() = (1.1) f() = f() = f() = f() = In Eercises 13-20, the graph of an eponential function of the form f() = b + c is shown. The dashed red line is a horizontal asmptote. Determine the range of the function. Epress our answer in interval notation

3 Section 8.2 Eponential Functions f() = (/2) ; p = f() = 9 ; p = f() = ; p = f() = 9 ; p = f() = (6/) ; p = f() = (3/) ; p = 0 In Eercises 33-40, use our calculator to evaluate the function at the given value p. Round our answer to the nearest hundredth. 33. f() = 10 ; p = f() = 10 ; p = f() = (2/) ; p = f() = 2 ; p = 3/ f() = 10 ; p = f() = 7 ; p = 4/ f() = 10 ; p = 1/. 40. f() = (4/3) ; p = 1.1. In Eercises 21-32, compute f(p) at the given value p. 21. f() = (1/3) ; p = f() = (3/4) ; p = f() = ; p = 24. f() = (1/3) ; p = 4 2. f() = 4 ; p = f() = ; p = This eercise eplores the propert that eponential growth functions eventuall increase rapidl as increases. Let f() = 1.0. Use our graphing calculator to graph f on the intervals (a) [0, 10] and (b) [0, 100]. For (a), use Ymin = 0 and Yma = 10. For (b), use Ymin = 0 and Yma = 100. Make accurate copies of the images in our viewing window on our homework paper. What do ou observe when ou compare the two graphs?

4 Chapter 8 Eponential and Logarithmic Functions 8.2 Solutions 1. a) Let P (t) represent the population t ears from now. Then, the initial population (at time t = 0) is P (0) = The population is growing at a rate of 4% per ear. Therefore, the population at the end of an ear will be 104% of the population at the end of the previous ear. Thus, at the end of the first ear, the population will be 104% of the initial population. In smbols, P (1) = 1.04P (0) = 1.04(10 000). At the end of the second ear, the population will be 104% of the population at the end of the first ear. In smbols, P (2) = 1.04P (1) = 1.04(1.04(10 000)) = (1.04) 2 (10 000). At the end of the third ear, the population will be 104% of the population at the end of the second ear. In smbols, P (3) = 1.04P (2) = 1.04(1.04) 2 (10 000) = (1.04) 3 (10 000). Thus, the pattern is formed. The population at the end of t ears is P (t) = (1.04) t (10 000), or equivalentl, P (t) = (1.04) t. b) To find the population at the end of 40 ears, set t = 40 and compute P (40) = (1.04) c) The graph of the population over the net 40 ears is displaed in the sequence of calculator snapshots that follow. Note especiall how we made appropriate settings for the domain and range in (b). (a) (b) (c)

5 Section 8.2 Eponential Functions 3. a) Let P (t) represent the population t ears from now. Then, the initial population (at time t = 0) is P (0) = The population is decaing at a rate of % per ear. Therefore, the population at the end of an ear will be 9% of the population at the end of the previous ear. Thus, at the end of the first ear, the population will be 9% of the initial population. In smbols, P (1) = 0.9P (0) = 0.9(1 000). At the end of the second ear, the population will be 9% of the population at the end of the first ear. In smbols, P (2) = 0.9P (1) = 0.9(0.9(1 000)) = (0.9) 2 (1 000). At the end of the third ear, the population will be 9% of the population at the end of the second ear. In smbols, P (3) = 0.9P (2) = 0.9(0.9) 2 (1 000) = (0.9) 3 (1 000). Thus, the pattern is formed. The population at the end of t ears is P (t) = (0.9) t (1 000), or equivalentl, P (t) = 1 000(0.9) t. b) To find the population at the end of 0 ears, set t = 0 and compute P (0) = 1 000(0.9) c) The graph of the population over the net 0 ears is displaed in the sequence of calculator snapshots that follow. Note especiall how we made appropriate settings for the domain and range in (b). (a) (b) (c)

6 Chapter 8 Eponential and Logarithmic Functions. a) The -intercept is (0, 1). Evaluate the function at = 1, 2, 1, 2 to obtain the points (1, 2.), (2, 6.2), ( 1, 0.4), ( 2, 0.16) (other answers are possible). b) See the graph in part (d). c) Since the base 1. is larger than 1, this is an eponential growth function. Therefore, = 0 is a horizontal asmptote on the left side of the graph. See the graph in part (d). d) 10 f()=(2.) =0 3 e) Domain = (, ), Range = (0, ) 7. a) The -intercept is (0, 1). Evaluate the function at = 1, 2, 1, 2 to obtain the points (1, 0.7), (2, 0.6), ( 1, 1.34), ( 2, 1.78) (other answers are possible). b) See the graph in part (d). c) Since the base 0.7 is smaller than 1, this is an eponential deca function. Therefore, = 0 is a horizontal asmptote on the right side of the graph. See the graph in part (d). d) f()=(0.7) =0

7 Section 8.2 Eponential Functions e) Domain = (, ), Range = (0, ) 9. a) The -intercept is (0, 2). Evaluate the function at = 1, 2, 1, 2 to obtain the points (1, 4), (2, 10), ( 1, 1.34), ( 2, 1.11) (other answers are possible). b) See the graph in part (d). c) The graph of f can be obtained from the graph of p() = 3 b a vertical shift up 1 unit. Therefore, the horizontal asmptote = 0 of the graph of p will also be shifted up 1 unit, so the graph of f has a horizontal asmptote = 1. See the graph in part (d). d) 20 f()=3 +1 =1 3 e) Domain = (, ), Range = (1, ) 11. a) The -intercept is (0, 2). Evaluate the function at = 1, 2, 1, 2 to obtain the points (1, 1), (2, 1), ( 1, 2.), ( 2, 2.7) (other answers are possible). b) See the graph in part (d). c) The graph of f can be obtained from the graph of p() = 2 b a vertical shift down 3 units. Therefore, the horizontal asmptote = 0 of the graph of p will also be shifted down 3 units, so the graph of f has a horizontal asmptote = 3. See the graph in part (d).

8 Chapter 8 Eponential and Logarithmic Functions d) f()=2 3 = 3 e) Domain = (, ), Range = ( 3, ) 13. Project all points on the graph onto the -ais. This is shaded in red in the figure below. Thus, the range is the set of all real numbers greater than 1. In interval notation, the range equals ( 1, ). 1. Project all points on the graph onto the -ais. This is shaded in red in the figure below. Thus, the range is the set of all real numbers greater than 2. In interval notation, the range equals (2, ).

9 Section 8.2 Eponential Functions 17. Project all points on the graph onto the -ais. This is shaded in red in the figure below. Thus, the range is the set of all real numbers greater than 2. In interval notation, the range equals (2, ). 19. Project all points on the graph onto the -ais. This is shaded in red in the figure below. Thus, the range is the set of all real numbers greater than 2. In interval notation, the range equals ( 2, ). 21. f( 4) = ( ) 1 4 = 1 3 ( = ) 81 = f() = = f( 4) = 4 4 = = f( 3) = ( ) 3 = 1 2 ( ) = = f( 4) = 4 = 1 4 = f( 4) = ( ) 6 4 = 1 ( 6 ) = = Using a calculator, f( 0.7) = Using a calculator, f(3.67) = (2/)

10 Chapter 8 Eponential and Logarithmic Functions 37. Using a calculator, f(2.07) = Using a calculator, f( 1/) = 10 1/ a) The graph on the interval [0, 10] increases ver slowl. In fact, the graph looks almost linear. b) The graph on the interval [0, 100] increases slowl at first, but then increases ver rapidl on the second half of the interval.