Exponential and logarithmic functions derivatives

derivatives of exponential and logarithmic functions and exponential and logarithmic functions answers and applications of exponential and logarithmic functions
MattGates Profile Pic
MattGates,United States,Professional
Published Date:02-08-2017
Your Website URL(Optional)
Comment
Exponential and Logarithmic Functions Figure 1 electron micrograph of E. Coli bacteria (credit: “mattosaurus,” Wikimedia Commons) Introduction Focus in on a square centimeter of your skin. Look closer. Closer still. If you could look closely enough, you would see hundreds of thousands of microscopic organisms. They are bacteria, and they are not only on your skin, but in your mouth, nose, and even your intestines. In fact, the bacterial cells in your body at any given moment outnumber your own cells. But that is no reason to feel bad about yourself. While some bacteria can cause illness, many are healthy and even essential to the body. Bacteria commonly reproduce through a process called binary fission, during which one bacterial cell splits into two. When conditions are right, bacteria can reproduce very quickly. Unlike humans and other complex organisms, the time 16 required to form a new generation of bacteria is oe ft n a matter of minutes or hours, as opposed to days or years. For simplicity’s sake, suppose we begin with a culture of one bacterial cell that can divide every hour. Table 1 shows the number of bacterial cells at the end of each subsequent hour. We see that the single bacterial cell leads to over one thousand bacterial cells in just ten hours And if we were to extrapolate the table to twenty-four hours, we would have over 16 million Hour 0 1 2 3 4 5 6 7 8 9 10 Bacteria 1 2 4 8 16 32 64 128 256 512 1024 Table 1 In this chapter, we will explore exponential functions, which can be used for, among other things, modeling growth patterns such as those found in bacteria. We will also investigate logarithmic functions, which are closely related to exponential functions. Both types of functions have numerous real-world applications when it comes to modeling and interpreting data. 16 Todar, PhD, Kenneth. Todar’s Online Textbook of Bacteriology. http://textbookofbacteriology.net/growth_3.html. 463464 CHAPTER 6 e x Po Ne Nti a al Nd l ogar ithmic f u Nctio Ns l eARnIng Obje CTIveS In this section, you will: • Evaluate exponential functions. • Find the equation of an exponential function. • Use compound interest formulas. • Evaluate exponential functions with base e. 6.1 exPOnen TIAl F UnCTIOnS India is the second most populous country in the world with a population of about 1.25 billion people in 2013. The 17 population is growing at a rate of about 1.2% each year . If this rate continues, the population of India will exceed China’s population by the year 2031. When populations grow rapidly, we often say that the growth is “exponential,” meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions, which model this kind of rapid growth. Identifying exponential Functions When exploring linear growth, we observed a constant rate of change—a constant number by which the output increased for each unit increase in input. For example, in the equation f (x) = 3x + 4, the slope tells us the output increases by 3 each time the input increases by 1. The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people. Defining an Exponential Function A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, 2.5% of the population was vegan, adhering to a diet that does not include any animal products—no meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up 10% of the U.S. population in 2015, 40% in 2019, and . 80% in 2021. What exactly does it mean to grow exponentially? What does the word double have in common with percent increase? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media. • Percent change refers to a change based on a percent of the original amount. • Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time. • Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time. For us to gain a clear understanding of exponential growth, let us contrast exponential growth with linear growth. We will construct two functions. The first function is exponential. We will start with an input of 0, and increase each input by 1. We will double the corresponding consecutive outputs. e Th second function is linear. We will start with an input of 0, and increase each input by 1. We will add 2 to the corresponding consecutive outputs. See Table 1. x f (x) = 2 g(x) = 2x x 0 1 0 1 2 2 2 4 4 3 8 6 4 16 8 5 32 10 6 64 12 Table 1 17 http://www.worldometers.info/world-population/. Accessed February 24, 2014.SECTION 6.1 e x Po Ne Ntial f u Ncti o Ns 465 From Table 2 we can infer that for these two functions, exponential growth dwarfs linear growth. • Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain. • Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain. Apparently, the die ff rence between “the same percentage” and “the same amount” is quite signic fi ant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding 2 to the output whenever the input was increased by one. x e g Th eneral form of the exponential function is f (x) = ab , where a is any nonzero number, b is a positive real number not equal to 1. • If b 1, the function grows at a rate proportional to its size. • If 0 b 1, the function decays at a rate proportional to its size. x Let’s look at the function f (x) = 2 from our example. We will create a table (Table 2) to determine the corresponding outputs over an interval in the domain from −3 to 3. −3 −2 −1 x 0 1 2 3 1 1 1 x _ _ _ −3 −2 −1 0 1 2 3 f (x) = 2 2 = 2 = 2 = 2 = 1 2 = 2 2 = 4 2 = 8 4 8 2 Table 2 Let us examine the graph of f by plotting the ordered pairs we observe on the table in Figure 1, and then make a few observations. y x f(x) = 2 9 8 (3, 8) 7 6 1 5 ) 2 4 (2, 4) 3 2 (1, 2) 1 y = 0 (0, 1) x –5 –4–3 –2 –1 14 2 3 5 –1 –2 Figure 1 x Let’s define the behavior of the graph of the exponential function f (x) = 2 and highlight some its key characteristics. • the domain is (−∞, ∞), • the range is (0, ∞), • as x → ∞, f (x) → ∞, • as x → −∞, f (x) → 0, • f (x) is always increasing, • the graph of f (x) will never touch the x-axis because base two raised to any exponent never has the result of zero. • y = 0 is the horizontal asymptote. • the y-intercept is 1.466 CHAPTER 6 e x Po Ne Nti al a N d l ogar ithmic f u Nctio Ns exponential function For any real number x, an exponential function is a function with the form x f (x) = ab where • a is the a non-zero real number called the initial value and • b is any positive real number such that b ≠ 1. • e do Th main of f is all real numbers. • e ra Th nge of f is all positive real numbers if a 0. • e ra Th nge of f is all negative real numbers if a 0. • The y-intercept is (0, a), and the horizontal asymptote is y = 0. Example 1 Identifying Exponential Functions Which of the following equations are not exponential functions? x 1 3(x − 2) 3 _ x f (x) = 4 g (x) = x h (x) = j (x) = (−2)   3 Solution B y definition, an exponential function has a constant as a base and an independent variable as an exponent. 3 3 us, Th g(x) = x does not represent an exponential function because the base is an independent variable. In fact, g(x) = x is a power function. x Recall that the base b of an exponential function is always a positive constant, and b ≠ 1. Thus, j(x) = (−2) does not represent an exponential function because the base, −2, is less than 0. Try It 1 Which of the following equations represent exponential functions? 2 • f (x) = 2x − 3x + 1 x • g(x) = 0.875 • h(x) = 1.75x + 2 −2x • j(x) = 1095.6 evaluating exponential Functions Recall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base b to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive: 1 __ — 1 1 _ _ 2 • Let b 9 and x . en Th f (x) f ( 9) √−9 , w hich is not a real number. = − = = = − =   2 2 Why do we limit the base to positive values other than 1? Because base 1 results in the constant function. Observe what happens if the base is 1: x • Let b = 1. Then f (x) = 1 = 1 for any value of x. x To evaluate an exponential function with the form f (x) = b , we simply substitute x with the given value, and calculate the resulting power. For example: x Let f (x) = 2 . What is f (3)? x f (x) = 2 3 f (3) = 2 Substitute x = 3. = 8 Evaluate the power. To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations.SECTION 6.1 e x Po Ne Ntial f u Ncti o Ns 467 For example: x Let f (x) = 30(2) . What is f (3)? x f (x) = 30(2) 3 f (3) = 30(2) Substitute x = 3. = 30(8) Simplify the power first. = 240 Multiply. Note that if the order of operations were not followed, the result would be incorrect: 3 3 f (3) = 30(2) ≠ 60 = 216,000 Example 2 Evaluating Exponential Functions x + 1 Let f (x) = 5(3) . Evaluate f (2) without using a calculator. Solution Follow the order of operations. Be sure to pay attention to the parentheses. x + 1 f (x) = 5(3) 2 + 1 f (2) = 5(3) Substitute x = 2. 3 = 5(3) Add the exponents. = 5(27) Simplify the power. = 135 Multiply. Try It 2 x − 5 Let f (x) = 8(1.2) . Evaluate f (3) using a calculator. Round to four decimal places. Defining Exponential Growth Because the output of exponential functions increases very rapidly, the term “exponential growth” is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth. exponential growth A function that models exponential growth grows by a rate proportional to the amount present. For any real number x and any positive real numbers a and b such that b ≠ 1, an exponential growth function has the form x f (x) = ab where • a is the initial or starting value of the function. • b is the growth factor or growth multiplier per unit x. In more general terms, we have an exponential function, in which a constant base is raised to a variable exponent. To die ff rentiate between linear and exponential functions, let’s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function A(x) = 100 + 50x. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be x represented by the function B(x) = 100(1 + 0.5) . A few years of growth for these companies are illustrated in Table 3. Year, x Stores, Company A Stores, Company B 0 0 100 + 50(0) = 100 100(1 + 0.5) = 100 1 1 100 + 50(1) = 150 100(1 + 0.5) = 150 2 2 100 + 50(2) = 200 100(1 + 0.5) = 225 3 3 100 + 50(3) = 250 100(1 + 0.5) = 337.5 x x A(x) = 100 + 50x B(x) = 100(1 + 0.5) Table 3x B(x) = 100(1 + 0.5) A(x) = 100 + 50x 468 CHAPTER 6 e x Po Ne Nti al a Nd l ogar ithmic f u Nctio Ns e g Th raphs comparing the number of stores for each company over a five-year period are shown in Figure 2. We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth. y 500 450 400 350 300 250 200 150 100 x 02 1 3 45 Years Figure 2 The graph shows the numbers of stores Companies A and b opened over a five-year period. Notice that the domain for both functions is 0, ∞), and the range for both functions is 100, ∞). Ae ft r year 1, Company B always has more stores than Company A. x Now we will turn our attention to the function representing the number of stores for Company B, B(x) = 100(1 + 0.5) . In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and x 1 + 0.5 = 1.5 represents the growth factor. Generalizing further, we can write this function as B(x) = 100(1.5) , where 100 is the initial value, 1.5 is called the base, and x is called the exponent. Example 3 Evaluating a Real-World Exponential Model At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with t an annual growth rate of about 1.2%. This situation is represented by the growth function P(t) = 1.25(1.012) , where t is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031? Solution To estimate the population in 2031, we evaluate the models for t = 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth, 18 P(18) = 1.25(1.012) ≈ 1.549 e Th re will be about 1.549 billion people in India in the year 2031. Try It 3 e p Th opulation of China was about 1.39 billion in the year 2013, with an annual growth rate of about 0.6%. This t situation is represented by the growth function P(t) = 1.39(1.006) , where t is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example 3? Finding equations of exponential Functions In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants a and b, and evaluate the function. How To… Given two data points, write an exponential model. 1. If one of the data points has the form (0, a), then a is the initial value. Using a, substitute the second point into the x equation f (x) = a(b) , and solve for b. x 2. If neither of the data points have the form (0, a), substitute both points into two equations with the form f (x) = a(b) . Solve the resulting system of two equations in two unknowns to find a and b. x 3. Using the a and b found in the steps above, write the exponential function in the form f (x) = a(b) . Number of StoresSECTION 6.1 e x Po Ne Ntial f u Ncti o Ns 469 Example 4 Writing an Exponential Model When the Initial Value Is Known In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population (N) of deer over time t. Solution We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute t the second point into the equation N(t) = 80b to find b: t N(t) = 80b 6 180 = 80b Substitute using point (6, 180). 9 __ 6 = b Divide and write in lowest terms. 4 1 __ 9 __ 6 b = Isolate b using properties of exponents.   4 b ≈ 1.1447 Round to 4 decimal places. NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the n fi al answer to four places for the remainder of this section. t e e Th xponential model for the population of deer is N(t) = 80(1.1447) . (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.) We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph in Figure 3 passes through the initial points given in the problem, (0, 80) and (6, 180). We can also see that the domain for the function is 0, ∞), and the range for the function is 80, ∞). N(t) 320 300 280 260 240 220 200 180 (6, 180) 160 140 120 100 80 (0, 80) t 02 1 3456789 10 Years t Figure 3 graph showing the population of deer over time, N(t) = 80(1.1447) , t years after 2006. Try It 4 A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013, the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population N of wolves over time t. Example 5 Writing an Exponential Model When the Initial Value is Not Known Find an exponential function that passes through the points (−2, 6) and (2, 1). x Solution Because we don’t have the initial value, we substitute both points into an equation of the form f (x) = ab , and then solve the system for a and b. −2 • Substituting (−2, 6) gives 6 = ab 2 • Substituting (2, 1) gives 1 = ab Deer Population470 CHAPTER 6 e x Po Ne Nti al a N d l ogar ithmic f u Nctio Ns Use the first equation to solve for a in terms of b: −2 6 = ab 6 _ = a Divide. −2 b 2 a = 6b Use properties of exponents to rewrite the denominator. Substitute a in the second equation, and solve for b: 2 1 = ab 2 2 4 1 = 6b b = 6b Substitute a. 1 __ 1 _ 4 b = Use properties of exponents to isolate b.   6 b ≈ 0.6389 Round 4 decimal places. Use the value of b in the first equation to solve for the value of a: 2 2 a = 6b ≈ 6(0.6389) ≈ 2.4492 x u Th s, the equation is f (x) = 2.4492(0.6389) . We can graph our model to check our work. Notice that the graph in Figure 4 passes through the initial points given in the problem, (−2, 6) and (2, 1). The graph is an example of an exponential decay function. f (x) 10 9 8 7 (−2, 6) 6 5 4 3 2 (2, 1) 1 x –5 –4–3 –2 –1 14 2 3 5 –1 –2 x Figure 4 The graph of f (x) = 2.4492(0.6389) models exponential decay. Try It 5 Given the two points (1, 3) and (2, 4.5), find the equation of the exponential function that passes through these two points. Q & A… Do two points always determine a unique exponential function? Yes, provided the two points are either both above the x-axis or both below the x-axis and have die ff rent x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in x, which in many real world cases involves time. How To… Given the graph of an exponential function, write its equation. 1. First, identify two points on the graph. Choose the y-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error. 2. If one of the data points is the y-intercept (0, a), then a is the initial value. Using a, substitute the second point into x the equation f (x) = a(b) , and solve for b. x 3. If neither of the data points have the form (0, a), substitute both points into two equations with the form f (x) = a(b) . Solve the resulting system of two equations in two unknowns to find a and b. x 4. Write the exponential function, f (x) = a(b) .SECTION 6.1 e x Po Ne Ntial f u Ncti o Ns 471 Example 6 Writing an Exponential Function Given Its Graph Find an equation for the exponential function graphed in Figure 5. f(x) 21 18 15 12 9 6 3 x –3.5–3–2.5–2–1.5–1–0.5 0.5 12 1.52.5 3 3.5 –3 Figure 5 Solution We can choose the y-intercept of the graph, (0, 3), as our first point. This gives us the initial value, a = 3. Next, choose a point on the curve some distance away from (0, 3) that has integer coordinates. One such point is (2, 12). x y = ab Write the general form of an exponential equation. x y = 3b Substitute the initial value 3 for a. 2 12 = 3b Substitute in 12 for y and 2 for x. 2 4 = b Divide by 3. b = ±2 Take the square root. Because we restrict ourselves to positive values of b, we will use b = 2. Substitute a and b into the standard form to x yield the equation f (x) = 3(2) . Try It 6 Find an equation for the exponential function graphed in Figure 6. f(x) 5 4 3 2 (0, √2 ) 1 x –5–4–3–2–1 14 2 3 5 –1 Figure 6 How To… Given two points on the curve of an exponential function, use a graphing calculator to find the equation. 1. Press STAT. 2. Clear any existing entries in columns L1 or L2. 3. In L1, enter the x-coordinates given. 4. In L2, enter the corresponding y-coordinates. 5. Press STAT again. Cursor right to CALC, scroll down to ExpReg (Exponential Regression), and press ENTER. x 6. e s Th creen displays the values of a and b in the exponential equation y = a ⋅ b Example 7 Using a Graphing Calculator to Find an Exponential Function Use a graphing calculator to find the exponential equation that includes the points (2, 24.8) and (5, 198.4). Solution Follow the guidelines above. First press STAT, EDIT, 1: Edit…, and clear the lists L1 and L2. Next, in the L1 column, enter the x-coordinates, 2 and 5. Do the same in the L2 column for the y-coordinates, 24.8 and 198.4. Now press STAT, CALC, 0: ExpReg and press ENTER. The values a = 6.2 and b = 2 will be displayed. The x exponential equation is y = 6.2 ⋅ 2 . Try It 7 Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6, 481.07).472 CHAPTER 6 e x Po Ne Nti al a Nd l ogar ithmic f u Nctio Ns Applying the Compound-Interest Formula Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest. The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account. The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the ee ff ctive interest rate ends up being greater than the nominal rate This is a powerful tool for investing. We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time t, principal P, APR r, and number of compounding periods in a year n: nt r _ A(t) = P 1 +   n For example, observe Table 4, which shows the result of investing 1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases. Frequency Value after 1 year Annually 1100 Semiannually 1102.50 Quarterly 1103.81 Monthly 1104.71 Daily 1105.16 Table 4 the compound interest formula Compound interest can be calculated using the formula nt r _ A(t) = P 1 +   n where • A(t) is the account value, • t is measured in years, • P is the starting amount of the account, oen c ft alled the principal, or more generally present value, • r is the annual percentage rate (APR) expressed as a decimal, and • n is the number of compounding periods in one year. Example 8 Calculating Compound Interest If we invest 3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years? Solution Because we are starting with 3,000, P = 3000. Our interest rate is 3%, so r = 0.03. Because we are compounding quarterly, we are compounding 4 times per year, so n = 4. We want to know the value of the account in 10 years, so we are looking for A(10), the value when t = 10. nt r _ A(t) P 1 + Use the compound interest formula. =   n 4 ⋅ 10 0.03 ____ A(10) 3000 1 + Substitute using given values. =   4 ≈ 4,045.05 Round to two decimal places. e ac Th count will be worth about 4,045.05 in 10 years. Try It 8 An initial investment of 100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What will the investment be worth in 30 years?SECTION 6.1 e x Po Ne Ntial f u Ncti o Ns 473 Example 9 Using the Compound Interest Formula to Solve for the Principal A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child’s future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to 40,000 over 18 years. She believes the account will earn 6% compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now? Solution e n Th ominal interest rate is 6%, so r = 0.06. Interest is compounded twice a year, so n = 2. We want to find the initial investment, P, needed so that the value of the account will be worth 40,000 in 18 years. Substitute the given values into the compound interest formula, and solve for P. nt r _ A(t) = P 1 + Use the compound interest formula.   n 2(18) 0.06 ____ 40,000 = P 1 + Substitute using given values A, r, n, and t.   2 36 40,000 = P(1.03) Simplify. 40,000 ______ = P Isolate P. 36 (1.03) P ≈ 13, 801 Di vide and round to the nearest dollar. Lily will need to invest 13,801 to have 40,000 in 18 years. Try It 9 Refer to Example 9. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly? evaluating Functions with base e As we saw earlier, the amount earned on an account increases as the compounding frequency increases. Table 5 shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue. Examine the value of 1 invested at 100% interest for 1 year, compounded at various frequencies, listed in Table 5. n 1 _ Frequency Value A(t) = 1 +   n 1 1 _ 1 + Annually 2   1 2 1 _ 1 + Semiannually 2.25   2 4 1 _ 1 + Quarterly 2.441406   4 12 1 _ 1 + Monthly 2.613035   12 365 1 _ 1 + Daily 2.714567   365 8760 1 _ 1 + Hourly 2.718127   8760 525600 1 _ 1 + Once per minute 2.718279   525600 31536000 1 ________ 1 + Once per second 2.718282   31536000 Table 5474 CHAPTER 6 e x Po Ne Nti a al Nd l ogar ithmic f u Nctio Ns e Th se values appear to be approaching a limit as n increases without bound. In fact, as n gets larger and larger, the n 1 _ expression 1 + approaches a number used so frequently in mathematics that it has its own name: the letter e.   n This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below. the number e e l Th etter e represents the irrational number n 1 __ 1 + , as n increases without bound   n e l Th etter e is used as a base for many real-world exponential models. To work with base e, we use the approximation, e ≈ 2.718282. The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties. Example 10 Using a Calculator to Find Powers of e 3.14 Calculate e . Round to five decimal places. x Solution On a calculator, press the button labeled e . The window shows e(. Type 3.14 and then close parenthesis, 3.14 ). Press ENTER. Rounding to 5 decimal places, e ≈ 23.10387. Caution: Many scientic c fi alculators have an “ Exp” button, which is used to enter numbers in scientic n fi otation. It is not used to find powers of e. Try It 10 −0.5 Use a calculator to find e . Round to five decimal places. Investigating Continuous growth So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, e is used as the base for exponential functions. Exponential models that use e as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics. the continuous growth/decay formula For all real numbers t, and all positive numbers a and r, continuous growth or decay is represented by the formula rt A(t) = ae where • a is the initial value, • r is the continuous growth rate per unit time, • and t is the elapsed time. If r 0, then the formula represents continuous growth. If r 0, then the formula represents continuous decay. For business applications, the continuous growth formula is called the continuous compounding formula and takes the form rt A(t) = Pe where • P is the principal or the initial invested, • r is the growth or interest rate per unit time, • and t is the period or term of the investment.SECTION 6.1 e x Po Ne Ntial f u Ncti o Ns 475 How To… Given the initial value, rate of growth or decay, and time t, solve a continuous growth or decay function. 1. Use the information in the problem to determine a, the initial value of the function. 2. Use the information in the problem to determine the growth rate r. a. If the problem refers to continuous growth, then r 0. b. If the problem refers to continuous decay, then r 0. 3. Use the information in the problem to determine the time t. 4. Substitute the given information into the continuous growth formula and solve for A(t). Example 11 Calculating Continuous Growth A person invested 1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year? Solution Since the account is growing in value, this is a continuous compounding problem with growth rate r = 0.10. e Th initial investment was 1,000, so P = 1000. We use the continuous compounding formula to find the value after t = 1 year: rt A(t) = Pe Use the continuous compounding formula. 0.1 = 1000(e) S ubstitute known values for P, r, and t. ≈ 1105.17 Use a calculator to approximate. e ac Th count is worth 1,105.17 after one year. Try It 11 A person invests 100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years? Example 12 Calculating Continuous Decay Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days? Solution Since the substance is decaying, the rate, 17.3%, is negative. So, r = −0.173. The initial amount of radon- 222 was 100 mg, so a = 100. We use the continuous decay formula to find the value after t = 3 days: rt A(t) = ae Use the continuous growth formula. −0.173(3) = 100e Substitute known values for a, r, and t. ≈ 59.5115 Use a calculator to approximate. So 59.5115 mg of radon-222 will remain. Try It 12 Using the data in Example 12, how much radon-222 will remain aer o ft ne year? Access these online resources for additional instruction and practice with exponential functions. • exponential growth Function (http://openstaxcollege.org/l/expgrowth) • Compound Interest (http://openstaxcollege.org/l/compoundint)476 CHAPTER 6 e x Po Ne Nti a al Nd l ogar ithmic f u Nctio Ns 6.1 SeCTIOn exe RCISeS veRbAl 1. Explain why the values of an increasing exponential 2. Given a formula for an exponential function, is it function will eventually overtake the values of an possible to determine whether the function grows or increasing linear function. decays exponentially just by looking at the formula? Explain. 3. e O Th xford Dictionary defines the word nominal as a value that is “stated or expressed but not 18 necessarily corresponding exactly to the real value.” Develop a reasonable argument for why the term nominal rate is used to describe the annual percentage rate of an investment account that compounds interest. Algeb RAIC For the following exercises, identify whether the statement represents an exponential function. Explain. 4. e a Th verage annual population increase of a pack 5. A population of bacteria decreases by a factor 1 __ of wolves is 25.of  every 24 hours. 8 6. e va Th lue of a coin collection has increased by 3.25% 7. For each training session, a personal trainer charges annually over the last 20 years. his clients 5 less than the previous training session. 8. e h Th eight of a projectile at time t is represented by 2 the function h(t) = −4.9t + 18t + 40. For the following exercises, consider this scenario: For each year t, the population of a forest of trees is represented t by the function A(t) = 115(1.025) . In a neighboring forest, the population of the same type of tree is represented t by the function B(t) = 82(1.029) . (Round answers to the nearest whole number.) 9. Which forest’s population is growing at a faster rate? 10. Which forest had a greater number of trees initially? By how many? 11. Assuming the population growth models continue 12. Assuming the population growth models continue to represent the growth of the forests, which forest to represent the growth of the forests, which forest will have a greater number of trees after 20 years? will have a greater number of trees after By how many? 100 years? By how many? 13. Discuss the above results from the previous four exercises. Assuming the population growth models continue to represent the growth of the forests, which forest will have the greater number of trees in the long run? Why? What are some factors that might influence the long-term validity of the exponential growth model? For the following exercises, determine whether the equation represents exponential growth, exponential decay, or neither. Explain. 5 x 14. y = 300(1 − t) 15. y = 220(1.06) 1 _ t x 16. y = 16.5(1.025) 17. y = 11,701(0.97) For the following exercises, find the formula for an exponential function that passes through the two points given. 3 _ 18. (0, 6) and (3, 750) 19. (0, 2000) and (2, 20) 20. −1, and (3, 24)   2 21. (−2, 6) and (3, 1) 22. (3, 1) and (5, 4) 18. Oxford Dictionary. http://oxforddictionaries.com/us/definition/american_english/nomina.SECTION 6.1 s ectio N e xercises 477 For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points. 23. 24. x 1 2 3 4 x 1 2 3 4 f (x) 70 40 10 −20 h(x) 70 49 34.3 24.01 25. 26. x 1 2 3 4 x 1 2 3 4 f (x) 10 20 40 80 m (x) 80 61 42.9 25.61 27. x 1 2 3 4 g (x) −3.25 2 7.25 12.5 nt r _ For the following exercises, use the compound interest formula, A(t) = P 1 + .   n 28. Aer a cer ft tain number of years, the value of an 29. What was the initial deposit made to the account in the previous exercise? investment account is represented by the equation 120 0.04 ____ 10, 250 1 + . What is the value of the   12 account? 30. How many years had the account from the previous 31. An account is opened with an initial deposit of 6,500 exercise been accumulating interest? and earns 3.6% interest compounded semi-annually. What will the account be worth in 20 years? 32. How much more would the account in the previous 33. Solve the compound interest formula for the exercise have been worth if the interest were principal, P. compounding weekly? 34. Use the formula found in Exercise 31 to calculate 35. How much more would the account in Exercises the initial deposit of an account that is worth 31 and 34 be worth if it were earning interest for 14,472.74 aer e ft arning 5.5% interest compounded 5 more years? monthly for 5 years. (Round to the nearest dollar.) 36. Use properties of rational exponents to solve the 37. Use the formula found in the previous exercise to compound interest formula for the interest rate, r. calculate the interest rate for an account that was compounded semi-annually, had an initial deposit of 9,000 and was worth 13,373.53 aer 10 y ft ears. 38. Use the formula found in the previous exercise to calculate the interest rate for an account that was compounded monthly, had an initial deposit of 5,500, and was worth 38,455 aer 30 y ft ears. For the following exercises, determine whether the equation represents continuous growth, continuous decay, or neither. Explain. 3.25 _ 0.75t −2t t 39. y = 3742(e) 40. y = 150( e) 41. y = 2.25(e) 42. Suppose an investment account is opened with 43. How much less would the account from Exercise an initial deposit of 12,000 earning 7.2% interest 42 be worth aer 30 y ft ears if it were compounded compounded continuously. How much will the monthly instead? account be worth aer 30 y ft ears? nUmeRIC For the following exercises, evaluate each function. Round answers to four decimal places, if necessary. x 2x + 3 x 44. f (x) = 2(5) , for f (−3) 45. f (x) = −4 , for f (−1) 46. f (x) = e , for f (3) x − 1 −x + 1 2x 47. f (x) = −2e , for f (−1) 48. f (x) = 2.7(4) + 1.5, for f (−2) 49. f (x) = 1.2e − 0.3, for f (3) 3 3 _ −x _ 50. f (x) = − (3) + , for f (2) 2 2478 CHAPTER 6 e x Po Ne Nti al a N d l ogar ithmic f u Nctio Ns TeChn Ol Ogy For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve. 51. (0, 3) and (3, 375) 52. (3, 222.62) and (10, 77.456) 53. (20, 29.495) and (150, 730.89) 54. (5, 2.909) and (13, 0.005) 55. (11,310.035) and (25,356.3652) exTen SIOnS 56. The annual percentage yield (APY) of an investment 57. Repeat the previous exercise to find the formula for account is a representation of the actual interest rate the APY of an account that compounds daily. Use earned on a compounding account. It is based on a the results from this and the previous exercise to compounding period of one year. Show that the APY develop a function I(n) for the APY of any account of an account that compounds monthly can be found that compounds n times per year. 12 r __ with the formula APY = 1 + − 1.   12 58. Recall that an exponential function is any equation 59. In an exponential decay function, the base of the x written in the form f (x) = a . b such that a and b are exponent is a value between 0 and 1. Thus, for some positive numbers and b ≠ 1. Any positive number number b 1, the exponential decay function can x n 1 b can be written as b = e for some value of n. Use _ be written as f (x) = a . . Use this formula, along   b this fact to rewrite the formula for an exponential n with the fact that b = e , to show that an exponential function that uses the number e as a base. −nx decay function takes the form f (x) = a(e) f or some positive number n. 60. e f Th ormula for the amount A in an investment account with a nominal interest rate r at any time rt t is given by A(t) = a(e) , where a is the amount of principal initially deposited into an account that compounds continuously. Prove that the percentage of interest earned to principal at any time t can be rt calculated with the formula I(t) = e − 1. ReAl-W ORld A PPl ICATIOnS 61. e f Th ox population in a certain region has an annual 62. A scientist begins with 100 milligrams of a growth rate of 9% per year. In the year 2012, there radioactive substance that decays exponentially. Aer ft were 23,900 fox counted in the area. What is the fox 35 hours, 50 mg of the substance remains. How many population predicted to be in the year 2020? milligrams will remain aer 54 h ft ours? 63. In the year 1985, a house was valued at 110,000. By 64. A car was valued at 38,000 in the year 2007. By 2013, the year 2005, the value had appreciated to 145,000. the value had depreciated to 11,000 If the car’s value What was the annual growth rate between 1985 and continues to drop by the same percentage, what will 2005? Assume that the value continued to grow by the it be worth by 2017? same percentage. What was the value of the house in the year 2010? 65. Jamal wants to save 54,000 for a down payment 66. Kyoko has 10,000 that she wants to invest. Her bank on a home. How much will he need to invest in an has several investment accounts to choose from, all account with 8.2% APR, compounding daily, in order compounding daily. Her goal is to have 15,000 by to reach his goal in 5 years? the time she finishes graduate school in 6 years. To the nearest hundredth of a percent, what should her minimum annual interest rate be in order to reach her goal? (Hint : solve the compound interest formula for the interest rate.) 67. Alyssa opened a retirement account with 7.25% APR 68. An investment account with an annual interest rate in the year 2000. Her initial deposit was 13,500. of 7% was opened with an initial deposit of 4,000 How much will the account be worth in 2025 if Compare the values of the account aer 9 y ft ears interest compounds monthly? How much more would when the interest is compounded annually, quarterly, she make if interest compounded continuously? monthly, and continuously.SECTION 6.2 g ra Phs o f e x Po Ne Ntial f u Ncti o Ns 479 l eARnIng Obje CTIveS In this section, you will: • Graph exponential functions. • Graph exponential functions using transformations. 6.2 gRAPhS OF exPOnen TIAl F UnCTIOnS As we discussed in the previous section, exponential functions are used for many real-world applications such as finance, forensics, computer science, and most of the life sciences. Working with an equation that describes a real- world situation gives us a method for making predictions. Most of the time, however, the equation itself is not enough. We learn a lot about things by seeing their pictorial representations, and that is exactly why graphing exponential equations is a powerful tool. It gives us another layer of insight for predicting future events. graphing exponential Functions Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a x x function of the form f (x) = b whose base is greater than one. We’ll use the function f (x) = 2 . Observe how the output values in Table 1 change as the input increases by 1. −3 −2 −1 x 0 1 2 3 1 1 1 _ _ _ f (x) = 2x 1 2 4 8 8 4 2 Table 1 Each output value is the product of the previous output and the base, 2. We call the base 2 the constant ratio. In fact, for x any exponential function with the form f (x) = ab , b is the constant ratio of the function. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of a. Notice from the table that • the output values are positive for all values of x; • as x increases, the output values increase without bound; and • as x decreases, the output values grow smaller, approaching zero. x Figure 1 shows the exponential growth function f (x) = 2 . f (x) 10 9 (3, 8) 8 7 x 6 f (x) = 2 1 −1, 5 2 4 (2, 4) 1 −2, 3 4 2 (1, 2) 1 −3, 8 1 (0, 1) x –5 –4 –3 –2 –1 14 2 3 5 –1 e x-axis is an asymptote. Figure 1 notice that the graph gets close to the x-axis, but never touches it. x e d Th omain of f (x) = 2 is all real numbers, the range is (0, ∞), and the horizontal asymptote is y = 0. x To get a sense of the behavior of exponential decay, we can create a table of values for a function of the form f (x) = b whose x 1 _ base is between zero and one. We’ll use the function g(x) = . Observe how the output values in Table 2 change as   2 the input increases by 1.480 CHAPTER 6 e x Po Ne Nti al a N d l ogar ithmic f u Nctio Ns x −3 −2 −1 0 1 2 3 x 1 1 1 1 _ _ _ __ 8 4 2 1 g(x) =   2 2 4 8 Table 2 Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or 1 _ constant ratio . 2 Notice from the table that • the output values are positive for all values of x; • as x increases, the output values grow smaller, approaching zero; and • as x decreases, the output values grow without bound. x 1 _ Figure 2 shows the exponential decay function, g(x) = .   2 log (x) –51 –4 –3 –2 –1 0 23 4 5 Figure 2 x 1 _ e d Th omain of g(x) = is all real numbers, the range is (0, ∞), and the horizontal asymptote is y = 0.   2 x characteristics of the graph of the parent function f (x) = b x An exponential function with the form f (x) = b , b 0, b ≠ 1, has these characteristics: f(x) f(x) • one-to-one function • horizontal asymptote: y = 0 x x f (x) = b f (x) = b • domain: (–∞, ∞) b 1 0 b 1 • range: (0, ∞) • x-intercept: none • y-intercept: (0, 1) (1, b) • increasing if b 1 (1, b) (0, 1) (0, 1) • decreasing if b 1 xx Figure 3 compares the graphs of exponential growth and decay functions. Figure 3 How To… x Given an exponential function of the form f (x) = b , graph the function. 1. Create a table of points. 2. Plot at least 3 point from the table, including the y-intercept (0, 1). 3. Draw a smooth curve through the points. 4. State the domain, (−∞, ∞), the range, (0, ∞), and the horizontal asymptote, y = 0.SECTION 6.2 g ra Ph os f e x Po Ne Ntial f u Ncti o Ns 481  x Example 1 Sketching the Graph of an Exponential Function of the For m f (x) = b x Sketch a graph of f (x) = 0.25 . State the domain, range, and asymptote. Solution Before graphing, identify the behavior and create a table of points for the graph. • Since b = 0.25 is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote y = 0. • Create a table of points as in Table 3. x −3 −2 −1 0 1 2 3 x f (x) = 0.25 64 16 4 1 0.25 0.0625 0.015625 Table 3 • Plot the y-intercept, (0, 1), along with two other points. We can use (−1, 4) and (1, 0.25). Draw a smooth curve connecting the points as in Figure 4. f(x) x f(x) = 0.25 6 5 4 (−1, 4) 3 2 (0, 1) 1 (1, 0.25) x –5 –4 –3 –2 –1 14 2 3 5 –1 –2 –3 –4 –5 –6 Figure 4 e d Th omain is ( −∞, ∞); the range is (0, ∞); the horizontal asymptote is y = 0. Try It 1 x Sketch the graph of f (x) = 4 . State the domain, range, and asymptote. graphing Transformations of exponential Functions Transformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, ree fl ctions, stretches, and compressions—to the parent function x f (x) = b without loss of shape. For instance, just as the quadratic function maintains its parabolic shape when shifted, ree fl cted, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied. Graphing a Vertical Shift x e fi Th rst transformation occurs when we add a constant d to the parent function f (x) = b , giving us a vertical shift d units x in the same direction as the sign. For example, if we begin by graphing a parent function, f (x) = 2 , we can then graph x x two vertical shifts alongside it, using d = 3: the upward shift, g(x) = 2 + 3 and the downward shift, h(x) = 2 − 3. Both vertical shifts are shown in Figure 5. y 12 10 8 6 4 x y = 3 g(x) = 2 + 3 2 x x y = 0 f (x) = 2 –6 –5 –4 –3 –2 –1 14 2 3 5 6 –2 x h(x) = 2 − 3 y = −3 –4 –6 Figure 5482 CHAPTER 6 e x Po Ne Nti a al N d l ogar ithmic f u Nctio Ns x Observe the results of shifting f (x) = 2 vertically: • e do Th main, ( −∞, ∞) remains unchanged. x • When the function is shifted up 3 units to g(x) = 2 + 3: ◦ The y-intercept shifts up 3 units to (0, 4). ◦ e a Th symptote shifts up 3 units to y = 3. ◦ e ra Th nge becomes (3, ∞). x • When the function is shifted down 3 units to h(x) = 2 − 3: ◦ The y-intercept shifts down 3 units to (0, −2). ◦ e a Th symptote also shifts down 3 units to y = −3. ◦ e ra Th nge becomes ( −3, ∞). Graphing a Horizontal Shift x The next transformation occurs when we add a constant c to the input of the parent function f (x) = b , giving us a horizontal shi ft c units in the opposite direction of the sign. For example, if we begin by graphing the parent function x x + 3 f (x) = 2 , we can then graph two horizontal shifts alongside it, using c = 3: the shift left, g(x) = 2 , and the shi ft x − 3 right, h (x) = 2 . Both horizontal shifts are shown in Figure 6. y x f (x) = 2 x − 3 h(x) = 2 10 8 6 x + 3 g(x) = 2 4 y = 0 2 x –10 –8 –6 –4 –2 28 4 6 10 –2 Figure 6 x Observe the results of shifting f (x) = 2 horizontally: • e do Th main, ( −∞, ∞), remains unchanged. • e Th asymptote, y = 0, remains unchanged. • The y-intercept shifts such that: x + 3 x + 3 x ◦ When the function is shifted left 3 units to g(x) = 2 , the y-intercept becomes (0, 8). This is because 2 = (8)2 , so the initial value of the function is 8. 1 1 x − 3 _ x − 3 _ x ◦ When the function is shifted right 3 units to h(x) = 2 , the y-intercept becomes 0, . A gain, see that 2 = 2 ,     8 8 1 _ so the initial value of the function is . 8 x shifts of the parent function f (x) = b x + c x For any constants c and d, the function f (x) = b + d shifts the parent function f (x) = b • vertically d units, in the same direction of the sign of d. • horizontally c units, in the opposite direction of the sign of c. c • The y-intercept becomes 0, b + d. • e h Th o rizontal asymptote becomes y = d. • Th e range becomes (d, ∞). • e do Th main, ( −∞, ∞), remains unchanged.