{"id":3503,"date":"2016-08-05T03:52:20","date_gmt":"2016-08-05T03:52:20","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=3503"},"modified":"2018-05-17T02:03:19","modified_gmt":"2018-05-17T02:03:19","slug":"introduction-exponential-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/introduction-exponential-functions\/","title":{"raw":"Exponential Functions","rendered":"Exponential Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Define and Evaluate Exponential Functions\r\n<ul>\r\n \t<li>Define an exponential function and it's domain and range<\/li>\r\n \t<li>Evaluate an exponential function<\/li>\r\n \t<li>Define and evaluate a compound interest formula<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Exponential Functions with Base e\r\n<ul>\r\n \t<li>Define the number\u00a0<em>e<\/em><\/li>\r\n \t<li>Define continuous growth as an exponential function with base\u00a0<em>e<\/em><\/li>\r\n \t<li>Evaluate exponential functions with base\u00a0<em>e<\/em><\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Graph Exponential Functions\r\n<ul>\r\n \t<li>Generate a table of values for an exponential function<\/li>\r\n \t<li>Plot an exponential function on Cartesian axes<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\nIndia is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year.[footnote]http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014.[\/footnote] If this rate continues, the population of India will exceed China\u2019s population by the year 2031. When populations grow rapidly, we often say that the growth is \"exponential,\" meaning that something is increasing very quickly. To a mathematician, however, the term <em>exponential growth <\/em>has a very specific meaning. In this section, we will take a look at <em>exponential functions<\/em>, which model this kind of rapid growth.\r\n\r\nLinear functions have a\u00a0constant rate of change \u2013 a constant number that the output increases for each increase in input. For example, in the equation [latex]f(x)=3x+4[\/latex]\u00a0, the slope tells us the output increases by three each time the input increases by one. Sometimes, on the other hand, quantities grow by a percent rate of change rather than by a fixed amount. In this lesson, we will define a function whose rate of change increases by a percent of the current value rather than a fixed quantity.\r\n\r\nTo illustrate\u00a0this difference consider two companies whose business is expanding: Company A has 100 stores, and expands by opening 50 new stores a year Company B has 100 stores, and expands by increasing the number of stores by 50% of their total each year.\r\n\r\nThe table below compares\u00a0the growth of each company where company A increases the number of stores linearly, and company B increases the number of stores by a rate of 50% each year.\r\n<table style=\"width: 60%;\">\r\n<thead>\r\n<tr>\r\n<td>Year<\/td>\r\n<td>Stores, Company A<\/td>\r\n<td>\u00a0Description of Growth<\/td>\r\n<td>Stores, Company B<\/td>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>0<\/td>\r\n<td>100<\/td>\r\n<td>Starting with 100 each<\/td>\r\n<td>100<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1<\/td>\r\n<td>100+50=150<\/td>\r\n<td>They both grow by 50 stores in the first year.<\/td>\r\n<td>100 + 50% of 100 100 + 0.50(100) = 150<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>150+50=200<\/td>\r\n<td>Store A grows by 50, Store B grows by 75<\/td>\r\n<td>150 + 50% of 150 150 + 0.50(150) = 225<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>200+50=250<\/td>\r\n<td>Store A grows by 50, Store B grows by 112.5<\/td>\r\n<td>\u00a0225 + 50% of 225 225 + 0.50(225) = 337.5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nCompany A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function [latex]A\\left(x\\right)=100+50x[\/latex]. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex].\r\n\r\nThe graphs comparing the number of stores for each company over a five-year period are shown in below<strong>.<\/strong> We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"338\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051913\/CNX_Precalc_Figure_04_01_0012.jpg\" alt=\"Graph of Companies A and B\u2019s functions, which values are found in the previous table.\" width=\"338\" height=\"586\" data-media-type=\"image\/jpg\" \/> The graph shows the numbers of stores Companies A and B opened over a five-year period.[\/caption]\r\n<p id=\"fs-id1165135209682\">Notice that the domain for both functions is [latex]\\left[0,\\infty \\right)[\/latex], and the range for both functions is [latex]\\left[100,\\infty \\right)[\/latex]. After year 1, Company B always has more stores than Company A.<\/p>\r\n<p id=\"fs-id1165137836429\">Consider\u00a0the function representing the number of stores for Company B<\/p>\r\n<p style=\"text-align: center;\">[latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]<\/p>\r\nIn this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex], where 100 is the initial value, 1.5 is called the <em data-effect=\"italics\">base<\/em>, and <em>x<\/em>\u00a0is called the <em data-effect=\"italics\">exponent<\/em>. This is an exponential function.\r\n<div id=\"fs-id1165137564690\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\r\n<h3 class=\"title\" data-type=\"title\">Exponential Growth<\/h3>\r\n<p id=\"fs-id1165137834019\">A function that models <strong>exponential growth<\/strong> grows by a rate proportional to the current amount. For any real number <em>x<\/em>\u00a0and any positive real numbers <em>a\u00a0<\/em>and <em>b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form<\/p>\r\n\r\n<div id=\"fs-id1165137851784\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\r\n<p id=\"eip-626\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137863819\">\r\n \t<li><em>a<\/em>\u00a0is the initial or starting value of the function.<\/li>\r\n \t<li><em>b<\/em>\u00a0is the growth factor or growth multiplier per unit <em>x<\/em>.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165137644244\">To evaluate an exponential function with the form [latex]f\\left(x\\right)={b}^{x}[\/latex], we simply substitute <em>x<\/em>\u00a0with the given value, and calculate the resulting power. For example:<\/p>\r\n<p id=\"fs-id1165135403544\">Let [latex]f\\left(x\\right)={2}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\r\n\r\n<div id=\"eip-id1165137643186\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill &amp; ={2}^{x}\\hfill &amp; \\hfill \\\\ f\\left(3\\right)\\hfill &amp; ={2}^{3}\\text{ }\\hfill &amp; \\text{Substitute }x=3.\\hfill \\\\ \\hfill &amp; =8\\text{ }\\hfill &amp; \\text{Evaluate the power}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137849020\">To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:<\/p>\r\n<p id=\"fs-id1165137849024\">Let [latex]f\\left(x\\right)=30{\\left(2\\right)}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\r\n\r\n<div id=\"eip-id1165134086025\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill &amp; =30{\\left(2\\right)}^{x}\\hfill &amp; \\hfill \\\\ f\\left(3\\right)\\hfill &amp; =30{\\left(2\\right)}^{3}\\hfill &amp; \\text{Substitute }x=3.\\hfill \\\\ \\hfill &amp; =30\\left(8\\right)\\text{ }\\hfill &amp; \\text{Simplify the power first}\\text{.}\\hfill \\\\ \\hfill &amp; =240\\hfill &amp; \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137841073\">Note that if the order of operations were not followed, the result would be incorrect:<\/p>\r\n\r\n<div id=\"eip-id1165135320147\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\"><\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">In our first example we will evaluate an exponential function without the aid of a calculator.<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLet [latex]f\\left(x\\right)=5{\\left(3\\right)}^{x+1}[\/latex]. Evaluate [latex]f\\left(2\\right)[\/latex] without using a calculator.\r\n[reveal-answer q=\"211228\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"211228\"]\r\n<p id=\"fs-id1165137598173\">Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\r\n\r\n<div id=\"eip-id1165135208555\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill &amp; =5{\\left(3\\right)}^{x+1}\\hfill &amp; \\hfill \\\\ f\\left(2\\right)\\hfill &amp; =5{\\left(3\\right)}^{2+1}\\hfill &amp; \\text{Substitute }x=2.\\hfill \\\\ \\hfill &amp; =5{\\left(3\\right)}^{3}\\hfill &amp; \\text{Add the exponents}.\\hfill \\\\ \\hfill &amp; =5\\left(27\\right)\\hfill &amp; \\text{Simplify the power}\\text{.}\\hfill \\\\ \\hfill &amp; =135\\hfill &amp; \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we present more examples of evaluating an exponential function at several different values.\r\n\r\nhttps:\/\/youtu.be\/QFFAoX0We34\r\n\r\nIn the next example we will revisit the population of India.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nAt the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with an annual growth rate of about 1.2%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t}[\/latex], where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?\r\n[reveal-answer q=\"385742\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"385742\"]\r\n<p id=\"fs-id1165137786635\">To estimate the population in 2031, we evaluate the models for <em>t\u00a0<\/em>= 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth,<\/p>\r\n\r\n<div id=\"eip-id1165135657117\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/div>\r\n<p id=\"fs-id1165135394343\">There will be about 1.549 billion people in India in the year 2031.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show another example of using an exponential function to predict the population of a small town.\r\n\r\nhttps:\/\/youtu.be\/SbIydBmJePE\r\n\r\nYou may have seen formulas that are used to calculate compound interest rates. \u00a0These formulas are another example of exponential growth.\u00a0The term <em data-effect=\"italics\">compounding<\/em> refers to interest earned not only on the original value, but on the accumulated value of the account.\r\n<p id=\"fs-id1165137447037\">The <strong>annual percentage rate (APR)<\/strong> of an account, also called the <strong>nominal rate<\/strong>, is the yearly interest rate earned by an investment account. The term\u00a0<em data-effect=\"italics\">nominal<\/em>\u00a0is 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 effective interest rate ends up being <em data-effect=\"italics\">greater<\/em> than the nominal rate! This is a powerful tool for investing.<\/p>\r\n<p id=\"fs-id1165135160118\">We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time <em>t<\/em>, principal <em>P<\/em>, APR <em>r<\/em>, and number of compounding periods in a year\u00a0<em>n<\/em>:<\/p>\r\n\r\n<div id=\"eip-986\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\\\[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3 class=\"title\" data-type=\"title\">The Compound Interest Formula<\/h3>\r\n<p id=\"fs-id1165135184167\"><strong>Compound interest<\/strong> can be calculated using the formula<\/p>\r\n\r\n<div id=\"fs-id1165135184172\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\\\[\/latex]<\/div>\r\n<p id=\"eip-237\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137448453\">\r\n \t<li><em>A<\/em>(<em>t<\/em>) is the account value,<\/li>\r\n \t<li><i>t<\/i> is measured in years,<\/li>\r\n \t<li><em>P<\/em>\u00a0is the starting amount of the account, often called the principal, or more generally present value,<\/li>\r\n \t<li><em>r<\/em>\u00a0is the annual percentage rate (APR) expressed as a decimal, and<\/li>\r\n \t<li><em>n<\/em>\u00a0is the number of compounding periods in one year.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn our next example we will calculate the value of an account after 10 years of interest compounded quarterly.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years?\r\n[reveal-answer q=\"689928\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"689928\"]\r\n<p id=\"fs-id1165137812832\">Because we are starting with $3,000, <em>P\u00a0<\/em>= 3000. Our interest rate is 3%, so <em>r<\/em>\u00a0=\u00a00.03. Because we are compounding quarterly, we are compounding 4 times per year, so <em>n\u00a0<\/em>= 4. We want to know the value of the account in 10 years, so we are looking for <em>A<\/em>(10), the value when <em>t <\/em>= 10.<\/p>\r\n\r\n<div id=\"eip-id1402796\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill &amp; =P\\left(1+\\frac{r}{n}\\right)^{nt}\\hfill &amp; \\text{Use the compound interest formula}. \\\\ A\\left(10\\right)\\hfill &amp; =3000\\left(1+\\frac{0.03}{4}\\right)^{4\\cdot 10}\\hfill &amp; \\text{Substitute using given values}. \\\\ \\text{ }\\hfill &amp; \\approx 4045.05\\hfill &amp; \\text{Round to two decimal places}.\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137694040\">The account will be worth about $4,045.05 in 10 years.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows an example of using an exponential growth to calculate interest compounded quarterly.\r\n\r\nhttps:\/\/youtu.be\/3az4AKvUmmI\r\n\r\nIn our next example we will use the compound interest formula to solve for the principal.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child\u2019s 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?\r\n[reveal-answer q=\"254680\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"254680\"]\r\n<p id=\"fs-id1165137664627\">The nominal interest rate is 6%, so <em>r\u00a0<\/em>= 0.06. Interest is compounded twice a year, so <em>k\u00a0<\/em>= 2.<\/p>\r\n<p id=\"fs-id1165135209414\">We want to find the initial investment, <em>P<\/em>, 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 <em>P<\/em>.<\/p>\r\n\r\n<div id=\"eip-id1165131884554\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill &amp; =P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\hfill &amp; \\text{Use the compound interest formula}.\\hfill \\\\ 40,000\\hfill &amp; =P{\\left(1+\\frac{0.06}{2}\\right)}^{2\\left(18\\right)}\\hfill &amp; \\text{Substitute using given values }A\\text{, }r, n\\text{, and }t.\\hfill \\\\ 40,000\\hfill &amp; =P{\\left(1.03\\right)}^{36}\\hfill &amp; \\text{Simplify}.\\hfill \\\\ \\frac{40,000}{{\\left(1.03\\right)}^{36}}\\hfill &amp; =P\\hfill &amp; \\text{Isolate }P.\\hfill \\\\ P\\hfill &amp; \\approx 13,801\\hfill &amp; \\text{Divide and round to the nearest dollar}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137937589\">Lily will need to invest $13,801 to have $40,000 in 18 years.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show another example of finding the deposit amount necessary to obtain a future value from compounded interest.\r\n\r\nhttps:\/\/youtu.be\/saq9dF7a4r8\r\n\r\n<section id=\"fs-id1165137724961\" data-depth=\"1\">\r\n<h2>Exponential functions with base e<\/h2>\r\nAs we saw earlier, the amount earned on an account increases as the compounding frequency increases. The table below\u00a0shows 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.\r\n<p id=\"fs-id1165135684377\">Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies.<\/p>\r\n\r\n<table id=\"Table_04_01_04\" summary=\"Nine rows and three columns. The first column is labeled,\">\r\n<thead>\r\n<tr>\r\n<th data-align=\"center\">Frequency<\/th>\r\n<th data-align=\"center\">[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\r\n<th data-align=\"center\">Value<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Annually<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\r\n<td>$2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Semiannually<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\r\n<td>$2.25<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Quarterly<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\r\n<td>$2.441406<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Monthly<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\r\n<td>$2.613035<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Daily<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\r\n<td>$2.714567<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Hourly<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\r\n<td>$2.718127<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Once per minute<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\r\n<td>$2.718279<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Once per second<\/td>\r\n<td>[latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\r\n<td>$2.718282<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165137828146\">These values appear to be approaching a limit as <em>n<\/em>\u00a0increases. In fact, as <em>n<\/em>\u00a0gets larger and larger, the expression [latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex] approaches a number used so frequently in mathematics that it has its own name: the letter [latex]e[\/latex]. 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.<\/p>\r\n\r\n<div id=\"fs-id1165135511324\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\r\n<h3 class=\"title\" data-type=\"title\">A General Note: The Number <em data-effect=\"italics\">e<\/em><\/h3>\r\n<p id=\"fs-id1165135511335\">The letter <em data-effect=\"italics\">e<\/em> represents the irrational number<\/p>\r\n\r\n<div id=\"eip-id1165135378658\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">as n increases without bound<\/div>\r\n<p id=\"fs-id1165135369344\">The letter <em data-effect=\"italics\">e <\/em>is used as a base for many real-world exponential models. To work with base <em data-effect=\"italics\">e<\/em>, we use the approximation, [latex]e\\approx 2.718282[\/latex]. The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.<\/p>\r\n\r\n<\/div>\r\nIn our first example we will use a calculator to find powers of\u00a0<em>e.<\/em>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nCalculate [latex]{e}^{3.14}[\/latex]. Round to five decimal places.\r\n[reveal-answer q=\"465847\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"465847\"]On a calculator, press the button labeled [latex]\\left[{e}^{x}\\right][\/latex]. The window shows [<em>e<\/em>^(]. Type 3.14 and then close parenthesis, (]). Press [ENTER]. Rounding to 5 decimal places, [latex]{e}^{3.14}\\approx 23.10387[\/latex]. Caution: Many scientific calculators have an \"Exp\" button, which is used to enter numbers in scientific notation. It is not used to find powers of <em>e<\/em>.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137827923\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Investigating Continuous Growth<\/h2>\r\n<p id=\"fs-id1165137827929\">So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, <em data-effect=\"italics\">e <\/em>is used as the base for exponential functions. Exponential models that use <em>e<\/em>\u00a0as the base are called <em data-effect=\"italics\">continuous growth or decay models<\/em>. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.<\/p>\r\n\r\n<div id=\"fs-id1165137664673\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\r\n<h3 class=\"title\" data-type=\"title\">The Continuous Growth\/Decay Formula<\/h3>\r\n<p id=\"fs-id1165135453868\">For all real numbers r,\u00a0<em>t<\/em>, and all positive numbers <em>a<\/em>, continuous growth or decay is represented by the formula<\/p>\r\n\r\n<div id=\"fs-id1165135536370\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]A\\left(t\\right)=a{e}^{rt}[\/latex]<\/div>\r\n<p id=\"eip-101\">where<\/p>\r\n\r\n<ul id=\"fs-id1165135152052\">\r\n \t<li><em>a<\/em>\u00a0is the initial value,<\/li>\r\n \t<li><em>r<\/em>\u00a0is the continuous growth or decay rate per unit time,<\/li>\r\n \t<li>and <em>t<\/em>\u00a0is the elapsed time.<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165135560686\">If <em>r\u00a0<\/em>&gt; 0, then the formula represents continuous growth. If <em>r\u00a0<\/em>&lt; 0, then the formula represents continuous decay.<\/p>\r\n<p id=\"fs-id1165137812323\">For business applications, the continuous growth formula is called the continuous compounding formula and takes the form<\/p>\r\n\r\n<div id=\"eip-id1165134324899\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A\\left(t\\right)=P{e}^{rt}[\/latex]<\/div>\r\n<p id=\"eip-962\">where<\/p>\r\n\r\n<ul id=\"fs-id1165137827330\">\r\n \t<li><em>P<\/em>\u00a0is the principal or the initial invested,<\/li>\r\n \t<li><em>r<\/em>\u00a0is the growth or interest rate per unit time,<\/li>\r\n \t<li>and <em>t<\/em>\u00a0is the period or term of the investment.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn our next example\u00a0we will calculate continuous growth of an investment. It is important to note the language that is used in the instructions for interest rate problems. \u00a0You will know to use the <em>continuous<\/em> growth or decay formula when you are asked to find an amount based on continuous compounding. \u00a0In previous examples we asked that you find an amount based on quarterly or monthly compounding, and for this you used the <em>compound<\/em> interest formula.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA 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?\r\n[reveal-answer q=\"33008\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"33008\"]\r\n\r\nSince the account is growing in value, this is a continuous compounding problem with growth rate <em>r\u00a0<\/em>= 0.10. The initial investment was $1,000, so <em>P\u00a0<\/em>= 1000. We use the continuous compounding formula to find the value after <em>t\u00a0<\/em>= 1 year:\r\n<div id=\"eip-id1165133351794\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill &amp; =P{e}^{rt}\\hfill &amp; \\text{Use the continuous compounding formula}.\\hfill \\\\ \\hfill &amp; =1000{\\left(e\\right)}^{0.1} &amp; \\text{Substitute known values for }P, r,\\text{ and }t.\\hfill \\\\ \\hfill &amp; \\approx 1105.17\\hfill &amp; \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137895288\">The account is worth $1,105.17 after one year.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show another example of continuous interest.\r\n\r\nhttps:\/\/youtu.be\/fEjrYCog_8w\r\n<div id=\"fs-id1165135411368\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\r\n<h3 id=\"fs-id1165135411373\">How To: Given the initial value, rate of growth or decay, and time <em>t<\/em>, solve a continuous growth or decay function.<\/h3>\r\n<ol id=\"fs-id1165135511371\" data-number-style=\"arabic\">\r\n \t<li>Use the information in the problem to determine <em>a<\/em>, the initial value of the function.<\/li>\r\n \t<li>Use the information in the problem to determine the growth rate <em>r<\/em>.\r\n<ol id=\"fs-id1165135188096\" data-number-style=\"lower-alpha\">\r\n \t<li>If the problem refers to continuous growth, then <em>r\u00a0<\/em>&gt; 0.<\/li>\r\n \t<li>If the problem refers to continuous decay, then <em>r\u00a0<\/em>&lt; 0.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use the information in the problem to determine the time <em>t<\/em>.<\/li>\r\n \t<li>Substitute the given information into the continuous growth formula and solve for <em>A<\/em>(<em>t<\/em>).<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn our next example we will calculate continuous decay. Pay attention to the rate - it is negative which means we are considering a situation where an amount decreases or decays.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nRadon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days?\r\n[reveal-answer q=\"995802\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"995802\"]\r\n\r\nSince the substance is decaying, the rate, 17.3%, is negative. So, <em>r\u00a0<\/em>=\u00a0\u20130.173. The initial amount of radon-222 was 100 mg, so <em>a\u00a0<\/em>= 100. We use the continuous decay formula to find the value after <em>t\u00a0<\/em>= 3 days:\r\n<div id=\"eip-id1165137779893\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill &amp; =a{e}^{rt}\\hfill &amp; \\text{Use the continuous growth formula}.\\hfill \\\\ \\hfill &amp; =100{e}^{-0.173\\left(3\\right)} &amp; \\text{Substitute known values for }a, r,\\text{ and }t.\\hfill \\\\ \\hfill &amp; \\approx 59.5115\\hfill &amp; \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137697132\">So 59.5115 mg of radon-222 will remain.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show an example of calculating remaining amount of a radioactive substance after it decays for a length of time.\r\n\r\nhttps:\/\/youtu.be\/Vyl3NcTGRAo\r\n<h2>Graph Exponential Functions<\/h2>\r\n<p id=\"fs-id1165137592823\">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. Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex] whose base is greater than one. We\u2019ll use the function [latex]f\\left(x\\right)={2}^{x}[\/latex]. Observe how the output values in the table below\u00a0change as the input increases by 1.<\/p>\r\n\r\n<table style=\"width: 60%;\" summary=\"Two rows and eight columns. The first row is labeled,\"><colgroup><col data-width=\"60\" \/><col \/><col \/><col \/><col \/><col \/><col \/><col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>\u20133<\/td>\r\n<td>\u20132<\/td>\r\n<td>\u20131<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\r\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165137432031\">Each output value is the product of the previous output and the base, 2. We call the base 2 the <em data-effect=\"italics\">constant ratio<\/em>. In fact, for any exponential function with the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex], <em>b<\/em>\u00a0is 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 <em>a<\/em>.<\/p>\r\n<p id=\"fs-id1165137585799\">Notice from the table that<\/p>\r\n\r\n<ul id=\"fs-id1165137658509\">\r\n \t<li>the output values are positive for all values of <em>x<\/em>;<\/li>\r\n \t<li>as <em>x<\/em>\u00a0increases, the output values increase without bound; and<\/li>\r\n \t<li>as <em>x<\/em>\u00a0decreases, the output values grow smaller, approaching zero.<\/li>\r\n<\/ul>\r\nFigure 1\u00a0shows the exponential growth function [latex]f\\left(x\\right)={2}^{x}[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051915\/CNX_Precalc_Figure_04_02_0012.jpg\" alt=\"Graph of the exponential function, 2^(x), with labeled points at (-3, 1\/8), (-2, \u00bc), (-1, \u00bd), (0, 1), (1, 2), (2, 4), and (3, 8). The graph notes that the x-axis is an asymptote.\" width=\"487\" height=\"520\" data-media-type=\"image\/jpg\" \/> Notice that the graph gets close to the x-axis, but never touches it.[\/caption]\r\n<p id=\"fs-id1165137459614\">The domain of [latex]f\\left(x\\right)={2}^{x}[\/latex] is all real numbers, the range is [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\r\n<p id=\"fs-id1165137838249\">To get a sense of the behavior of <strong>exponential decay<\/strong>, we can create a table of values for a function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex] whose base is between zero and one. We\u2019ll use the function [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex]. Observe how the output values in the table below\u00a0change as the input increases by 1.<\/p>\r\n\r\n<table style=\"width: 60%;\" summary=\"Two rows and eight columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>\u20133<\/td>\r\n<td>\u20132<\/td>\r\n<td>\u20131<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]g\\left(x\\right)=\\left(\\frac{1}{2}\\right)^{x}[\/latex]<\/strong><\/td>\r\n<td>8<\/td>\r\n<td>4<\/td>\r\n<td>2<\/td>\r\n<td>1<\/td>\r\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\r\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165135347846\">Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or constant ratio [latex]\\frac{1}{2}[\/latex].<\/p>\r\n<p id=\"fs-id1165137452063\">Notice from the table that<\/p>\r\n\r\n<ul id=\"fs-id1165135499992\">\r\n \t<li>the output values are positive for all values of <em>x<\/em>;<\/li>\r\n \t<li>as <em>x<\/em>\u00a0increases, the output values grow smaller, approaching zero; and<\/li>\r\n \t<li>as <em>x<\/em>\u00a0decreases, the output values grow without bound.<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137405421\">The graph shows the exponential decay function, [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex].<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051917\/CNX_Precalc_Figure_04_02_0022.jpg\" alt=\"Graph of decreasing exponential function, (1\/2)^x, with labeled points at (-3, 8), (-2, 4), (-1, 2), (0, 1), (1, 1\/2), (2, 1\/4), and (3, 1\/8). The graph notes that the x-axis is an asymptote.\" width=\"487\" height=\"520\" data-media-type=\"image\/jpg\" \/>\r\n<p id=\"fs-id1165137723586\" style=\"text-align: center;\"><strong>\u00a0<\/strong>The domain of [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex] is all real numbers, the range is [latex]\\left(0,\\infty \\right)[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165135571835\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\r\n<h3 class=\"title\" data-type=\"title\">Characteristics of the Graph of \u00a0<em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = <em data-effect=\"italics\">b<\/em><sup><em data-effect=\"italics\">x<\/em><\/sup><\/h3>\r\n<p id=\"fs-id1165137848929\">An exponential function with the form [latex]f\\left(x\\right)={b}^{x}[\/latex], [latex]b&gt;0[\/latex], [latex]b\\ne 1[\/latex], has these characteristics:<\/p>\r\n\r\n<ul id=\"fs-id1165135186684\">\r\n \t<li><strong>one-to-one<\/strong> function<\/li>\r\n \t<li>domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\r\n \t<li>range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\r\n \t<li><em data-effect=\"italics\">x-<\/em>intercept: none<\/li>\r\n \t<li><em data-effect=\"italics\">y-<\/em>intercept: [latex]\\left(0,1\\right)[\/latex]<\/li>\r\n \t<li>increasing if [latex]b&gt;1[\/latex]<\/li>\r\n \t<li>decreasing if [latex]b&lt;1[\/latex]<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137471878\">Compare the graphs of <strong>exponential growth<\/strong> and decay functions.<\/p>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051919\/CNX_Precalc_Figure_04_02_003new2.jpg\" alt=\"Graph of two functions where the first graph is of a function of f(x) = b^x when b&gt;1 and the second graph is of the same function when b is 0&lt;b&lt;1. Both graphs have the points (0, 1) and (1, b) labeled.\" width=\"731\" height=\"407\" data-media-type=\"image\/jpg\" \/>\r\n\r\n<\/div>\r\nIn our first example we will plot an exponential decay function where the base is between 0 and 1.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSketch a graph of [latex]f\\left(x\\right)={0.25}^{x}[\/latex]. State the domain, range.\r\n[reveal-answer q=\"203605\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"203605\"]\r\n<p id=\"fs-id1165137734539\">Before graphing, identify the behavior and create a table of points for the graph.<\/p>\r\n\r\n<ul>\r\n \t<li>Since <em>b\u00a0<\/em>= 0.25 is between zero and one, we know the function is decreasing, and we can verify this by creating a table of values. The left tail of the graph will increase without bound and the right tail will get really close to the x-axis.<\/li>\r\n \t<li>Create a table of points.\r\n<table id=\"Table_04_02_03\" summary=\"Two rows and eight columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>\u20133<\/td>\r\n<td>\u20132<\/td>\r\n<td>\u20131<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)={0.25}^{x}[\/latex]<\/strong><\/td>\r\n<td>64<\/td>\r\n<td>16<\/td>\r\n<td>4<\/td>\r\n<td>1<\/td>\r\n<td>0.25<\/td>\r\n<td>0.0625<\/td>\r\n<td>0.015625<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>Plot the <em data-effect=\"italics\">y<\/em>-intercept, [latex]\\left(0,1\\right)[\/latex], along with two other points. We can use [latex]\\left(-1,4\\right)[\/latex] and [latex]\\left(1,0.25\\right)[\/latex].<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137482830\">Draw a smooth curve connecting the points.<span id=\"fs-id1165137940681\" data-type=\"media\" data-alt=\"Graph of the decaying exponential function f(x) = 0.25^x with labeled points at (-1, 4), (0, 1), and (1, 0.25).\">\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051920\/CNX_Precalc_Figure_04_02_0042.jpg\" alt=\"Graph of the decaying exponential function f(x) = 0.25^x with labeled points at (-1, 4), (0, 1), and (1, 0.25).\" data-media-type=\"image\/jpg\" \/><\/span><\/p>\r\n<p id=\"fs-id1165137548870\" style=\"text-align: center;\"><strong>\u00a0<\/strong>The domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]; the range is [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show another example of graphing an exponential function. The base of the exponential term is between 0 and 1, so this graph will represent decay.\r\n\r\nhttps:\/\/youtu.be\/FMzZB9Ve-1U\r\n<div id=\"fs-id1165134195243\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\r\n<h3 id=\"fs-id1165135194093\">How To: Given an exponential function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex], graph the function.<\/h3>\r\n<ol id=\"fs-id1165137435782\" data-number-style=\"arabic\">\r\n \t<li>Create a table of points.<\/li>\r\n \t<li>Plot at least 3\u00a0point from the table, including the <em data-effect=\"italics\">y<\/em>-intercept [latex]\\left(0,1\\right)[\/latex].<\/li>\r\n \t<li>Draw a smooth curve through the points.<\/li>\r\n \t<li>State the domain, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], the range, [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn our next example we will plot an exponential growth function where the base is greater than 1.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSketch a graph of [latex]f(x)={\\sqrt{2}(\\sqrt{2})}^{x}[\/latex].\u00a0State the domain, range.\r\n[reveal-answer q=\"334418\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"334418\"]\r\n<p id=\"fs-id1165137734539\">Before graphing, identify the behavior and create a table of points for the graph.<\/p>\r\n\r\n<ul>\r\n \t<li>Since <em>b\u00a0<\/em>= [latex]\\sqrt{2}[\/latex] which is greater than\u00a0one, we know the function is increasing, and we can verify this by creating a table of values. The left tail of the graph will\u00a0get really close to the x-axis and the right tail will increase without bound.<\/li>\r\n \t<li>Create a table of points.\r\n<table id=\"Table_04_02_03\" summary=\"Two rows and eight columns. The first row is labeled,\">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>\u20133<\/td>\r\n<td>\u20132<\/td>\r\n<td>\u20131<\/td>\r\n<td>0<\/td>\r\n<td>1<\/td>\r\n<td>2<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong>[latex]f\\left(x\\right)=\\sqrt{2}{(\\sqrt{2})}^{x}[\/latex]<\/strong><\/td>\r\n<td>0.5<\/td>\r\n<td>0.71<\/td>\r\n<td>1<\/td>\r\n<td>1.41<\/td>\r\n<td>2<\/td>\r\n<td>2.83<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t<li>Plot the <em data-effect=\"italics\">y<\/em>-intercept, [latex]\\left(0,1.41\\right)[\/latex], along with two other points. We can use [latex]\\left(-1,1\\right)[\/latex] and [latex]\\left(1,2\\right)[\/latex].<\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165137482830\">Draw a smooth curve connecting the points.<\/p>\r\n<img class=\" wp-image-3623 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05185425\/Screen-Shot-2016-08-05-at-11.53.45-AM.png\" alt=\"Screen Shot 2016-08-05 at 11.53.45 AM\" width=\"326\" height=\"231\" \/>\r\n\r\nThe domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]; the range is [latex]\\left(0,\\infty \\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nOur next video example includes graphing an exponential growth function and defining the domain and range of the function.\r\n\r\nhttps:\/\/youtu.be\/M6bpp0BRIf0\r\n<h2>Summary<\/h2>\r\n<\/section>Exponential growth grows by a rate proportional to the current amount.\u00a0For any real number <em>x<\/em>\u00a0and any positive real numbers <em>a\u00a0<\/em>and <em>b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form\u00a0[latex]f\\left(x\\right)=a{b}^{x}[\/latex]. \u00a0Evaluating exponential functions requires careful attention to the order of operations. Compound interest is an example of exponential growth.\r\n\r\nContinuous growth or decay functions are of the form\u00a0[latex]A\\left(t\\right)=a{e}^{rt}[\/latex].\u00a0If <em>r\u00a0<\/em>&gt; 0, then the formula represents continuous growth. If <em>r\u00a0<\/em>&lt; 0, then the formula represents continuous decay.\u00a0For business applications, the continuous growth formula is called the continuous compounding formula and takes the form\u00a0[latex]A\\left(t\\right)=P{e}^{rt}[\/latex].\r\n\r\nGraphs of exponential growth functions will have a right tail that increases without bound, and a left tail that gets really close to the x-axis. On the other hand, graph so exponential decay functions will have a left tail that increases without bound and a right tail that gets really close to the x-axis. Points can be generated with a table of values and plotted on a set of cartesian axes.\r\n<h2><\/h2>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Define and Evaluate Exponential Functions\n<ul>\n<li>Define an exponential function and it&#8217;s domain and range<\/li>\n<li>Evaluate an exponential function<\/li>\n<li>Define and evaluate a compound interest formula<\/li>\n<\/ul>\n<\/li>\n<li>Exponential Functions with Base e\n<ul>\n<li>Define the number\u00a0<em>e<\/em><\/li>\n<li>Define continuous growth as an exponential function with base\u00a0<em>e<\/em><\/li>\n<li>Evaluate exponential functions with base\u00a0<em>e<\/em><\/li>\n<\/ul>\n<\/li>\n<li>Graph Exponential Functions\n<ul>\n<li>Generate a table of values for an exponential function<\/li>\n<li>Plot an exponential function on Cartesian axes<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<p>India is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year.<a class=\"footnote\" title=\"http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014.\" id=\"return-footnote-3503-1\" href=\"#footnote-3503-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> If this rate continues, the population of India will exceed China\u2019s population by the year 2031. When populations grow rapidly, we often say that the growth is &#8220;exponential,&#8221; meaning that something is increasing very quickly. To a mathematician, however, the term <em>exponential growth <\/em>has a very specific meaning. In this section, we will take a look at <em>exponential functions<\/em>, which model this kind of rapid growth.<\/p>\n<p>Linear functions have a\u00a0constant rate of change \u2013 a constant number that the output increases for each increase in input. For example, in the equation [latex]f(x)=3x+4[\/latex]\u00a0, the slope tells us the output increases by three each time the input increases by one. Sometimes, on the other hand, quantities grow by a percent rate of change rather than by a fixed amount. In this lesson, we will define a function whose rate of change increases by a percent of the current value rather than a fixed quantity.<\/p>\n<p>To illustrate\u00a0this difference consider two companies whose business is expanding: Company A has 100 stores, and expands by opening 50 new stores a year Company B has 100 stores, and expands by increasing the number of stores by 50% of their total each year.<\/p>\n<p>The table below compares\u00a0the growth of each company where company A increases the number of stores linearly, and company B increases the number of stores by a rate of 50% each year.<\/p>\n<table style=\"width: 60%;\">\n<thead>\n<tr>\n<td>Year<\/td>\n<td>Stores, Company A<\/td>\n<td>\u00a0Description of Growth<\/td>\n<td>Stores, Company B<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>0<\/td>\n<td>100<\/td>\n<td>Starting with 100 each<\/td>\n<td>100<\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>100+50=150<\/td>\n<td>They both grow by 50 stores in the first year.<\/td>\n<td>100 + 50% of 100 100 + 0.50(100) = 150<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>150+50=200<\/td>\n<td>Store A grows by 50, Store B grows by 75<\/td>\n<td>150 + 50% of 150 150 + 0.50(150) = 225<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>200+50=250<\/td>\n<td>Store A grows by 50, Store B grows by 112.5<\/td>\n<td>\u00a0225 + 50% of 225 225 + 0.50(225) = 337.5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function [latex]A\\left(x\\right)=100+50x[\/latex]. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function [latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex].<\/p>\n<p>The graphs comparing the number of stores for each company over a five-year period are shown in below<strong>.<\/strong> We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.<\/p>\n<div style=\"width: 348px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051913\/CNX_Precalc_Figure_04_01_0012.jpg\" alt=\"Graph of Companies A and B\u2019s functions, which values are found in the previous table.\" width=\"338\" height=\"586\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">The graph shows the numbers of stores Companies A and B opened over a five-year period.<\/p>\n<\/div>\n<p id=\"fs-id1165135209682\">Notice that the domain for both functions is [latex]\\left[0,\\infty \\right)[\/latex], and the range for both functions is [latex]\\left[100,\\infty \\right)[\/latex]. After year 1, Company B always has more stores than Company A.<\/p>\n<p id=\"fs-id1165137836429\">Consider\u00a0the function representing the number of stores for Company B<\/p>\n<p style=\"text-align: center;\">[latex]B\\left(x\\right)=100{\\left(1+0.5\\right)}^{x}[\/latex]<\/p>\n<p>In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and [latex]1+0.5=1.5[\/latex] represents the growth factor. Generalizing further, we can write this function as [latex]B\\left(x\\right)=100{\\left(1.5\\right)}^{x}[\/latex], where 100 is the initial value, 1.5 is called the <em data-effect=\"italics\">base<\/em>, and <em>x<\/em>\u00a0is called the <em data-effect=\"italics\">exponent<\/em>. This is an exponential function.<\/p>\n<div id=\"fs-id1165137564690\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\n<h3 class=\"title\" data-type=\"title\">Exponential Growth<\/h3>\n<p id=\"fs-id1165137834019\">A function that models <strong>exponential growth<\/strong> grows by a rate proportional to the current amount. For any real number <em>x<\/em>\u00a0and any positive real numbers <em>a\u00a0<\/em>and <em>b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form<\/p>\n<div id=\"fs-id1165137851784\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(x\\right)=a{b}^{x}[\/latex]<\/div>\n<p id=\"eip-626\">where<\/p>\n<ul id=\"fs-id1165137863819\">\n<li><em>a<\/em>\u00a0is the initial or starting value of the function.<\/li>\n<li><em>b<\/em>\u00a0is the growth factor or growth multiplier per unit <em>x<\/em>.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165137644244\">To evaluate an exponential function with the form [latex]f\\left(x\\right)={b}^{x}[\/latex], we simply substitute <em>x<\/em>\u00a0with the given value, and calculate the resulting power. For example:<\/p>\n<p id=\"fs-id1165135403544\">Let [latex]f\\left(x\\right)={2}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\n<div id=\"eip-id1165137643186\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill & ={2}^{x}\\hfill & \\hfill \\\\ f\\left(3\\right)\\hfill & ={2}^{3}\\text{ }\\hfill & \\text{Substitute }x=3.\\hfill \\\\ \\hfill & =8\\text{ }\\hfill & \\text{Evaluate the power}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137849020\">To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example:<\/p>\n<p id=\"fs-id1165137849024\">Let [latex]f\\left(x\\right)=30{\\left(2\\right)}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\n<div id=\"eip-id1165134086025\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill & =30{\\left(2\\right)}^{x}\\hfill & \\hfill \\\\ f\\left(3\\right)\\hfill & =30{\\left(2\\right)}^{3}\\hfill & \\text{Substitute }x=3.\\hfill \\\\ \\hfill & =30\\left(8\\right)\\text{ }\\hfill & \\text{Simplify the power first}\\text{.}\\hfill \\\\ \\hfill & =240\\hfill & \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137841073\">Note that if the order of operations were not followed, the result would be incorrect:<\/p>\n<div id=\"eip-id1165135320147\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\"><\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">In our first example we will evaluate an exponential function without the aid of a calculator.<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Let [latex]f\\left(x\\right)=5{\\left(3\\right)}^{x+1}[\/latex]. Evaluate [latex]f\\left(2\\right)[\/latex] without using a calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q211228\">Show Answer<\/span><\/p>\n<div id=\"q211228\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137598173\">Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\n<div id=\"eip-id1165135208555\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}f\\left(x\\right)\\hfill & =5{\\left(3\\right)}^{x+1}\\hfill & \\hfill \\\\ f\\left(2\\right)\\hfill & =5{\\left(3\\right)}^{2+1}\\hfill & \\text{Substitute }x=2.\\hfill \\\\ \\hfill & =5{\\left(3\\right)}^{3}\\hfill & \\text{Add the exponents}.\\hfill \\\\ \\hfill & =5\\left(27\\right)\\hfill & \\text{Simplify the power}\\text{.}\\hfill \\\\ \\hfill & =135\\hfill & \\text{Multiply}\\text{.}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we present more examples of evaluating an exponential function at several different values.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Determine Exponential Function Values and Graph the Function\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/QFFAoX0We34?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next example we will revisit the population of India.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with an annual growth rate of about 1.2%. This situation is represented by the growth function [latex]P\\left(t\\right)=1.25{\\left(1.012\\right)}^{t}[\/latex], where <em>t<\/em>\u00a0is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q385742\">Show Answer<\/span><\/p>\n<div id=\"q385742\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137786635\">To estimate the population in 2031, we evaluate the models for <em>t\u00a0<\/em>= 18, because 2031 is 18 years after 2013. Rounding to the nearest thousandth,<\/p>\n<div id=\"eip-id1165135657117\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]P\\left(18\\right)=1.25{\\left(1.012\\right)}^{18}\\approx 1.549[\/latex]<\/div>\n<p id=\"fs-id1165135394343\">There will be about 1.549 billion people in India in the year 2031.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show another example of using an exponential function to predict the population of a small town.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Evaluate a Given Exponential Function to Predict a Future Population\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/SbIydBmJePE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You may have seen formulas that are used to calculate compound interest rates. \u00a0These formulas are another example of exponential growth.\u00a0The term <em data-effect=\"italics\">compounding<\/em> refers to interest earned not only on the original value, but on the accumulated value of the account.<\/p>\n<p id=\"fs-id1165137447037\">The <strong>annual percentage rate (APR)<\/strong> of an account, also called the <strong>nominal rate<\/strong>, is the yearly interest rate earned by an investment account. The term\u00a0<em data-effect=\"italics\">nominal<\/em>\u00a0is 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 effective interest rate ends up being <em data-effect=\"italics\">greater<\/em> than the nominal rate! This is a powerful tool for investing.<\/p>\n<p id=\"fs-id1165135160118\">We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time <em>t<\/em>, principal <em>P<\/em>, APR <em>r<\/em>, and number of compounding periods in a year\u00a0<em>n<\/em>:<\/p>\n<div id=\"eip-986\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\\\[\/latex]<\/div>\n<div class=\"textbox\">\n<h3 class=\"title\" data-type=\"title\">The Compound Interest Formula<\/h3>\n<p id=\"fs-id1165135184167\"><strong>Compound interest<\/strong> can be calculated using the formula<\/p>\n<div id=\"fs-id1165135184172\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\\\[\/latex]<\/div>\n<p id=\"eip-237\">where<\/p>\n<ul id=\"fs-id1165137448453\">\n<li><em>A<\/em>(<em>t<\/em>) is the account value,<\/li>\n<li><i>t<\/i> is measured in years,<\/li>\n<li><em>P<\/em>\u00a0is the starting amount of the account, often called the principal, or more generally present value,<\/li>\n<li><em>r<\/em>\u00a0is the annual percentage rate (APR) expressed as a decimal, and<\/li>\n<li><em>n<\/em>\u00a0is the number of compounding periods in one year.<\/li>\n<\/ul>\n<\/div>\n<p>In our next example we will calculate the value of an account after 10 years of interest compounded quarterly.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q689928\">Show Answer<\/span><\/p>\n<div id=\"q689928\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137812832\">Because we are starting with $3,000, <em>P\u00a0<\/em>= 3000. Our interest rate is 3%, so <em>r<\/em>\u00a0=\u00a00.03. Because we are compounding quarterly, we are compounding 4 times per year, so <em>n\u00a0<\/em>= 4. We want to know the value of the account in 10 years, so we are looking for <em>A<\/em>(10), the value when <em>t <\/em>= 10.<\/p>\n<div id=\"eip-id1402796\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =P\\left(1+\\frac{r}{n}\\right)^{nt}\\hfill & \\text{Use the compound interest formula}. \\\\ A\\left(10\\right)\\hfill & =3000\\left(1+\\frac{0.03}{4}\\right)^{4\\cdot 10}\\hfill & \\text{Substitute using given values}. \\\\ \\text{ }\\hfill & \\approx 4045.05\\hfill & \\text{Round to two decimal places}.\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137694040\">The account will be worth about $4,045.05 in 10 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows an example of using an exponential growth to calculate interest compounded quarterly.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 1:  Compounded Interest Formula - Quarterly\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/3az4AKvUmmI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In our next example we will use the compound interest formula to solve for the principal.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child\u2019s 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?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q254680\">Show Answer<\/span><\/p>\n<div id=\"q254680\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137664627\">The nominal interest rate is 6%, so <em>r\u00a0<\/em>= 0.06. Interest is compounded twice a year, so <em>k\u00a0<\/em>= 2.<\/p>\n<p id=\"fs-id1165135209414\">We want to find the initial investment, <em>P<\/em>, 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 <em>P<\/em>.<\/p>\n<div id=\"eip-id1165131884554\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =P{\\left(1+\\frac{r}{n}\\right)}^{nt}\\hfill & \\text{Use the compound interest formula}.\\hfill \\\\ 40,000\\hfill & =P{\\left(1+\\frac{0.06}{2}\\right)}^{2\\left(18\\right)}\\hfill & \\text{Substitute using given values }A\\text{, }r, n\\text{, and }t.\\hfill \\\\ 40,000\\hfill & =P{\\left(1.03\\right)}^{36}\\hfill & \\text{Simplify}.\\hfill \\\\ \\frac{40,000}{{\\left(1.03\\right)}^{36}}\\hfill & =P\\hfill & \\text{Isolate }P.\\hfill \\\\ P\\hfill & \\approx 13,801\\hfill & \\text{Divide and round to the nearest dollar}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137937589\">Lily will need to invest $13,801 to have $40,000 in 18 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show another example of finding the deposit amount necessary to obtain a future value from compounded interest.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex: Compounded Interest Formula - Determine Deposit Needed (Present Value)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/saq9dF7a4r8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<section id=\"fs-id1165137724961\" data-depth=\"1\">\n<h2>Exponential functions with base e<\/h2>\n<p>As we saw earlier, the amount earned on an account increases as the compounding frequency increases. The table below\u00a0shows 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.<\/p>\n<p id=\"fs-id1165135684377\">Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies.<\/p>\n<table id=\"Table_04_01_04\" summary=\"Nine rows and three columns. The first column is labeled,\">\n<thead>\n<tr>\n<th data-align=\"center\">Frequency<\/th>\n<th data-align=\"center\">[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\n<th data-align=\"center\">Value<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Annually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\n<td>$2<\/td>\n<\/tr>\n<tr>\n<td>Semiannually<\/td>\n<td>[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\n<td>$2.25<\/td>\n<\/tr>\n<tr>\n<td>Quarterly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\n<td>$2.441406<\/td>\n<\/tr>\n<tr>\n<td>Monthly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\n<td>$2.613035<\/td>\n<\/tr>\n<tr>\n<td>Daily<\/td>\n<td>[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\n<td>$2.714567<\/td>\n<\/tr>\n<tr>\n<td>Hourly<\/td>\n<td>[latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\n<td>$2.718127<\/td>\n<\/tr>\n<tr>\n<td>Once per minute<\/td>\n<td>[latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\n<td>$2.718279<\/td>\n<\/tr>\n<tr>\n<td>Once per second<\/td>\n<td>[latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\n<td>$2.718282<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137828146\">These values appear to be approaching a limit as <em>n<\/em>\u00a0increases. In fact, as <em>n<\/em>\u00a0gets larger and larger, the expression [latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex] approaches a number used so frequently in mathematics that it has its own name: the letter [latex]e[\/latex]. 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.<\/p>\n<div id=\"fs-id1165135511324\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\n<h3 class=\"title\" data-type=\"title\">A General Note: The Number <em data-effect=\"italics\">e<\/em><\/h3>\n<p id=\"fs-id1165135511335\">The letter <em data-effect=\"italics\">e<\/em> represents the irrational number<\/p>\n<div id=\"eip-id1165135378658\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">as n increases without bound<\/div>\n<p id=\"fs-id1165135369344\">The letter <em data-effect=\"italics\">e <\/em>is used as a base for many real-world exponential models. To work with base <em data-effect=\"italics\">e<\/em>, we use the approximation, [latex]e\\approx 2.718282[\/latex]. The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.<\/p>\n<\/div>\n<p>In our first example we will use a calculator to find powers of\u00a0<em>e.<\/em><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Calculate [latex]{e}^{3.14}[\/latex]. Round to five decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q465847\">Show Answer<\/span><\/p>\n<div id=\"q465847\" class=\"hidden-answer\" style=\"display: none\">On a calculator, press the button labeled [latex]\\left[{e}^{x}\\right][\/latex]. The window shows [<em>e<\/em>^(]. Type 3.14 and then close parenthesis, (]). Press [ENTER]. Rounding to 5 decimal places, [latex]{e}^{3.14}\\approx 23.10387[\/latex]. Caution: Many scientific calculators have an &#8220;Exp&#8221; button, which is used to enter numbers in scientific notation. It is not used to find powers of <em>e<\/em>.<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137827923\" data-depth=\"1\">\n<h2 data-type=\"title\">Investigating Continuous Growth<\/h2>\n<p id=\"fs-id1165137827929\">So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, <em data-effect=\"italics\">e <\/em>is used as the base for exponential functions. Exponential models that use <em>e<\/em>\u00a0as the base are called <em data-effect=\"italics\">continuous growth or decay models<\/em>. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.<\/p>\n<div id=\"fs-id1165137664673\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\n<h3 class=\"title\" data-type=\"title\">The Continuous Growth\/Decay Formula<\/h3>\n<p id=\"fs-id1165135453868\">For all real numbers r,\u00a0<em>t<\/em>, and all positive numbers <em>a<\/em>, continuous growth or decay is represented by the formula<\/p>\n<div id=\"fs-id1165135536370\" class=\"equation\" style=\"text-align: center;\" data-type=\"equation\">[latex]A\\left(t\\right)=a{e}^{rt}[\/latex]<\/div>\n<p id=\"eip-101\">where<\/p>\n<ul id=\"fs-id1165135152052\">\n<li><em>a<\/em>\u00a0is the initial value,<\/li>\n<li><em>r<\/em>\u00a0is the continuous growth or decay rate per unit time,<\/li>\n<li>and <em>t<\/em>\u00a0is the elapsed time.<\/li>\n<\/ul>\n<p id=\"fs-id1165135560686\">If <em>r\u00a0<\/em>&gt; 0, then the formula represents continuous growth. If <em>r\u00a0<\/em>&lt; 0, then the formula represents continuous decay.<\/p>\n<p id=\"fs-id1165137812323\">For business applications, the continuous growth formula is called the continuous compounding formula and takes the form<\/p>\n<div id=\"eip-id1165134324899\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A\\left(t\\right)=P{e}^{rt}[\/latex]<\/div>\n<p id=\"eip-962\">where<\/p>\n<ul id=\"fs-id1165137827330\">\n<li><em>P<\/em>\u00a0is the principal or the initial invested,<\/li>\n<li><em>r<\/em>\u00a0is the growth or interest rate per unit time,<\/li>\n<li>and <em>t<\/em>\u00a0is the period or term of the investment.<\/li>\n<\/ul>\n<\/div>\n<p>In our next example\u00a0we will calculate continuous growth of an investment. It is important to note the language that is used in the instructions for interest rate problems. \u00a0You will know to use the <em>continuous<\/em> growth or decay formula when you are asked to find an amount based on continuous compounding. \u00a0In previous examples we asked that you find an amount based on quarterly or monthly compounding, and for this you used the <em>compound<\/em> interest formula.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>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?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q33008\">Show Answer<\/span><\/p>\n<div id=\"q33008\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the account is growing in value, this is a continuous compounding problem with growth rate <em>r\u00a0<\/em>= 0.10. The initial investment was $1,000, so <em>P\u00a0<\/em>= 1000. We use the continuous compounding formula to find the value after <em>t\u00a0<\/em>= 1 year:<\/p>\n<div id=\"eip-id1165133351794\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =P{e}^{rt}\\hfill & \\text{Use the continuous compounding formula}.\\hfill \\\\ \\hfill & =1000{\\left(e\\right)}^{0.1} & \\text{Substitute known values for }P, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 1105.17\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137895288\">The account is worth $1,105.17 after one year.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show another example of continuous interest.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex 1:  Continuous Interest Formula\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/fEjrYCog_8w?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div id=\"fs-id1165135411368\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\n<h3 id=\"fs-id1165135411373\">How To: Given the initial value, rate of growth or decay, and time <em>t<\/em>, solve a continuous growth or decay function.<\/h3>\n<ol id=\"fs-id1165135511371\" data-number-style=\"arabic\">\n<li>Use the information in the problem to determine <em>a<\/em>, the initial value of the function.<\/li>\n<li>Use the information in the problem to determine the growth rate <em>r<\/em>.\n<ol id=\"fs-id1165135188096\" data-number-style=\"lower-alpha\">\n<li>If the problem refers to continuous growth, then <em>r\u00a0<\/em>&gt; 0.<\/li>\n<li>If the problem refers to continuous decay, then <em>r\u00a0<\/em>&lt; 0.<\/li>\n<\/ol>\n<\/li>\n<li>Use the information in the problem to determine the time <em>t<\/em>.<\/li>\n<li>Substitute the given information into the continuous growth formula and solve for <em>A<\/em>(<em>t<\/em>).<\/li>\n<\/ol>\n<\/div>\n<p>In our next example we will calculate continuous decay. Pay attention to the rate &#8211; it is negative which means we are considering a situation where an amount decreases or decays.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>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?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q995802\">Show Answer<\/span><\/p>\n<div id=\"q995802\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the substance is decaying, the rate, 17.3%, is negative. So, <em>r\u00a0<\/em>=\u00a0\u20130.173. The initial amount of radon-222 was 100 mg, so <em>a\u00a0<\/em>= 100. We use the continuous decay formula to find the value after <em>t\u00a0<\/em>= 3 days:<\/p>\n<div id=\"eip-id1165137779893\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =a{e}^{rt}\\hfill & \\text{Use the continuous growth formula}.\\hfill \\\\ \\hfill & =100{e}^{-0.173\\left(3\\right)} & \\text{Substitute known values for }a, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 59.5115\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137697132\">So 59.5115 mg of radon-222 will remain.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show an example of calculating remaining amount of a radioactive substance after it decays for a length of time.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Determine a Continuous Exponential Decay Function and Make a Prediction\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Vyl3NcTGRAo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Graph Exponential Functions<\/h2>\n<p id=\"fs-id1165137592823\">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. Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex] whose base is greater than one. We\u2019ll use the function [latex]f\\left(x\\right)={2}^{x}[\/latex]. Observe how the output values in the table below\u00a0change as the input increases by 1.<\/p>\n<table style=\"width: 60%;\" summary=\"Two rows and eight columns. The first row is labeled,\">\n<colgroup>\n<col data-width=\"60\" \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>\u20133<\/td>\n<td>\u20132<\/td>\n<td>\u20131<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>8<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165137432031\">Each output value is the product of the previous output and the base, 2. We call the base 2 the <em data-effect=\"italics\">constant ratio<\/em>. In fact, for any exponential function with the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex], <em>b<\/em>\u00a0is 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 <em>a<\/em>.<\/p>\n<p id=\"fs-id1165137585799\">Notice from the table that<\/p>\n<ul id=\"fs-id1165137658509\">\n<li>the output values are positive for all values of <em>x<\/em>;<\/li>\n<li>as <em>x<\/em>\u00a0increases, the output values increase without bound; and<\/li>\n<li>as <em>x<\/em>\u00a0decreases, the output values grow smaller, approaching zero.<\/li>\n<\/ul>\n<p>Figure 1\u00a0shows the exponential growth function [latex]f\\left(x\\right)={2}^{x}[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051915\/CNX_Precalc_Figure_04_02_0012.jpg\" alt=\"Graph of the exponential function, 2^(x), with labeled points at (-3, 1\/8), (-2, \u00bc), (-1, \u00bd), (0, 1), (1, 2), (2, 4), and (3, 8). The graph notes that the x-axis is an asymptote.\" width=\"487\" height=\"520\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">Notice that the graph gets close to the x-axis, but never touches it.<\/p>\n<\/div>\n<p id=\"fs-id1165137459614\">The domain of [latex]f\\left(x\\right)={2}^{x}[\/latex] is all real numbers, the range is [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<p id=\"fs-id1165137838249\">To get a sense of the behavior of <strong>exponential decay<\/strong>, we can create a table of values for a function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex] whose base is between zero and one. We\u2019ll use the function [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex]. Observe how the output values in the table below\u00a0change as the input increases by 1.<\/p>\n<table style=\"width: 60%;\" summary=\"Two rows and eight columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>\u20133<\/td>\n<td>\u20132<\/td>\n<td>\u20131<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]g\\left(x\\right)=\\left(\\frac{1}{2}\\right)^{x}[\/latex]<\/strong><\/td>\n<td>8<\/td>\n<td>4<\/td>\n<td>2<\/td>\n<td>1<\/td>\n<td>[latex]\\frac{1}{2}[\/latex]<\/td>\n<td>[latex]\\frac{1}{4}[\/latex]<\/td>\n<td>[latex]\\frac{1}{8}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165135347846\">Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or constant ratio [latex]\\frac{1}{2}[\/latex].<\/p>\n<p id=\"fs-id1165137452063\">Notice from the table that<\/p>\n<ul id=\"fs-id1165135499992\">\n<li>the output values are positive for all values of <em>x<\/em>;<\/li>\n<li>as <em>x<\/em>\u00a0increases, the output values grow smaller, approaching zero; and<\/li>\n<li>as <em>x<\/em>\u00a0decreases, the output values grow without bound.<\/li>\n<\/ul>\n<p id=\"fs-id1165137405421\">The graph shows the exponential decay function, [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051917\/CNX_Precalc_Figure_04_02_0022.jpg\" alt=\"Graph of decreasing exponential function, (1\/2)^x, with labeled points at (-3, 8), (-2, 4), (-1, 2), (0, 1), (1, 1\/2), (2, 1\/4), and (3, 1\/8). The graph notes that the x-axis is an asymptote.\" width=\"487\" height=\"520\" data-media-type=\"image\/jpg\" \/><\/p>\n<p id=\"fs-id1165137723586\" style=\"text-align: center;\"><strong>\u00a0<\/strong>The domain of [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex] is all real numbers, the range is [latex]\\left(0,\\infty \\right)[\/latex]<\/p>\n<div id=\"fs-id1165135571835\" class=\"note textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"A General Note\">\n<h3 class=\"title\" data-type=\"title\">Characteristics of the Graph of \u00a0<em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = <em data-effect=\"italics\">b<\/em><sup><em data-effect=\"italics\">x<\/em><\/sup><\/h3>\n<p id=\"fs-id1165137848929\">An exponential function with the form [latex]f\\left(x\\right)={b}^{x}[\/latex], [latex]b>0[\/latex], [latex]b\\ne 1[\/latex], has these characteristics:<\/p>\n<ul id=\"fs-id1165135186684\">\n<li><strong>one-to-one<\/strong> function<\/li>\n<li>domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\n<li>range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\n<li><em data-effect=\"italics\">x-<\/em>intercept: none<\/li>\n<li><em data-effect=\"italics\">y-<\/em>intercept: [latex]\\left(0,1\\right)[\/latex]<\/li>\n<li>increasing if [latex]b>1[\/latex]<\/li>\n<li>decreasing if [latex]b<1[\/latex]<\/li>\n<\/ul>\n<p id=\"fs-id1165137471878\">Compare the graphs of <strong>exponential growth<\/strong> and decay functions.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051919\/CNX_Precalc_Figure_04_02_003new2.jpg\" alt=\"Graph of two functions where the first graph is of a function of f(x) = b^x when b&gt;1 and the second graph is of the same function when b is 0&lt;b&lt;1. Both graphs have the points (0, 1) and (1, b) labeled.\" width=\"731\" height=\"407\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<p>In our first example we will plot an exponential decay function where the base is between 0 and 1.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Sketch a graph of [latex]f\\left(x\\right)={0.25}^{x}[\/latex]. State the domain, range.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q203605\">Show Answer<\/span><\/p>\n<div id=\"q203605\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137734539\">Before graphing, identify the behavior and create a table of points for the graph.<\/p>\n<ul>\n<li>Since <em>b\u00a0<\/em>= 0.25 is between zero and one, we know the function is decreasing, and we can verify this by creating a table of values. The left tail of the graph will increase without bound and the right tail will get really close to the x-axis.<\/li>\n<li>Create a table of points.<br \/>\n<table id=\"Table_04_02_03\" summary=\"Two rows and eight columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>\u20133<\/td>\n<td>\u20132<\/td>\n<td>\u20131<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)={0.25}^{x}[\/latex]<\/strong><\/td>\n<td>64<\/td>\n<td>16<\/td>\n<td>4<\/td>\n<td>1<\/td>\n<td>0.25<\/td>\n<td>0.0625<\/td>\n<td>0.015625<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Plot the <em data-effect=\"italics\">y<\/em>-intercept, [latex]\\left(0,1\\right)[\/latex], along with two other points. We can use [latex]\\left(-1,4\\right)[\/latex] and [latex]\\left(1,0.25\\right)[\/latex].<\/li>\n<\/ul>\n<p id=\"fs-id1165137482830\">Draw a smooth curve connecting the points.<span id=\"fs-id1165137940681\" data-type=\"media\" data-alt=\"Graph of the decaying exponential function f(x) = 0.25^x with labeled points at (-1, 4), (0, 1), and (1, 0.25).\"><br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05051920\/CNX_Precalc_Figure_04_02_0042.jpg\" alt=\"Graph of the decaying exponential function f(x) = 0.25^x with labeled points at (-1, 4), (0, 1), and (1, 0.25).\" data-media-type=\"image\/jpg\" \/><\/span><\/p>\n<p id=\"fs-id1165137548870\" style=\"text-align: center;\"><strong>\u00a0<\/strong>The domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]; the range is [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we show another example of graphing an exponential function. The base of the exponential term is between 0 and 1, so this graph will represent decay.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Graph a Basic Exponential Function Using a Table of Values\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/FMzZB9Ve-1U?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div id=\"fs-id1165134195243\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\n<h3 id=\"fs-id1165135194093\">How To: Given an exponential function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex], graph the function.<\/h3>\n<ol id=\"fs-id1165137435782\" data-number-style=\"arabic\">\n<li>Create a table of points.<\/li>\n<li>Plot at least 3\u00a0point from the table, including the <em data-effect=\"italics\">y<\/em>-intercept [latex]\\left(0,1\\right)[\/latex].<\/li>\n<li>Draw a smooth curve through the points.<\/li>\n<li>State the domain, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], the range, [latex]\\left(0,\\infty \\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>In our next example we will plot an exponential growth function where the base is greater than 1.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Sketch a graph of [latex]f(x)={\\sqrt{2}(\\sqrt{2})}^{x}[\/latex].\u00a0State the domain, range.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q334418\">Show Answer<\/span><\/p>\n<div id=\"q334418\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137734539\">Before graphing, identify the behavior and create a table of points for the graph.<\/p>\n<ul>\n<li>Since <em>b\u00a0<\/em>= [latex]\\sqrt{2}[\/latex] which is greater than\u00a0one, we know the function is increasing, and we can verify this by creating a table of values. The left tail of the graph will\u00a0get really close to the x-axis and the right tail will increase without bound.<\/li>\n<li>Create a table of points.<br \/>\n<table id=\"Table_04_02_03\" summary=\"Two rows and eight columns. The first row is labeled,\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>\u20133<\/td>\n<td>\u20132<\/td>\n<td>\u20131<\/td>\n<td>0<\/td>\n<td>1<\/td>\n<td>2<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td><strong>[latex]f\\left(x\\right)=\\sqrt{2}{(\\sqrt{2})}^{x}[\/latex]<\/strong><\/td>\n<td>0.5<\/td>\n<td>0.71<\/td>\n<td>1<\/td>\n<td>1.41<\/td>\n<td>2<\/td>\n<td>2.83<\/td>\n<td>4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<li>Plot the <em data-effect=\"italics\">y<\/em>-intercept, [latex]\\left(0,1.41\\right)[\/latex], along with two other points. We can use [latex]\\left(-1,1\\right)[\/latex] and [latex]\\left(1,2\\right)[\/latex].<\/li>\n<\/ul>\n<p id=\"fs-id1165137482830\">Draw a smooth curve connecting the points.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3623 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/08\/05185425\/Screen-Shot-2016-08-05-at-11.53.45-AM.png\" alt=\"Screen Shot 2016-08-05 at 11.53.45 AM\" width=\"326\" height=\"231\" \/><\/p>\n<p>The domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]; the range is [latex]\\left(0,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Our next video example includes graphing an exponential growth function and defining the domain and range of the function.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Graph an Exponential Function Using a Table of Values\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/M6bpp0BRIf0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<\/section>\n<p>Exponential growth grows by a rate proportional to the current amount.\u00a0For any real number <em>x<\/em>\u00a0and any positive real numbers <em>a\u00a0<\/em>and <em>b<\/em>\u00a0such that [latex]b\\ne 1[\/latex], an exponential growth function has the form\u00a0[latex]f\\left(x\\right)=a{b}^{x}[\/latex]. \u00a0Evaluating exponential functions requires careful attention to the order of operations. Compound interest is an example of exponential growth.<\/p>\n<p>Continuous growth or decay functions are of the form\u00a0[latex]A\\left(t\\right)=a{e}^{rt}[\/latex].\u00a0If <em>r\u00a0<\/em>&gt; 0, then the formula represents continuous growth. If <em>r\u00a0<\/em>&lt; 0, then the formula represents continuous decay.\u00a0For business applications, the continuous growth formula is called the continuous compounding formula and takes the form\u00a0[latex]A\\left(t\\right)=P{e}^{rt}[\/latex].<\/p>\n<p>Graphs of exponential growth functions will have a right tail that increases without bound, and a left tail that gets really close to the x-axis. On the other hand, graph so exponential decay functions will have a left tail that increases without bound and a right tail that gets really close to the x-axis. Points can be generated with a table of values and plotted on a set of cartesian axes.<\/p>\n<h2><\/h2>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3503\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Determine Exponential Function Values and Graph the Function. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/QFFAoX0We34\">https:\/\/youtu.be\/QFFAoX0We34<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Evaluate a Given Exponential Function to Predict a Future Population. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/SbIydBmJePE\">https:\/\/youtu.be\/SbIydBmJePE<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Determine a Continuous Exponential Decay Function and Make a Prediction. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Vyl3NcTGRAo\">https:\/\/youtu.be\/Vyl3NcTGRAo<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Graph a Basic Exponential Function Using a Table of Values. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/FMzZB9Ve-1U\">https:\/\/youtu.be\/FMzZB9Ve-1U<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Graph an Exponential Function Using a Table of Values. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/M6bpp0BRIf0\">https:\/\/youtu.be\/M6bpp0BRIf0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 1: Compounded Interest Formula - Quarterly. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/3az4AKvUmmI\">https:\/\/youtu.be\/3az4AKvUmmI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Compounded Interest Formula - Determine Deposit Needed (Present Value). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/saq9dF7a4r8\">https:\/\/youtu.be\/saq9dF7a4r8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at : http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/li><li>Ex 1: Continuous Interest Formula. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/fEjrYCog_8w\">https:\/\/youtu.be\/fEjrYCog_8w<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-3503-1\">http:\/\/www.worldometers.info\/world-population\/. Accessed February 24, 2014. <a href=\"#return-footnote-3503-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Determine Exponential Function Values and Graph the Function\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/QFFAoX0We34\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Evaluate a Given Exponential Function to Predict a Future Population\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/SbIydBmJePE\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Compounded Interest Formula - Quarterly\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/3az4AKvUmmI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Compounded Interest Formula - Determine Deposit Needed (Present Value)\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/saq9dF7a4r8\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay, et al.\",\"organization\":\"OpenStax\",\"url\":\" http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at : http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Continuous Interest Formula\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/fEjrYCog_8w\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Determine a Continuous Exponential Decay Function and Make a Prediction\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Vyl3NcTGRAo\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Graph a Basic Exponential Function Using a Table of Values\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/FMzZB9Ve-1U\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Graph an Exponential Function Using a Table of Values\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/M6bpp0BRIf0\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3503","chapter","type-chapter","status-publish","hentry"],"part":1709,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3503","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":17,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3503\/revisions"}],"predecessor-version":[{"id":4375,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3503\/revisions\/4375"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1709"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/3503\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/media?parent=3503"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=3503"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/contributor?post=3503"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/wp-json\/wp\/v2\/license?post=3503"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}