{"id":1066,"date":"2025-07-22T00:05:49","date_gmt":"2025-07-22T00:05:49","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1066"},"modified":"2025-09-09T17:02:47","modified_gmt":"2025-09-09T17:02:47","slug":"exponential-functions-learn-it-3-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/exponential-functions-learn-it-3-2\/","title":{"raw":"Exponential Functions: Learn It 3","rendered":"Exponential Functions: Learn It 3"},"content":{"raw":"

Evaluating Exponential Functions with Base [latex]e[\/latex]<\/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\r\nExamine the value of [latex]$1[\/latex] invested at [latex]100 \\%[\/latex] interest for [latex]1[\/latex] year compounded at various frequencies.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Frequency<\/th>\r\n[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\r\nValue<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
Annually<\/td>\r\n[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\r\n[latex]$2[\/latex]<\/td>\r\n<\/tr>\r\n
Semiannually<\/td>\r\n[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\r\n[latex]$2.25[\/latex]<\/td>\r\n<\/tr>\r\n
Quarterly<\/td>\r\n[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\r\n[latex]$2.441406[\/latex]<\/td>\r\n<\/tr>\r\n
Monthly<\/td>\r\n[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\r\n[latex]$2.613035[\/latex]<\/td>\r\n<\/tr>\r\n
Daily<\/td>\r\n[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\r\n[latex]$2.714567[\/latex]<\/td>\r\n<\/tr>\r\n
Hourly<\/td>\r\n[latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\r\n[latex]$2.718127[\/latex]<\/td>\r\n<\/tr>\r\n
Once per minute<\/td>\r\n[latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\r\n[latex]\u00a0$2.718279[\/latex]<\/td>\r\n<\/tr>\r\n
Once per second<\/td>\r\n[latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\r\n[latex]$2.718282[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThese values appear to be approaching a limit as [latex]n[\/latex]\u00a0increases without bound. In fact, as [latex]n[\/latex] gets 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.\r\n\r\n
\r\n

the number [latex]e[\/latex]<\/h3>\r\nThe letter [latex]e[\/latex] represents the irrational number\r\n

[latex]{\\left(1+\\frac{1}{n}\\right)}^{n},\\text{as }n\\text{ increases without bound}[\/latex]<\/p>\r\n \r\n\r\nThe letter [latex]e[\/latex] is used as a base for many real-world exponential models. To work with base [latex]e[\/latex], we use the approximation, [latex]e\\approx 2.718282[\/latex].\r\n\r\n \r\n\r\nThe constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.\r\n\r\n<\/section>

Calculate [latex]{e}^{3.14}[\/latex]. Round to five decimal places.[reveal-answer q=\"201253\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"201253\"]On a calculator, press the button labeled [latex]\\left[{e}^{x}\\right][\/latex]. The window shows [e<\/em>^( ]. Type 3.14 and then close parenthesis, [ e^(3.14) ]. 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 e<\/em>.[\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]291146[\/ohm_question]<\/section>
Graph and state the characteristics of [latex]f(x) = e^x[\/latex].\r\n\r\n
\r\n\r\nTo graph the function [latex]f(x) = e^x[\/latex], let's identify and plot key points on the graph.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
[latex]x[\/latex]<\/td>\r\n[latex]f(x) = e^x[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]-1[\/latex]<\/td>\r\n[latex]f(-1) = e^{-1} \\approx 0.37[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]0[\/latex]<\/td>\r\n[latex]f(-1) = e^{0} \\approx 1[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]1[\/latex]<\/td>\r\n[latex]f(-1) = e^{1} \\approx 2.72[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]2[\/latex]<\/td>\r\n[latex]f(-1) = e^{2} \\approx 7.39[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n\r\nCharacteristics:\r\n\r\n[latex]\\begin{align*} \\text{Function:} & \\quad f(x) = e^x \\\\ \\text{Domain:} & \\quad (-\\infty, \\infty) \\\\ \\text{Range:} & \\quad (0, \\infty) \\\\ \\text{Y-intercept:} & \\quad (0, 1) \\\\ \\text{Horizontal Asymptote:} & \\quad y = 0 \\\\ \\text{Behavior as } x \\rightarrow \\infty: & \\quad f(x) \\rightarrow \\infty \\\\ \\text{Behavior as } x \\rightarrow -\\infty: & \\quad f(x) \\rightarrow 0 \\end{align*}[\/latex]\r\n\r\n<\/section>
Write the equation for the function described below. Give the horizontal asymptote, domain, and range.[latex]f\\left(x\\right)={e}^{x}[\/latex] is vertically stretched by a factor of [latex]2[\/latex], reflected across the [latex]y[\/latex]-axis, and then shifted up [latex]4[\/latex]\u00a0units.[reveal-answer q=\"290621\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"290621\"]We want to find an equation of the general form [latex] f\\left(x\\right)=a{b}^{x+c}+d[\/latex]. We use the description provided to find [latex]a, b, c[\/latex], and [latex]d[\/latex].\r\n
    \r\n \t
  • We are given the parent function [latex]f\\left(x\\right)={e}^{x}[\/latex], so [latex]b\u00a0= e[\/latex].<\/li>\r\n \t
  • The function is stretched by a factor of [latex]2[\/latex], so [latex]a\u00a0= 2[\/latex].<\/li>\r\n \t
  • The function is reflected about the [latex]y[\/latex]-axis. We replace [latex]x[\/latex]\u00a0with [latex]\u2013x[\/latex]\u00a0to get: [latex]{e}^{-x}[\/latex].<\/li>\r\n \t
  • The graph is shifted vertically [latex]4[\/latex] units, so [latex]d\u00a0= 4[\/latex].<\/li>\r\n<\/ul>\r\nSubstituting in the general form, we get:\r\n

    [latex]\\begin{array}{llll}f\\left(x\\right)\\hfill & =a{b}^{x+c}+d\\hfill \\\\ \\hfill & =2{e}^{-x+0}+4\\hfill \\\\ \\hfill & =2{e}^{-x}+4\\hfill \\end{array}[\/latex]<\/p>\r\nThe domain is [latex]\\left(-\\infty ,\\infty \\right)[\/latex]; the range is [latex]\\left(4,\\infty \\right)[\/latex]; the horizontal asymptote is [latex]y=4[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

    [ohm2_question hide_question_numbers=1]24717[\/ohm2_question]<\/section>","rendered":"

    Evaluating Exponential Functions with Base [latex]e[\/latex]<\/h2>\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>\n

    Examine the value of [latex]$1[\/latex] invested at [latex]100 \\%[\/latex] interest for [latex]1[\/latex] year compounded at various frequencies.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
    Frequency<\/th>\n[latex]A\\left(t\\right)={\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/th>\nValue<\/th>\n<\/tr>\n<\/thead>\n
    Annually<\/td>\n[latex]{\\left(1+\\frac{1}{1}\\right)}^{1}[\/latex]<\/td>\n[latex]$2[\/latex]<\/td>\n<\/tr>\n
    Semiannually<\/td>\n[latex]{\\left(1+\\frac{1}{2}\\right)}^{2}[\/latex]<\/td>\n[latex]$2.25[\/latex]<\/td>\n<\/tr>\n
    Quarterly<\/td>\n[latex]{\\left(1+\\frac{1}{4}\\right)}^{4}[\/latex]<\/td>\n[latex]$2.441406[\/latex]<\/td>\n<\/tr>\n
    Monthly<\/td>\n[latex]{\\left(1+\\frac{1}{12}\\right)}^{12}[\/latex]<\/td>\n[latex]$2.613035[\/latex]<\/td>\n<\/tr>\n
    Daily<\/td>\n[latex]{\\left(1+\\frac{1}{365}\\right)}^{365}[\/latex]<\/td>\n[latex]$2.714567[\/latex]<\/td>\n<\/tr>\n
    Hourly<\/td>\n[latex]{\\left(1+\\frac{1}{\\text{8766}}\\right)}^{\\text{8766}}[\/latex]<\/td>\n[latex]$2.718127[\/latex]<\/td>\n<\/tr>\n
    Once per minute<\/td>\n[latex]{\\left(1+\\frac{1}{\\text{525960}}\\right)}^{\\text{525960}}[\/latex]<\/td>\n[latex]\u00a0$2.718279[\/latex]<\/td>\n<\/tr>\n
    Once per second<\/td>\n[latex]{\\left(1+\\frac{1}{31557600}\\right)}^{31557600}[\/latex]<\/td>\n[latex]$2.718282[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    These values appear to be approaching a limit as [latex]n[\/latex]\u00a0increases without bound. In fact, as [latex]n[\/latex] gets 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>\n

    \n

    the number [latex]e[\/latex]<\/h3>\n

    The letter [latex]e[\/latex] represents the irrational number<\/p>\n

    [latex]{\\left(1+\\frac{1}{n}\\right)}^{n},\\text{as }n\\text{ increases without bound}[\/latex]<\/p>\n

     <\/p>\n

    The letter [latex]e[\/latex] is used as a base for many real-world exponential models. To work with base [latex]e[\/latex], we use the approximation, [latex]e\\approx 2.718282[\/latex].<\/p>\n

     <\/p>\n

    The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.<\/p>\n<\/section>\n

    Calculate [latex]{e}^{3.14}[\/latex]. Round to five decimal places.<\/p>\n