Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 34 pps

9.2 Exponential: The Bare Bones 311 Exponentiation does not distribute over addition or subtraction. Enough parentheses must be used when entering an exponential into your calculator or computer so that the typed expression is interpreted as you intended. These points are illustrated below. bc 2 =bcc =(bc) 2 e.g., for x = 0, −x 2 is negative while (−x) 2 is positive. (ab) n = a n b n e.g., (−2x) 2 =(−2) 2 ·x 2 =4x 2 =−2x 2 =−(2x 2 )  a b  n = a n b n e.g.,  4 3 =  4 3  1/2 = 4 1/2 3 1/2 = 2 3 1/2 = 2 √ 3 But (a + b) n =a n +b n e.g., (2 +3) 2 = 5 2 = 25 but 2 2 + 3 2 = 4 + 9 = 13. (a − b) n =a n −b n e.g., √ 25 − 9 =(25 − 9) 1/2 = 16 1/2 = 4 but 25 1/2 − 9 1/2 = 5 − 3 = 2. Entering 2 * x ∧ 5 + √ 2 in a calculator gives 2x 5 + √ 2 not 2x 5+ √ 2 or (2x) 5+ √ 2 . To enter (2x) 5+ √ 2 you can type ( 2 x ) ∧ ( 5 + √ 2 ) . Different calculators may have slightly different conventions. Check yours out either by using the instruction manual or by playing around with numbers you know. Rounding off the bases of exponentials can cause substantial inaccuracy. Let your calculator work for you; do not round off early. EXERCISE 9.2 Use a calculator or computer to evaluate the following to two decimal places. (a) π 3 (1 + 2 3 ) 50/3 (b) 100(1 + 0.05 12 ) 12·60 (c) (3.001) 3.001 −3 3 0.001 Answer (a) 154500.10 (b) 1996.07 (c) 56.73 EXERCISE 9.3 Each statement that follows is incorrect. Find the incorrect step and correct it. 312 CHAPTER 9 Exponential Functions i) 3  −8x −2 = 3 √ −8 3  x −2 = 1 2 x −2/3 ii)  B 3 C −1 2B  −3 =  B 2 C −1 2  −3 = B −6 C 3 2 −3 = 2 −3 C 3 B 6 = C 3 8B 6 iii)  4x 2 + 16y 2 =  4(x 2 + 4y 2 ) = √ 4  x 2 + 4y 2 = 2(x + 2y) =2x +4y iv) A n B n + C n D n = A n D n + C n B n B n D n = (AD) n + (CB) n (BD) n =  AD + CB BD  n v)  1 2  −1  1 x + 1 y  −1 = 2  1 x + 1 y  −1 = 2(x + y) =2x +2y Answers are provided at the end of the chapter. EXERCISE 9.4 Simplify the following. i) √ 2 √ x 3 (2x) −1/2 ii) 4(9y −x ) 1/2 iii) b x+w − b x b x iv) ( 3 √ 64x 3 ) 1/2  1 2  −2 v) Q 3R + Q R+1 Q 2R Answers are provided at the end of the chapter. The Exponential Function The function f is called an exponential function if it can be written in the form f(t)=Cb t , where C is a constant and b is a positive constant. This function is called exponential because the variable, t, is in the exponent. The domain of the exponential function is all real numbers, and the range is all positive real numbers. If a quantity Q grows or decays exponentially with time t, then Q(t) =Cb t for some constants C and b. There are two unknown constants, so two data points are necessary in order to find them. At t = 0, Q(0) = Cb 0 so C =Q(0). Sometimes the quantity Q(0) is denoted by Q 0 , indicating the initial quantity; we write Q(t) = Q 0 b t . In Figure 9.3(a) on the following page are the graphs of f(t)=b t for b = 1, 2, 3, and 10. In Figure 9.3(b) are graphs of f(t)=b −t for b = 2, 3, and 10 or, equivalently, of f(t)=b t for b = 1 2 , 1 3 , and 1 10 . The key notion is that numbers greater than 1 increase when squared, cubed, etc. 2 2 = 4; 2 3 = 8; 2 4 = 16; 2 5 = 32; while numbers between 0 and 1 decrease when squared, cubed, etc.  1 2  2 = 1 4 ;  1 2  3 = 1 8 ;  1 2  4 = 1 16 ;  1 2  5 = 1 32 . Recall that x −k = 1 x k . Therefore,  1 2  t = 1 t 2 t = 1 2 t = 2 −t ; the graph of ( 1 2 ) t = 2 −t can be obtained by reflecting the graph of 2 t across the y-axis. Similarly,  1 3  t = 3 −t and  1 10  t = 10 −t . Therefore, the functions graphed in Figure 9.3(b) can also be written as f(t)=  1 2  t , f(t)=  1 3  t , and f(t)=  1 10  t , respectively. 9.2 Exponential: The Bare Bones 313 123 4 6 8 10 12 2 f=2 t f=3 t f=10 t t f f=1 t (a) (b) –1–2–3123 2 4 6 8 10 12 t f f=3 -t f=10 -t f=2 -t Figure 9.3 Observe that b t is always positive. Suppose b>1. i. The graph of b t is increasing and concave up. b t increases at an ever-increasing rate. ii. The more negative t becomes the smaller b t becomes. In fact, by taking t negative enough, b t can be made arbitrarily close to zero. lim t→−∞ b t = 0. Suppose 0 <b<1. i. The graph of b t is decreasing and concave up. ii. As t increases without bound b t gets arbitrarily close to 0. We can write this as t →∞,b t →0;equivalently, lim t→∞ b t = 0. The graph of b t has a horizontal asymptote at y = 0. EXERCISE 9.5 (a) Using your graphing calculator or computer, graph f(t)=C·2 t for different values of C. (Be sure to include both positive and negative values of C.) How does the value of C affect the graph? Does it alter the y-intercept? The horizontal asymptote? If something changes, explain how it changes. Do your answers make sense to you? (b) Using your graphing calculator or computer, graph f(t)=2 −t +D. How does the value of D affect the graph? Does it alter the y-intercept? The horizontal asymptote? If something changes, explain how it changes. Do your answers make sense to you? (c) Given your answers to parts (a) and (b), take a guess at what the graph of f(x)=3−2 x looks like. (Note: This function is the same as g(t) =−2 t +3.) 6 Check your answer using your graphing calculator. Then try h(t) =−3−2 −t . (d) Using your graphing calculator, graph f(t)=2 kt for different values of k. (Be sure to include both positive and negative values of k.) How does the value of k affect the 6 Did changing the order in which the terms were written make the problem easier for you? If so, you need to practice fiddling around with the form in which expressions are written so you can help yourself by transforming an expression that looks unfamiliar into one that looks familiar. This is a standard problem-solving strategy. 314 CHAPTER 9 Exponential Functions graph? Does it alter the y-intercept? The horizontal asymptote? If something changes, explain how it changes. Rewriting f(t)as  2 k  t helps in interpreting the results. Manipulating Exponents: Taking Control The three exponent laws given at the beginning of this section are identities; they can be read from left to right or from right to left. Flexibility and a sense of your own goals helps in applying them when working with exponential functions. ◆ EXAMPLE 9.4 In the beautiful desolate desert of the Sinai, garbage disposal is becoming a growing problem. An observer hypothesizes that in a certain region of the Sinai the number of plastic bags littering the desert is increasing exponentially. Two months after he arrived he estimates there are 100 plastic bags littering the region, and 7 months after he arrived he estimates there are 120 plastic bags. He sets the date of his arrival as a benchmark time of t = 0. Using the two data points and the observer’s hypothesis of exponential increase, find an equation for B(t), the number of bags in this region t months after the observer arrived. Approach 1 The hypothesis of exponential growth implies B(t) = Cb t for constants C and b. We know that when t = 2, B = 100 and when t = 7, B = 120. Therefore,  100 = Cb 2 and 120 = Cb 7 . We have a pair of simultaneous equations with two unknowns. Our goal is to eliminate a variable. One strategy is to divide one equation by the other, eliminating C. Equivalently, we could solve each equation for C and set them equal. We’ll do this.        C = 100 b 2 and C = 120 b 7 So 100 b 2 = 120 b 7 Multiplying both sides by b 7 /100 leaves b’s on one side. b 7 b 2 = 120 100 b 5 = 1.2 Raise both sides of the equation to 1/5 to solve for b. (b 5 ) 1/5 = (1.2) 1/5 b = (1.2) 1/5 . 9.2 Exponential: The Bare Bones 315 We now have B(t) = C[(1.2) 1/5 ] t B(t) = C(1.2) t/5 . We can use either of the data points to solve for C. 100 = C(1.2) 2/5 100 (1.2) 2/5 = C C = 100 (6/5) 0.4 = 100(5/6) 0.4 Therefore, B(t) = 100(5/6) 0.4 (1.2) t/5 . Approach 2 We have two data points: when t = 2, B = 100 and when t = 7, B = 120. In 5 months the number of bags has increased by 20. Because we’re hypothesizing exponential growth, the percent increase is the important bit of information, not the actual increase itself. Twenty is 20% of 100, so every 5 months the number of bags increases by 20%. B(t) = C(1.2) t/5 Now find C as done above. Note that if we were willing to adjust our benchmark time of t = 0 from the observer’s day of arrival in the Sinai to the day he made the first bag count we would have B(t) = 100(1.2) t/5 , where t = 0 corresponds to 2 months after his arrival. ◆ ◆ EXAMPLE 9.5 Suppose f(x)=Ca 2x , where C and a are constants, a>0.The percent change in f over the interval [x, x + 3] is given by final value − initial value initial value = f(x+3)−f(x) f(x) ,or Ca 2(x+3) − Ca 2x Ca 2x . Simplify this expression. SOLUTION C[a 2(x+3) − a 2x ] Ca 2x Factor out the C. = a 2x+6 − a 2x a 2x We want to factor out a 2x , so we write a 2x+6 as a 2x · a 6 . = a 2x · a 6 − a 2x a 2x Factor out a 2x . = a 2x [a 6 − 1] a 2x = a 6 − 1 316 CHAPTER 9 Exponential Functions Notice that the percent change in f over any interval 3 units in length is a constant. The percent change does not depend upon the particular endpoints of the interval. This is a defining characteristic of exponential functions. ◆ PROBLEMS FOR SECTION 9.2 For Problems 1 through 9, simplify the following expressions. 1. x −1 + z −1 (z + x)x −2 2. (xy) −3 xy −3 3. √ 2x 3 +  12y 3 x 4  6y 2 x 5 4. x −1 zx −1 + z −1 5. z 0 x −1 y −2 z −2 x −1 y 2 6. x n+1 y 2n  x y  n 7. (ab x ) −2 (ab) −x 8. (ab) −x a −x + b −y 9. (a −x+1 b) 3 (a 2 b 3 ) x For Problems 10 through 15, factor b x out of the following expressions. Check your answer by multiplying out. 10. b x+y + b x 11. b 2x + b x+1 12. 3b 2x+1 − 4b 2x−1 13. (ab 2 ) x +  a b  −x 9.2 Exponential: The Bare Bones 317 14. b 3x − (2b) −1+2x 15. √ ab x −a √ b 3x For Problems 16 through 21, if the statement is always true, write “True;” if the statement is not always true, produce a counterexample. In these problems, a and b are positive constants. 16. √ a 2x = a x 17. (a x + b x ) 1 x = a + b 18.  a 2x b −2x = (ab) x 19. a −x + a x = 1 20. a x bc x = (a x b)(ac) x 21. 3 2x + 2 3x = 9 x + 8 x For Problems 22 through 24, simplify as much as possible. 22. a x+y − a 2x a x 23. a 2x − b 4x a x + b x 24. A 4+p − A 5p A 2+p − A 3p 25. The graphs below are graphs of functions of the form f(t)=Ca t . For each graph, determine the sign of C and whether a ∈ (0, 1) or a>1. ttt t f ff f (d) (b)(a) (c) 318 CHAPTER 9 Exponential Functions For Problems 26 through 29, find a function of the form f(x)=Ca x + D to fit the graph given. 26. x f (1, 5) 3 27. x y = 2 f (–1, 3) 3 4 2 1 28. x y = –1 f (1, – 4) –2 –1 –3 – 4 29. x f (2, .5) 1.5 30. True or False: If the statement is always true, write “True;” if the statement is not always true, produce a counterexample. 9.2 Exponential: The Bare Bones 319 (a) (a 2 + b 2 ) 1/2 = a + b (b) (a +b) −1 = 1 a+b , a, b = 0 (c) (a +b) −1 = 1 a + 1 b , a, b = 0 (d) R −1/2 =− 1 √ R , R>0 (e) x z + x z = 2x z (f) x z x z = x (z 2 ) (g) x z x z = x 2z 31. Mix and Match: Below are functions. To each function match the graph that best fits the function. Since there are no units on the graphs, you may match one graph to several different functions. Note: First do this problem without using a graphing calculator. You can then check your answers with the calculator. If you made a mismatch, figure out what your mistake was and learn from it. (a) f(x)=3( 3 2 ) x (b) f(x)=−2(0.4) x (c) f(x)=2−3 −x (d) f(x)=4( 2 3 ) x (e) f(x)=4 −x (f) f(x)=1+2 x xxxx xx x f ff f fff (i) (ii) (iii) (iv) (vii) (v) (vi) 32. Simplify as much as possible: (a) x 2y + x y+2 x y (b) √ x x −1/2 y − 1 y − x 2 y (c) A B+4 − A 3B A B (A 2 − A B ) (d) y 3w − y w+4 y w (y w + y 2 ) 33. Let f(x)=a x . Weknow that replacing x by x + k shifts the graph of f left k units. For exponential functions this is equivalent to vertical stretching or shrinking. Explain. In Problems 34 through 37, evaluate the limits. 34. (a) lim x→∞ −2(1.1) −x (b) lim x→−∞ −2(1.1) x 35. (a) lim x→∞ (0.89 x − 1) (b) lim x→−∞ (0.89 x − 1) 36. (a) lim x→∞ ( 3 5 ) −x (b) lim x→−∞ ( 3 5 ) −x 320 CHAPTER 9 Exponential Functions 37. (a) lim x→∞ −5( 2 3 ) x + 7 (b) lim x→−∞ −5( 2 3 ) x + 7 38. Factor b x out of each of the following expressions. (a) 3b x − b 2x (b) (3b) x − b x+2 (c) b 3x/2 − b 2x−1 9.3 APPLICATIONS OF THE EXPONENTIAL FUNCTION In Section 9.2 we looked at exponential growth in the context of population growth under ideal conditions. In this section we’ll look at the growth of money in an interest-earning account and the decay of radioactive materials. The Growth of Money in a Bank Account Exponential functions arise in questions dealing with money earning interest in a bank account and in calculations of loan paybacks. If money is deposited in a bank and left there (with no additional deposits), then the growth of the amount of money in the account is due to the interest earned. The amount of interest earned during each period is directly proportional to the amount of money in the account. Let’s look at an example. ◆ EXAMPLE 9.6 Suppose we put $5000 in a bank account earning 4% interest per year compounded annually. (a) Write a function that gives the amount of money in the account after t years. (b) How long does it take for the money to double? SOLUTION (a) At the end of the first year, the interest added to the $5000 we already had is 4% of $5000, or $200 (= 0.04 · $5000). We start the next year with $5200, and at the end of that year get interest amounting to 4% of $5200, or $208, giving a total of $5408, and so on. Each year the amount in the account is 104% of the previous year’s amount. Recall that if a population grows at a rate of 4% per year, then P(t)=P 0 (1.04) t , where P 0 is the population at t = 0. Similarly, if money in a bank account grows at a rate of 4% per year, then M(t) = M 0 (1.04) t , where M 0 is the amount of money in the account at t = 0 and t is measured in years. Our function is M(t) = 5000 (1.04) t . (b) To find out how long it takes for the money to double we want to solve 10,000 = 5000 (1.04) t , or, equivalently, 2 = (1.04) t . Observe that doubling time is independent of the amount of money we have in the account. Suppose the amount in the bank initially is denoted by B 0 . The doubling time is the solution to the equation 2B 0 = B 0 (1.04) t . Notice that we are back to the equation 2 = (1.04) t . This equation can be solved for t analytically using logarithms to obtain an exact answer. 7 The solution can be approximated using a graphing calculator. Solutions to an equation of the form f(x)=k,where k is a constant, can be approximated graphically in two ways. 8 7 We will take this up in Chapter 13. 8 You might also have an “equation solver” on your calculator that allows you to approximate the solution. . t analytically using logarithms to obtain an exact answer. 7 The solution can be approximated using a graphing calculator. Solutions to an equation of the form f(x)=k,where k is a constant, can. interest-earning account and the decay of radioactive materials. The Growth of Money in a Bank Account Exponential functions arise in questions dealing with money earning interest in a bank account and in calculations. beginning of this section are identities; they can be read from left to right or from right to left. Flexibility and a sense of your own goals helps in applying them when working with exponential functions. ◆ EXAMPLE