Stitz-Zeager_College_Algebra_e-book

Louis talmans mathematics animated website 71

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Unformatted text preview: xponential function of time t, to A = 100(1.0125) 1 1 A(t). To check this against our previous calculations, we find A 4 = 100(1.0125)4( 4 ) = 101.25, 1 A 2 ≈ $102.51, A 3 ≈ $103.79, and A(1) ≈ $105.08. 4 Example 6.5.1. Suppose $2000 is invested in an account which offers 7.125% compounded monthly. 1. Express the amount A in the account as a function of the term of the investment t in years. 2. How much is in the account after 5 years? 3. How long will it take for the initial investment to double? 4. Find and interpret the average rate of change5 of the amount in the account from the end of the fourth year to the end of the fifth year, and from the end of the thirty-fourth year to the end of the thirty-fifth year. Solution. 1. Substituting P = 2000, r = 0.07125, and n = 12 (monthly) into Equation 6.2 yields A = 12t 2000 1 + 0.07125 . Using function notation, we get A(t) = 2000(1.0059375)12t . 12 2. Since t represents the length of the investment, we substitute t = 5 into A(t) to find A(5) = 2000(1.0059375)12(5) ≈ 2852.92. After 5 years, we have approximately $2852.92. 5 See Definition 2.3 in Section 2.1. 380 Exp...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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