![\"Escherichia](\"http:\/\/cnx.org\/resources\/f4bed7000de1219b5e7633366abef4a76617da73\/CNX_Precalc_Figure_04_00_001.jpg\")
Figure 1.<\/b> Electron micrograph of E.Coli bacteria (credit: \u201cMattosaurus,\u201d Wikimedia Commons)[\/caption]\r\nExponential Functions<\/h2>\r\n
In this chapter, we will explore exponential functions, which can be used for, among other things, modeling growth patterns such as those found in bacteria. We will also investigate logarithmic functions, which are closely related to exponential functions. Both types of functions have numerous real-world applications when it comes to modeling and interpreting data.<\/p>\r\n\r\n
Learning Outcomes<\/h3>\r\n\r\n \t- Review Laws of Exponents<\/li>\r\n \t
- Evaluate exponential functions.<\/li>\r\n \t
- Find the equation of an exponential function.<\/li>\r\n \t
- Evaluate exponential functions with base e.<\/li>\r\n \t
- Graph exponential functions<\/li>\r\n \t
- Graph exponential functions using transformations.<\/li>\r\n<\/ul>\r\n<\/div>\r\n
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.[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 growing very rapidly. To a mathematician, however, the term exponential growth <\/em>has a very specific meaning. In this section, we will take a look at exponential functions<\/em>, which model this kind of rapid growth.<\/p>\r\nFirst, we will briefly review laws of exponents.\r\n\r\nThe Product Rule for Exponents<\/h3>\r\nFor any number a<\/i> and any integers n<\/i> and m<\/i>, [latex]\\left(a^{n}\\right)\\left(a^{m}\\right) = a^{n+m}[\/latex].\r\n\r\n<\/div>\r\n\r\nThe Quotient (Division) Rule for Exponents<\/h3>\r\nFor any non-zero number a<\/i> and any integers n<\/i> and n<\/i>: [latex] \\displaystyle \\frac{{{a}^{n}}}{{{a}^{m}}}={{a}^{n-m}}[\/latex]\r\n\r\n<\/div>\r\n\r\nThe Power Rule for Exponents<\/h3>\r\nFor any positive number a<\/i> and integers n<\/i> and m<\/i>: [latex]\\left(a^{n}\\right)^{m}=a^{n\\cdot{m}}[\/latex].\r\n\r\nTake a moment to contrast how this is different from the product rule for exponents.\r\n\r\n<\/div>\r\n\r\nA Product Raised to a Power<\/h3>\r\nFor any nonzero numbers a<\/i> and b<\/i> and any integer n<\/i>, [latex]\\left(ab\\right)^{n}=a^{n}\\cdot{b^{n}}[\/latex].\r\n\r\n<\/div>\r\n\r\nA Quotient Raised to a Power<\/h3>\r\nFor any number a<\/i>, any non-zero number b<\/i>, and any integer n<\/i>, [latex] \\displaystyle {\\left(\\frac{a}{b}\\right)}^{n}=\\frac{a^{n}}{b^{n}}[\/latex].\r\n\r\n<\/div>\r\n\r\nExponents of 0 or 1<\/h3>\r\nAny number or variable raised to a power of 1 is the number itself.\r\n
[latex]a^{1}=a[\/latex]<\/p>\r\nAny non-zero number or variable raised to a power of 0 is equal to 1\r\n
[latex]a^{0}=1[\/latex]<\/p>\r\nThe quantity [latex]0^{0}[\/latex] is undefined.\r\n\r\n<\/div>\r\n
\r\nThe Negative Rule of Exponents<\/h3>\r\nFor any nonzero real number [latex]a[\/latex] and natural number [latex]n[\/latex], the negative rule of exponents states that\r\n[latex]{a}^{-n}=\\dfrac{1}{{a}^{n}}[\/latex]<\/div>\r\nand<\/p>\r\n\r\n
[latex]\\left(\\dfrac{a}{b}\\right)^{-n}=\\left(\\dfrac{b}{a}\\right)^{n}[\/latex]<\/div>\r\n<\/div>\r\n\r\n\r\n\r\nExample 1: Practice Exponent Rules<\/h3>\r\n
Simplify [latex]\\frac{\\left(t^{3}\\right)^2}{\\left(t^2\\right)^{-8}}[\/latex]\r\nWrite your answer with positive exponents.<\/p>\r\n[reveal-answer q=\"847890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847890\"]\r\n\r\nWe can either rewrite this expression with positive exponents first or use the Product Raised to a Power Rule first.\r\n\r\nLet\u2019s start by simplifying the numerator and denominator using the Product Raised to a Power Rule.\r\n\r\nNumerator: [latex]\\left(t^{3}\\right)^2=t^{3\\cdot{2}}=t^6[\/latex]\r\n\r\nDenominator: [latex]\\left(t^2\\right)^{-8}=t^{2\\cdot{-8}}=t^{-16}[\/latex]\r\n\r\nNow the expression looks like this:\r\n
[latex]\\frac{t^6}{t^{-16}}[\/latex]<\/p>\r\nWe can use the quotient rule because we have the same base.\r\n\r\nQuotient Rule: [latex]\\frac{t^6}{t^{-16}}=t^{6-\\left(-16\\right)}=t^{6+16}=t^{22}[\/latex]\r\n
Answer<\/h4>\r\n[latex]\\frac{\\left(t^{3}\\right)^2}{\\left(t^2\\right)^{-8}}=t^{22}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n\r\n\r\nExample 2: Practice Exponent Rules<\/h3>\r\n
Simplify [latex]\\frac{\\left(5x\\right)^{-2}y}{x^3y^{-1}}[\/latex]<\/p>\r\nWrite your answer with positive exponents.\r\n\r\n[reveal-answer q=\"847880\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847880\"]\r\n\r\nThis time, let\u2019s start by rewriting the terms in the expression so they have positive exponents. The terms with negative exponents in the top will go to the bottom of the fraction, and the terms with negative exponents in the bottom will go to the top.\r\n
[latex]\\begin{array}{c}\\frac{\\left(5x\\right)^{-2}y}{x^3y^{-1}}\\\\\\text{ }\\\\=\\frac{\\left({y^{1}}\\right)y}{x^3\\left(5x\\right)^{2}}\\end{array}[\/latex]<\/p>\r\nNote how we left the single y term in the top because it did not have a negative exponent on it, and we left the [latex]x^3[\/latex] term in the bottom because it did not have a negative exponent on it.\r\n\r\nNow we can apply the Product Raised to a Power Rule:\r\n
[latex]\\frac{yy^{1}}{5^{2}x^3x^{2}}[\/latex]<\/p>\r\nUse the product rule to simplify further:\r\n
[latex]\\frac{yy^{1}}{5^{2}x^3x^{2}}=\\frac{y^2}{25x^{3+2}}=\\frac{y^2}{25x^5}[\/latex]<\/p>\r\n
We can\u2019t simplify any further, so our answer is<\/p>\r\n\r\n
Answer<\/h4>\r\n[latex]\\frac{\\left(5x\\right)^{-2}y}{x^3y^{-1}}=\\frac{y^2}{25x^5}[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nEvaluate Exponential Functions<\/h2>\r\n
Recall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base b<\/em>\u00a0to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:<\/p>\r\n\r\n\r\n \t- Let b\u00a0<\/em>= \u20139 and [latex]x=\\frac{1}{2}[\/latex]. Then [latex]f\\left(x\\right)=f\\left(\\frac{1}{2}\\right)={\\left(-9\\right)}^{\\frac{1}{2}}=\\sqrt{-9}[\/latex], which is not a real number.<\/li>\r\n<\/ul>\r\n
Why do we limit the base to positive values other than 1? Because base 1\u00a0results in the constant function. Observe what happens if the base is\u00a01:<\/p>\r\n\r\n
\r\n \t- Let b\u00a0<\/em>= 1. Then [latex]f\\left(x\\right)={1}^{x}=1[\/latex] for any value of x<\/em>.<\/li>\r\n<\/ul>\r\n
To evaluate an exponential function with the form [latex]f\\left(x\\right)={b}^{x}[\/latex], we simply substitute x<\/em>\u00a0with the given value, and calculate the resulting power. For example:<\/p>\r\nLet [latex]f\\left(x\\right)={2}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\r\n\r\n
[latex]\\begin{align}f\\left(x\\right)& ={2}^{x} \\\\ f\\left(3\\right)& ={2}^{3} && \\text{Substitute }x=3. \\\\ & =8 && \\text{Evaluate the power.} \\end{align}[\/latex]<\/div>\r\nTo 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
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
[latex]\\begin{align}f\\left(x\\right)& =30{\\left(2\\right)}^{x} \\\\ f\\left(3\\right)& =30{\\left(2\\right)}^{3} && \\text{Substitute }x=3. \\\\ & =30\\left(8\\right) && \\text{Simplify the power first.} \\\\ & =240 && \\text{Multiply.} \\end{align}[\/latex]<\/div>\r\nNote that if the order of operations were not followed, the result would be incorrect:<\/p>\r\n
[latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/p>\r\n\r\n
\r\n\r\n\r\nExample 3: Evaluating Exponential Functions<\/h3>\r\n
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>\r\n[reveal-answer q=\"847875\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847875\"]\r\n
Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\r\n
[latex]\\begin{align}f\\left(x\\right) & =5{\\left(3\\right)}^{x+1} \\\\ f\\left(2\\right) & =5{\\left(3\\right)}^{2+1} && \\text{Substitute }x=2. \\\\ & =5{\\left(3\\right)}^{3} && \\text{Add the exponents}. \\\\ & =5\\left(27\\right) && \\text{Simplify the power}. \\\\ & =135 && \\text{Multiply}. \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\nTry It<\/h3>\r\n
Let [latex]f\\left(x\\right)=8{\\left(1.2\\right)}^{x - 5}[\/latex]. Evaluate [latex]f\\left(3\\right)[\/latex] using a calculator. Round to four decimal places.<\/p>\r\n[reveal-answer q=\"292533\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"292533\"]\r\n\r\n5.5556\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
\u00a0Use Compound Interest Formulas<\/h2>\r\n
Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest<\/strong>. The term compounding<\/em> refers to interest earned not only on the original value, but on the accumulated value of the account.<\/p>\r\nThe annual percentage rate (APR)<\/strong> of an account, also called the nominal rate<\/strong>, is the yearly interest rate earned by an investment account. The term\u00a0nominal<\/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 greater<\/em> than the nominal rate! This is a powerful tool for investing.<\/p>\r\nWe can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time t<\/em>, principal P<\/em>, APR r<\/em>, and number of compounding periods in a year\u00a0n<\/em>:<\/p>\r\n\r\n[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/div>\r\nFor example, observe the table below, which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases.<\/p>\r\n\r\n
\r\n\r\n\r\nFrequency<\/th>\r\n Value after 1 year<\/th>\r\n<\/tr>\r\n<\/thead>\r\n \r\n\r\nAnnually<\/td>\r\n $1100<\/td>\r\n<\/tr>\r\n \r\nSemiannually<\/td>\r\n $1102.50<\/td>\r\n<\/tr>\r\n \r\nQuarterly<\/td>\r\n $1103.81<\/td>\r\n<\/tr>\r\n \r\nMonthly<\/td>\r\n $1104.71<\/td>\r\n<\/tr>\r\n \r\nDaily<\/td>\r\n $1105.16<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\nA General Note: The Compound Interest Formula<\/h3>\r\n
Compound interest<\/strong> can be calculated using the formula<\/p>\r\n\r\n[latex]A\\left(t\\right)=P{\\left(1+\\frac{r}{n}\\right)}^{nt}[\/latex]<\/div>\r\nwhere<\/p>\r\n\r\n
\r\n \t- A<\/em>(t<\/em>) is the account value,<\/li>\r\n \t
- t<\/i> is measured in years,<\/li>\r\n \t
- P<\/em>\u00a0is the starting amount of the account, often called the principal, or more generally present value,<\/li>\r\n \t
- r<\/em>\u00a0is the annual percentage rate (APR) expressed as a decimal, and<\/li>\r\n \t
- n<\/em>\u00a0is the number of compounding periods in one year.<\/li>\r\n<\/ul>\r\n<\/div>\r\n\r\n\r\n\r\n
Example 4: Calculating Compound Interest<\/h3>\r\n
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>\r\n[reveal-answer q=\"162183\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"162183\"]\r\n
Because we are starting with $3,000, P\u00a0<\/em>= 3000. Our interest rate is 3%, so r<\/em>\u00a0=\u00a00.03. Because we are compounding quarterly, we are compounding 4 times per year, so n\u00a0<\/em>= 4. We want to know the value of the account in 10 years, so we are looking for A<\/em>(10), the value when t <\/em>= 10.<\/p>\r\n[latex]\\begin{align}A\\left(t\\right) & =P\\left(1+\\frac{r}{n}\\right)^{nt}&& \\text{Use the compound interest formula}. \\\\ A\\left(10\\right)& =3000\\left(1+\\frac{0.03}{4}\\right)^{4\\cdot 10}&& \\text{Substitute using given values}. \\\\ & \\approx 4045.05&& \\text{Round to two decimal places}.\\end{align}[\/latex]<\/p>\r\n
The account will be worth about $4,045.05 in 10 years.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\nTry It<\/h3>\r\n
An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What will the investment be worth in 30 years?<\/p>\r\n[reveal-answer q=\"474337\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"474337\"]\r\n\r\nabout $3,644,675.88\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
\r\nTry It<\/h3>\r\n[ohm_question hide_question_numbers=1]14375[\/ohm_question]\r\n\r\n<\/div>\r\nEvaluate exponential functions with base e<\/h2>\r\n\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.<\/p>\r\n
Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies.<\/p>\r\n\r\n
\r\n\r\n\r\nFrequency<\/th>\r\n [latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\r\n Value<\/th>\r\n<\/tr>\r\n<\/thead>\r\n \r\n\r\nAnnually<\/td>\r\n [latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\r\n $2<\/td>\r\n<\/tr>\r\n \r\nSemiannually<\/td>\r\n [latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\r\n $2.25<\/td>\r\n<\/tr>\r\n \r\nQuarterly<\/td>\r\n [latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\r\n $2.441406<\/td>\r\n<\/tr>\r\n \r\nMonthly<\/td>\r\n [latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\r\n $2.613035<\/td>\r\n<\/tr>\r\n \r\nDaily<\/td>\r\n [latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\r\n $2.714567<\/td>\r\n<\/tr>\r\n \r\nHourly<\/td>\r\n [latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\r\n $2.718127<\/td>\r\n<\/tr>\r\n \r\nOnce per minute<\/td>\r\n [latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\r\n $2.718279<\/td>\r\n<\/tr>\r\n \r\nOnce per second<\/td>\r\n [latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\r\n $2.718282<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThese values appear to be approaching a limit as n<\/em>\u00a0increases without bound. In fact, as 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.<\/p>\r\nAs [latex]n[\/latex] approaches infinity, the expression\u00a0[latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex] approaches\u00a0[latex]e[\/latex].\u00a0 Continuous compounding occurs when [latex]n[\/latex] approaches infinity.\u00a0 Below is the formula for continuous compounding.\u00a0 Notice there is an\u00a0[latex]e[\/latex] in the formula instead of\u00a0[latex]n[\/latex].\u00a0 The [latex]e[\/latex] on a calculator is typically above the LN<\/strong> button.\r\n\r\nA General Note: continuous compounding<\/h3>\r\n
Continuous compounding interest<\/strong> can be calculated using the formula<\/p>\r\n\r\n[latex]A\\left(t\\right)=Pe^{rt}[\/latex]<\/div>\r\nwhere<\/p>\r\n\r\n
\r\n \t- A<\/em>(t<\/em>) is the account value,<\/li>\r\n \t
- t<\/i> is measured in years,<\/li>\r\n \t
- P<\/em>\u00a0is the starting amount of the account, often called the principal, or more generally present value,<\/li>\r\n \t
- r<\/em>\u00a0is the annual percentage rate (APR) expressed as a decimal<\/li>\r\n<\/ul>\r\n<\/div>\r\n\r\n\r\n\r\n
Example 5: Calculating Continuous Compounding Interest<\/h3>\r\n
If we invest $3,000 in an investment account paying 3% interest compounded continuously, how much will the account be worth in 10 years?<\/p>\r\n[reveal-answer q=\"162180\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"162180\"]\r\n
Because we are starting with $3,000, P\u00a0<\/em>= 3000. Our interest rate is 3%, so r<\/em>\u00a0=\u00a00.03. We want to know the value of the account in 10 years, so we are looking for A<\/em>(10), the value when t <\/em>= 10.<\/p>\r\n[latex]\\begin{align}A\\left(t\\right) & =Pe^{rt}&& \\text{Use the continuous compounding interest formula}. \\\\ A\\left(10\\right)& =3000e^{0.03\\cdot 10}&& \\text{Substitute using given values}. \\\\ & \\approx 4049.58&& \\text{Round to two decimal places}.\\end{align}[\/latex]<\/p>\r\n
The account will be worth about $4,049.58 in 10 years.\u00a0 We see this is slightly more than compounding semiannually as in the previous example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\nTry It<\/h3>\r\n
An initial investment of $100,000 at 12% interest is compounded continuously. What will the investment be worth in 30 years?<\/p>\r\n[reveal-answer q=\"474330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"474330\"]\r\n\r\nabout $3,659,823.44\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n
Graph exponential functions<\/h2>\r\n\r\nAn exponential function<\/strong> has 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 \r\n\r\nx<\/strong><\/em><\/td>\r\n\u20133<\/td>\r\n \u20132<\/td>\r\n \u20131<\/td>\r\n 0<\/td>\r\n 1<\/td>\r\n 2<\/td>\r\n 3<\/td>\r\n<\/tr>\r\n \r\n[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\r\n[latex]\\frac{1}{8}[\/latex]<\/td>\r\n [latex]\\frac{1}{4}[\/latex]<\/td>\r\n [latex]\\frac{1}{2}[\/latex]<\/td>\r\n 1<\/td>\r\n 2<\/td>\r\n 4<\/td>\r\n 8<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nEach output value is the product of the previous output and the base, 2. We call the base 2 the constant ratio<\/em>. In fact, for any exponential function with the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex], 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 a<\/em>.<\/p>\r\nNotice from the table that<\/p>\r\n\r\n
\r\n \t- the output values are positive for all values of x<\/em>;<\/li>\r\n \t
- as x<\/em>\u00a0increases, the output values increase without bound; and<\/li>\r\n \t
- as x<\/em>\u00a0decreases, the output values grow smaller, approaching zero.<\/li>\r\n<\/ul>\r\nFigure 2\u00a0shows the exponential growth function [latex]f\\left(x\\right)={2}^{x}[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]
![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010811\/CNX_Precalc_Figure_04_02_0012.jpg\")
Figure 2.<\/b> Notice that the graph gets close to the x-axis, but never touches it.[\/caption]\r\nThe domain of [latex]f\\left(x\\right)={2}^{x}[\/latex] is all real numbers, the range is [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote is [latex]y=0[\/latex].<\/p>\r\n
Now let's look at an example where the base is a fraction. 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
\r\n\r\n\r\nx<\/strong><\/em><\/td>\r\n\u20133<\/td>\r\n \u20132<\/td>\r\n \u20131<\/td>\r\n 0<\/td>\r\n 1<\/td>\r\n 2<\/td>\r\n 3<\/td>\r\n<\/tr>\r\n \r\n[latex]g\\left(x\\right)=\\left(\\frac{1}{2}\\right)^{x}[\/latex]<\/strong><\/td>\r\n8<\/td>\r\n 4<\/td>\r\n 2<\/td>\r\n 1<\/td>\r\n [latex]\\frac{1}{2}[\/latex]<\/td>\r\n [latex]\\frac{1}{4}[\/latex]<\/td>\r\n [latex]\\frac{1}{8}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAgain, 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
Notice from the table that<\/p>\r\n\r\n
\r\n \t- the output values are positive for all values of x<\/em>;<\/li>\r\n \t
- as x<\/em>\u00a0increases, the output values grow smaller, approaching zero; and<\/li>\r\n \t
- as x<\/em>\u00a0decreases, the output values grow without bound.<\/li>\r\n<\/ul>\r\n
The graph shows the exponential decay function, [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex].<\/p>\r\n![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010812\/CNX_Precalc_Figure_04_02_0022.jpg\")
\r\n
Figure 3.\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], and the horizontal asymptote is [latex]y=0[\/latex].<\/p>\r\n\r\n\r\nA General Note: Characteristics of the Graph of the Parent Function f<\/em>(x<\/em>) = b<\/em>x<\/em><\/sup><\/h3>\r\nAn 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>\r\n\r\n
\r\n \t- one-to-one<\/strong> function<\/li>\r\n \t
- horizontal asymptote: [latex]y=0[\/latex]<\/li>\r\n \t
- domain: [latex]\\left(-\\infty , \\infty \\right)[\/latex]<\/li>\r\n \t
- range: [latex]\\left(0,\\infty \\right)[\/latex]<\/li>\r\n \t
- x-<\/em>intercept: none<\/li>\r\n \t
- y-<\/em>intercept: [latex]\\left(0,1\\right)[\/latex]<\/li>\r\n \t
- increasing if [latex]b>1[\/latex]<\/li>\r\n \t
- decreasing if [latex]b<1[\/latex]<\/li>\r\n<\/ul>\r\n
Compare the graphs of exponential growth<\/strong> and decay functions.<\/p>\r\n![\""Graph\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010812\/CNX_Precalc_Figure_04_02_003new2.jpg\")
\r\n\r\n<\/div>\r\n\r\nHow To: Given an exponential function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex], graph the function.<\/h3>\r\n\r\n \t- Create a table of points.<\/li>\r\n \t
- Plot at least 3\u00a0point from the table, including the y<\/em>-intercept [latex]\\left(0,1\\right)[\/latex].<\/li>\r\n \t
- Draw a smooth curve through the points.<\/li>\r\n \t
- State the domain, [latex]\\left(-\\infty ,\\infty \\right)[\/latex], the range, [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote, [latex]y=0[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n\r\n\r\n\r\n
Example 6: Sketching the Graph of an Exponential Function of the Form f<\/em>(x<\/em>) = b<\/em>x<\/em><\/sup><\/h3>\r\nSketch a graph of [latex]f\\left(x\\right)={0.25}^{x}[\/latex]. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"170706\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"170706\"]\r\n
Before graphing, identify the behavior and create a table of points for the graph.<\/p>\r\n\r\n
\r\n \t- Since b\u00a0<\/em>= 0.25 is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote y\u00a0<\/em>= 0.<\/li>\r\n \t
- Create a table of points.\r\n
\r\n\r\n\r\nx<\/strong><\/em><\/td>\r\n\u20133<\/td>\r\n \u20132<\/td>\r\n \u20131<\/td>\r\n 0<\/td>\r\n 1<\/td>\r\n 2<\/td>\r\n 3<\/td>\r\n<\/tr>\r\n \r\n[latex]f\\left(x\\right)={0.25}^{x}[\/latex]<\/strong><\/td>\r\n64<\/td>\r\n 16<\/td>\r\n 4<\/td>\r\n 1<\/td>\r\n 0.25<\/td>\r\n 0.0625<\/td>\r\n 0.015625<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t- Plot the 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
Draw a smooth curve connecting the points.\r\n![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010812\/CNX_Precalc_Figure_04_02_0042.jpg\")
<\/span><\/p>\r\nFigure 4.\u00a0<\/strong>The domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]; the range is [latex]\\left(0,\\infty \\right)[\/latex]; the horizontal asymptote is [latex]y=0[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\nTry It<\/h3>\r\n
Sketch the graph of [latex]f\\left(x\\right)={4}^{x}[\/latex]. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"680272\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"680272\"]\r\n\r\nThe domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]; the range is [latex]\\left(0,\\infty \\right)[\/latex]; the horizontal asymptote is [latex]y=0[\/latex].\r\n![\"Graph](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010812\/CNX_Precalc_Figure_04_02_0052.jpg\")
<\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nGraph exponential functions using transformations<\/h2>\r\n\r\nTransformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations\u2014shifts, reflections, stretches, and compressions\u2014to the parent function [latex]f\\left(x\\right)={b}^{x}[\/latex] without loss of shape. For instance, just as the quadratic function maintains its parabolic shape when shifted, reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied.<\/p>\r\n\r\n\r\nGraphing a Vertical Shift<\/h2>\r\nThe first transformation occurs when we add a constant
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.[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 growing very rapidly. To a mathematician, however, the term exponential growth <\/em>has a very specific meaning. In this section, we will take a look at exponential functions<\/em>, which model this kind of rapid growth.<\/p>\r\nFirst, we will briefly review laws of exponents.\r\n [latex]a^{1}=a[\/latex]<\/p>\r\nAny non-zero number or variable raised to a power of 0 is equal to 1\r\n [latex]a^{0}=1[\/latex]<\/p>\r\nThe quantity [latex]0^{0}[\/latex] is undefined.\r\n\r\n<\/div>\r\n and<\/p>\r\n\r\n Simplify [latex]\\frac{\\left(t^{3}\\right)^2}{\\left(t^2\\right)^{-8}}[\/latex]\r\nWrite your answer with positive exponents.<\/p>\r\n[reveal-answer q=\"847890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847890\"]\r\n\r\nWe can either rewrite this expression with positive exponents first or use the Product Raised to a Power Rule first.\r\n\r\nLet\u2019s start by simplifying the numerator and denominator using the Product Raised to a Power Rule.\r\n\r\nNumerator: [latex]\\left(t^{3}\\right)^2=t^{3\\cdot{2}}=t^6[\/latex]\r\n\r\nDenominator: [latex]\\left(t^2\\right)^{-8}=t^{2\\cdot{-8}}=t^{-16}[\/latex]\r\n\r\nNow the expression looks like this:\r\n [latex]\\frac{t^6}{t^{-16}}[\/latex]<\/p>\r\nWe can use the quotient rule because we have the same base.\r\n\r\nQuotient Rule: [latex]\\frac{t^6}{t^{-16}}=t^{6-\\left(-16\\right)}=t^{6+16}=t^{22}[\/latex]\r\n Simplify [latex]\\frac{\\left(5x\\right)^{-2}y}{x^3y^{-1}}[\/latex]<\/p>\r\nWrite your answer with positive exponents.\r\n\r\n[reveal-answer q=\"847880\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847880\"]\r\n\r\nThis time, let\u2019s start by rewriting the terms in the expression so they have positive exponents. The terms with negative exponents in the top will go to the bottom of the fraction, and the terms with negative exponents in the bottom will go to the top.\r\n [latex]\\begin{array}{c}\\frac{\\left(5x\\right)^{-2}y}{x^3y^{-1}}\\\\\\text{ }\\\\=\\frac{\\left({y^{1}}\\right)y}{x^3\\left(5x\\right)^{2}}\\end{array}[\/latex]<\/p>\r\nNote how we left the single y term in the top because it did not have a negative exponent on it, and we left the [latex]x^3[\/latex] term in the bottom because it did not have a negative exponent on it.\r\n\r\nNow we can apply the Product Raised to a Power Rule:\r\n [latex]\\frac{yy^{1}}{5^{2}x^3x^{2}}[\/latex]<\/p>\r\nUse the product rule to simplify further:\r\n [latex]\\frac{yy^{1}}{5^{2}x^3x^{2}}=\\frac{y^2}{25x^{3+2}}=\\frac{y^2}{25x^5}[\/latex]<\/p>\r\n We can\u2019t simplify any further, so our answer is<\/p>\r\n\r\n Recall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base b<\/em>\u00a0to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:<\/p>\r\n\r\n Why do we limit the base to positive values other than 1? Because base 1\u00a0results in the constant function. Observe what happens if the base is\u00a01:<\/p>\r\n\r\n To evaluate an exponential function with the form [latex]f\\left(x\\right)={b}^{x}[\/latex], we simply substitute x<\/em>\u00a0with the given value, and calculate the resulting power. For example:<\/p>\r\n Let [latex]f\\left(x\\right)={2}^{x}[\/latex]. What is [latex]f\\left(3\\right)[\/latex]?<\/p>\r\n\r\n 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 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 Note that if the order of operations were not followed, the result would be incorrect:<\/p>\r\n [latex]f\\left(3\\right)=30{\\left(2\\right)}^{3}\\ne {60}^{3}=216,000[\/latex]<\/p>\r\n\r\n 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>\r\n[reveal-answer q=\"847875\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"847875\"]\r\n Follow the order of operations. Be sure to pay attention to the parentheses.<\/p>\r\n [latex]\\begin{align}f\\left(x\\right) & =5{\\left(3\\right)}^{x+1} \\\\ f\\left(2\\right) & =5{\\left(3\\right)}^{2+1} && \\text{Substitute }x=2. \\\\ & =5{\\left(3\\right)}^{3} && \\text{Add the exponents}. \\\\ & =5\\left(27\\right) && \\text{Simplify the power}. \\\\ & =135 && \\text{Multiply}. \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n Let [latex]f\\left(x\\right)=8{\\left(1.2\\right)}^{x - 5}[\/latex]. Evaluate [latex]f\\left(3\\right)[\/latex] using a calculator. Round to four decimal places.<\/p>\r\n[reveal-answer q=\"292533\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"292533\"]\r\n\r\n5.5556\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest<\/strong>. The term compounding<\/em> refers to interest earned not only on the original value, but on the accumulated value of the account.<\/p>\r\n The annual percentage rate (APR)<\/strong> of an account, also called the nominal rate<\/strong>, is the yearly interest rate earned by an investment account. The term\u00a0nominal<\/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 greater<\/em> than the nominal rate! This is a powerful tool for investing.<\/p>\r\n We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time t<\/em>, principal P<\/em>, APR r<\/em>, and number of compounding periods in a year\u00a0n<\/em>:<\/p>\r\n\r\n For example, observe the table below, which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases.<\/p>\r\n\r\n Compound interest<\/strong> can be calculated using the formula<\/p>\r\n\r\n where<\/p>\r\n\r\n 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>\r\n[reveal-answer q=\"162183\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"162183\"]\r\n Because we are starting with $3,000, P\u00a0<\/em>= 3000. Our interest rate is 3%, so r<\/em>\u00a0=\u00a00.03. Because we are compounding quarterly, we are compounding 4 times per year, so n\u00a0<\/em>= 4. We want to know the value of the account in 10 years, so we are looking for A<\/em>(10), the value when t <\/em>= 10.<\/p>\r\n [latex]\\begin{align}A\\left(t\\right) & =P\\left(1+\\frac{r}{n}\\right)^{nt}&& \\text{Use the compound interest formula}. \\\\ A\\left(10\\right)& =3000\\left(1+\\frac{0.03}{4}\\right)^{4\\cdot 10}&& \\text{Substitute using given values}. \\\\ & \\approx 4045.05&& \\text{Round to two decimal places}.\\end{align}[\/latex]<\/p>\r\n The account will be worth about $4,045.05 in 10 years.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What will the investment be worth in 30 years?<\/p>\r\n[reveal-answer q=\"474337\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"474337\"]\r\n\r\nabout $3,644,675.88\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n 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>\r\n Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies.<\/p>\r\n\r\n These values appear to be approaching a limit as n<\/em>\u00a0increases without bound. In fact, as 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.<\/p>\r\nAs [latex]n[\/latex] approaches infinity, the expression\u00a0[latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex] approaches\u00a0[latex]e[\/latex].\u00a0 Continuous compounding occurs when [latex]n[\/latex] approaches infinity.\u00a0 Below is the formula for continuous compounding.\u00a0 Notice there is an\u00a0[latex]e[\/latex] in the formula instead of\u00a0[latex]n[\/latex].\u00a0 The [latex]e[\/latex] on a calculator is typically above the LN<\/strong> button.\r\n Continuous compounding interest<\/strong> can be calculated using the formula<\/p>\r\n\r\n where<\/p>\r\n\r\n If we invest $3,000 in an investment account paying 3% interest compounded continuously, how much will the account be worth in 10 years?<\/p>\r\n[reveal-answer q=\"162180\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"162180\"]\r\n Because we are starting with $3,000, P\u00a0<\/em>= 3000. Our interest rate is 3%, so r<\/em>\u00a0=\u00a00.03. We want to know the value of the account in 10 years, so we are looking for A<\/em>(10), the value when t <\/em>= 10.<\/p>\r\n [latex]\\begin{align}A\\left(t\\right) & =Pe^{rt}&& \\text{Use the continuous compounding interest formula}. \\\\ A\\left(10\\right)& =3000e^{0.03\\cdot 10}&& \\text{Substitute using given values}. \\\\ & \\approx 4049.58&& \\text{Round to two decimal places}.\\end{align}[\/latex]<\/p>\r\n The account will be worth about $4,049.58 in 10 years.\u00a0 We see this is slightly more than compounding semiannually as in the previous example.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n An initial investment of $100,000 at 12% interest is compounded continuously. What will the investment be worth in 30 years?<\/p>\r\n[reveal-answer q=\"474330\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"474330\"]\r\n\r\nabout $3,659,823.44\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n An exponential function<\/strong> has 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 Each output value is the product of the previous output and the base, 2. We call the base 2 the constant ratio<\/em>. In fact, for any exponential function with the form [latex]f\\left(x\\right)=a{b}^{x}[\/latex], 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 a<\/em>.<\/p>\r\n Notice from the table that<\/p>\r\n\r\n The domain of [latex]f\\left(x\\right)={2}^{x}[\/latex] is all real numbers, the range is [latex]\\left(0,\\infty \\right)[\/latex], and the horizontal asymptote is [latex]y=0[\/latex].<\/p>\r\n Now let's look at an example where the base is a fraction. 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 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 Notice from the table that<\/p>\r\n\r\n The graph shows the exponential decay function, [latex]g\\left(x\\right)={\\left(\\frac{1}{2}\\right)}^{x}[\/latex].<\/p>\r\n Figure 3.\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], and the horizontal asymptote is [latex]y=0[\/latex].<\/p>\r\n\r\n 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>\r\n\r\n Compare the graphs of exponential growth<\/strong> and decay functions.<\/p>\r\n Sketch a graph of [latex]f\\left(x\\right)={0.25}^{x}[\/latex]. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"170706\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"170706\"]\r\n Before graphing, identify the behavior and create a table of points for the graph.<\/p>\r\n\r\n Draw a smooth curve connecting the points.\r\n Figure 4.\u00a0<\/strong>The domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]; the range is [latex]\\left(0,\\infty \\right)[\/latex]; the horizontal asymptote is [latex]y=0[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n Sketch the graph of [latex]f\\left(x\\right)={4}^{x}[\/latex]. State the domain, range, and asymptote.<\/p>\r\n[reveal-answer q=\"680272\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"680272\"]\r\n\r\nThe domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]; the range is [latex]\\left(0,\\infty \\right)[\/latex]; the horizontal asymptote is [latex]y=0[\/latex].\r\n Transformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations\u2014shifts, reflections, stretches, and compressions\u2014to the parent function [latex]f\\left(x\\right)={b}^{x}[\/latex] without loss of shape. For instance, just as the quadratic function maintains its parabolic shape when shifted, reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied.<\/p>\r\n\r\nThe Product Rule for Exponents<\/h3>\r\nFor any number a<\/i> and any integers n<\/i> and m<\/i>, [latex]\\left(a^{n}\\right)\\left(a^{m}\\right) = a^{n+m}[\/latex].\r\n\r\n<\/div>\r\n
The Quotient (Division) Rule for Exponents<\/h3>\r\nFor any non-zero number a<\/i> and any integers n<\/i> and n<\/i>: [latex] \\displaystyle \\frac{{{a}^{n}}}{{{a}^{m}}}={{a}^{n-m}}[\/latex]\r\n\r\n<\/div>\r\n
The Power Rule for Exponents<\/h3>\r\nFor any positive number a<\/i> and integers n<\/i> and m<\/i>: [latex]\\left(a^{n}\\right)^{m}=a^{n\\cdot{m}}[\/latex].\r\n\r\nTake a moment to contrast how this is different from the product rule for exponents.\r\n\r\n<\/div>\r\n
A Product Raised to a Power<\/h3>\r\nFor any nonzero numbers a<\/i> and b<\/i> and any integer n<\/i>, [latex]\\left(ab\\right)^{n}=a^{n}\\cdot{b^{n}}[\/latex].\r\n\r\n<\/div>\r\n
A Quotient Raised to a Power<\/h3>\r\nFor any number a<\/i>, any non-zero number b<\/i>, and any integer n<\/i>, [latex] \\displaystyle {\\left(\\frac{a}{b}\\right)}^{n}=\\frac{a^{n}}{b^{n}}[\/latex].\r\n\r\n<\/div>\r\n
Exponents of 0 or 1<\/h3>\r\nAny number or variable raised to a power of 1 is the number itself.\r\n
The Negative Rule of Exponents<\/h3>\r\nFor any nonzero real number [latex]a[\/latex] and natural number [latex]n[\/latex], the negative rule of exponents states that\r\n
Example 1: Practice Exponent Rules<\/h3>\r\n
Answer<\/h4>\r\n[latex]\\frac{\\left(t^{3}\\right)^2}{\\left(t^2\\right)^{-8}}=t^{22}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
Example 2: Practice Exponent Rules<\/h3>\r\n
Answer<\/h4>\r\n[latex]\\frac{\\left(5x\\right)^{-2}y}{x^3y^{-1}}=\\frac{y^2}{25x^5}[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
Evaluate Exponential Functions<\/h2>\r\n
\r\n \t
\r\n \t
Example 3: Evaluating Exponential Functions<\/h3>\r\n
Try It<\/h3>\r\n
\u00a0Use Compound Interest Formulas<\/h2>\r\n
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\r\n \r\nFrequency<\/th>\r\n Value after 1 year<\/th>\r\n<\/tr>\r\n<\/thead>\r\n \r\n Annually<\/td>\r\n $1100<\/td>\r\n<\/tr>\r\n \r\n Semiannually<\/td>\r\n $1102.50<\/td>\r\n<\/tr>\r\n \r\n Quarterly<\/td>\r\n $1103.81<\/td>\r\n<\/tr>\r\n \r\n Monthly<\/td>\r\n $1104.71<\/td>\r\n<\/tr>\r\n \r\n Daily<\/td>\r\n $1105.16<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n A General Note: The Compound Interest Formula<\/h3>\r\n
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Example 4: Calculating Compound Interest<\/h3>\r\n
Try It<\/h3>\r\n
Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]14375[\/ohm_question]\r\n\r\n<\/div>\r\n
Evaluate exponential functions with base e<\/h2>\r\n
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\r\n \r\nFrequency<\/th>\r\n [latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\r\n Value<\/th>\r\n<\/tr>\r\n<\/thead>\r\n \r\n Annually<\/td>\r\n [latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\r\n $2<\/td>\r\n<\/tr>\r\n \r\n Semiannually<\/td>\r\n [latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\r\n $2.25<\/td>\r\n<\/tr>\r\n \r\n Quarterly<\/td>\r\n [latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\r\n $2.441406<\/td>\r\n<\/tr>\r\n \r\n Monthly<\/td>\r\n [latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\r\n $2.613035<\/td>\r\n<\/tr>\r\n \r\n Daily<\/td>\r\n [latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\r\n $2.714567<\/td>\r\n<\/tr>\r\n \r\n Hourly<\/td>\r\n [latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\r\n $2.718127<\/td>\r\n<\/tr>\r\n \r\n Once per minute<\/td>\r\n [latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\r\n $2.718279<\/td>\r\n<\/tr>\r\n \r\n Once per second<\/td>\r\n [latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\r\n $2.718282<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n A General Note: continuous compounding<\/h3>\r\n
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Example 5: Calculating Continuous Compounding Interest<\/h3>\r\n
Try It<\/h3>\r\n
Graph exponential functions<\/h2>\r\n
\r\n x<\/strong><\/em><\/td>\r\n \u20133<\/td>\r\n \u20132<\/td>\r\n \u20131<\/td>\r\n 0<\/td>\r\n 1<\/td>\r\n 2<\/td>\r\n 3<\/td>\r\n<\/tr>\r\n \r\n [latex]f\\left(x\\right)={2}^{x}[\/latex]<\/strong><\/td>\r\n [latex]\\frac{1}{8}[\/latex]<\/td>\r\n [latex]\\frac{1}{4}[\/latex]<\/td>\r\n [latex]\\frac{1}{2}[\/latex]<\/td>\r\n 1<\/td>\r\n 2<\/td>\r\n 4<\/td>\r\n 8<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n \t
Figure 2.<\/b> Notice that the graph gets close to the x-axis, but never touches it.[\/caption]\r\n\r\n\r\n
\r\n x<\/strong><\/em><\/td>\r\n \u20133<\/td>\r\n \u20132<\/td>\r\n \u20131<\/td>\r\n 0<\/td>\r\n 1<\/td>\r\n 2<\/td>\r\n 3<\/td>\r\n<\/tr>\r\n \r\n [latex]g\\left(x\\right)=\\left(\\frac{1}{2}\\right)^{x}[\/latex]<\/strong><\/td>\r\n 8<\/td>\r\n 4<\/td>\r\n 2<\/td>\r\n 1<\/td>\r\n [latex]\\frac{1}{2}[\/latex]<\/td>\r\n [latex]\\frac{1}{4}[\/latex]<\/td>\r\n [latex]\\frac{1}{8}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n \t
\r\nA General Note: Characteristics of the Graph of the Parent Function f<\/em>(x<\/em>) = b<\/em>x<\/em><\/sup><\/h3>\r\n
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\r\n\r\n<\/div>\r\nHow To: Given an exponential function of the form [latex]f\\left(x\\right)={b}^{x}[\/latex], graph the function.<\/h3>\r\n
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Example 6: Sketching the Graph of an Exponential Function of the Form f<\/em>(x<\/em>) = b<\/em>x<\/em><\/sup><\/h3>\r\n
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\r\n x<\/strong><\/em><\/td>\r\n \u20133<\/td>\r\n \u20132<\/td>\r\n \u20131<\/td>\r\n 0<\/td>\r\n 1<\/td>\r\n 2<\/td>\r\n 3<\/td>\r\n<\/tr>\r\n \r\n [latex]f\\left(x\\right)={0.25}^{x}[\/latex]<\/strong><\/td>\r\n 64<\/td>\r\n 16<\/td>\r\n 4<\/td>\r\n 1<\/td>\r\n 0.25<\/td>\r\n 0.0625<\/td>\r\n 0.015625<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n \t <\/span><\/p>\r\nTry It<\/h3>\r\n
<\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nGraph exponential functions using transformations<\/h2>\r\n
Graphing a Vertical Shift<\/h2>\r\nThe first transformation occurs when we add a constant