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1 In the previous lesson we introduced Eponential Functions and their graphs, and covered an application of Eponential Functions (Compound Interest). We saw that when interest is compounded n times per year for some number of years, the accumulated value of that investment can be found by using the formula A = P (1 + r n )nt. Using that formula we found the accumulated value of investments when interest is compounded bi-annually (n = 2), quarterly (n = 4), monthly (n = 12), weekly (n = 52), etc. But what would happen if the number of compounding periods per year (n) continued to increase? For instance what if interest were compounded every day (n = 365), or every hour (n = 8760), or every minute (n = 525,000), or every second (n = 31,536, 000)? For a simple eample, imagine that one dollar is invested for one year at a rate of 100% (P, t, and r = 1); what would that investment accumulate to if the number of compounding periods continued to increase? How often is interest compounded? n A = P (1 + r n ) nt monthly 12 A = 1 ( ) weekly 52 A = 1 ( ) daily 365 A = 1 ( ) every hour 8760 A every minute 525,000 A every second 31,536,000 A continuously n A

2 What we see is that as the number of compounding periods per year increase (n ), the accumulated value A continues to get closer and closer to the value This value that we get closer and closer to is known as the natural number e, and it is used etensively in finance (as we ll see later with our second compound interest formula), as well as in other disciplines such as statistics and engineering. The Natural Number e: - a non-terminating constant o to find the value on your calculator press the number 1, then the 2 nd function button, then the LN button - it is NOT a variable like or y, but rather a number like π The way we will use the natural number e is either as the base of an eponential function, which is known as the natural eponential function, or as the base of a logarithmic function, which is known as the natural logarithmic function g() = ln(). We will work with the Natural Eponential Function in this lesson, and we ll cover the Natural Logarithmic Function g() = ln() in Lesson 32. Natural Eponential Functions: -, where the eponent is a variable and base e is the natural number Just like any other eponential function, the domain of the natural eponential function is unrestricted (, ) and the range is only positive numbers (0, ). And just like any other eponential function, the natural eponential function can be transformed, which in the case of a vertical transformation would alter the range (more on this later).

3 Eample 1: Given the input/output table for the function, as well as its graph, find its domain, range, zeros, positive/negative intervals, increasing/decreasing intervals, and intercepts. f() 0 5 f( 5) = e 5 = 1 e f( 4) = e 4 = 1 e f( 3) = e 3 = 1 e f( 2) = e 2 = 1 e Notice that the graph of is increasing throughout its domain. When a function is always increasing or always decreasing that function is one-to-one, and it will have an inverse function. All eponential functions are one-to-one, therefore all eponential functions have an inverse (we ll discuss this further in Lessons 31 & 32). f() 1 f( 1) = e 1 = 1 e f(0) = e 0 = 1 1 f(1) = e f(2) = e f(3) = e f(4) = e f(5) = e f() Domain: (, ) Range: (0, ) Zeros: f() = 0 when = NONE Positive intervals: f() > 0 when is (, ) Negative intervals: f() < 0 when is NONE Increasing intervals: f() is rising when is (, ) Decreasing intervals: f() is falling when is NONE Intercepts: intercept: NONE y intercept: (0, 1)

4 Notice that just like the graph of 2, the graph of e has a horizontal asymptote at y = 0 (the -ais). That is because the graphs of both (2 and e ) approach the -ais, but they never touch it or cross it. Eample 2: Re-write the following function in terms of. Then find its y-intercept and sketch its graph using transformations and the y-intercept. Enter eact answers only, no approimations. a. g() = e f() Re write g() in terms of f(): g() = y intercept: g(0) = (0, )

5 Eample 3: Re-write each of the following functions in terms of, then match the transformation with the appropriate graph. Also, find the y-intercept for each function. Enter eact answers only, no approimations. f() f() e e e e f() a. h() = e b. j() = e +2 c. k() = e + 2 b. c. f() = F( ) h() = j() = k() = d. y int: y int: y int: e. f. l() = 2e e. m() = e 2 g.f() = F( ) l() = m() = h. y int: y int:

6 A. B. C. D. E. F.

7 Answers to Eamples: 3a. A, (0, 1) ; 3b. C, (0, e 2 ) ; 3c. F, (0, 3) ; 3d. B, (0, 2) ; 3e. E, (0, 1) ;

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