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L5.3 Exponential Functions

L5.3 Exponential Functions. Graphs, including transforms Applications. L5.3 Warm up. Let y = 3 x be the base function Exponential functions can also undergo transformations. For each new function below, describe the transformation and identify the new y-intercept:

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L5.3 Exponential Functions

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  1. L5.3 Exponential Functions • Graphs, including transforms • Applications

  2. L5.3 Warm up • Let y = 3x be the base function • Exponential functions can also undergo transformations. For each new function below, describe the transformation and identify the new y-intercept: • y = 3x+1 • y = 3x – 2 • y = 3-x • y = –3x 1/9 1/3 Exponential growth 1 3 (0, 1) 9 Horizontal shift left; y-int: (0, 3) Vertical shift down; y-int: (0, –1) Horizontal reflection; y-int: (0, 1). Becomes exp’l decay! Vertical reflection; y-int: (0, –1). No longer exponential growth.

  3. (0, 1) (0, 1) e.g., and are defined. Exponential Functions y = (⅓)x = 3−x Ex: y = 3x 9 3 1 1/3 1/9 Exponential Growth, b > 1 Exponential Decay, 0 < b < 1 Exponential functions: Are continuous and one-to-one Domain: {x | x is Real} Range: {y | y > 0 } [if a > 0] Contain the point (0, a)

  4. Find an exponential function having the given values. Use the given values to find a and b: f(x) = a·bx. • f(0) = 5, f(3) = 40 f(0) = 5 → a·b0 = 5 → a·1 = 5 → a = 5 Thus, f(x) = 5bx [y-intercept gives you a] f(3) = 40 → 5b3 = 40 → b3 = 8 → b = 2 Thus, f(x) = 5·2x • f(0) = 3, f(2) = 75, Find f(x) and f(-3) f(x) = 3·5x; f(-3) = 3/125 Note: This only works if one of the given values is the y-intercept → a

  5. b = (1 + r) Applications of Exponential Functions • Exponential functions are used to describe exponential growth and decay as a function of time, t. • Multiple forms: ─ f(t) = a·bt, where a > 0 • If b > 1 : growth [f↑ as t↑] • 0 < b < 1 : decay [f↓ as t↑] ─ A(t) = A0(1+ r)t, where • if r > 0: growth • −1 < r < 0 : decay ─ A(t) = A0·bt/k, where • k is the time needed to multiply A0 by b ─ or A = Pert • Compound interest L5.1 & 5.2 L5.3 ** This lesson! L5.4

  6. 5.3 Example Problems: A(t) = A0·bt/k • Let A(t) be the mass of a particular radioactive element whose half-life is 25 years. After t years, the mass (in grams) is given by A(t) = 10(1/2)t/25. a) What was the initial mass (when t = 0)? b) How much of the initial mass is present after 80 years? A(0) = 10(1/2)0 = 10 grams A(80) = 10(1/2)80/25 = ~1.088 grams • Suppose a culture of 100 bacteria are put in a petri dish and the culture doubles every 12.5 hours. How many bacteria will there be in 1 week (assuming adequate nutrients/space)? # hrs in a week: t = 7 * 24 = 168; b = 2, k = 12.5 P(168) = 100(2)168/12.5 = over 1.1 million bacteria • The half-life of radium is about 1600 years. If 1kg is present now, how much will be present after: • a. 3200 years • b. t years A(3200) = 1000(1/2)3200/1600 = 250g A(t) = 1000(1/2)t/1600

  7. Rule of 72 The Rule of 72 provides an estimate of the time it takes for a quantity to double. Its called the Rule of 72 because at r = 10%, the quantity will double in 7.2 years. To use this rule, just divide 72 by the growth rate. For example. If you get 6% on an investment and that rate stays constant, your money will double in 72 / 6 = ~12 years. • How long does it take to double an investment that grows at a rate of 9% per year? • If your money has to double in 2 years so that you can buy a new car, what kind of rate of return do you need to get? 72/9 = ~8 years 72/r = 2 years r = 36% (good luck finding a rate like that!)

  8. Wrap up Domain: {x | x is all Reals }, Range: {f(x) | f(x) < 0 } • What are the domain and range of f(x) = −3∙4x? Does this function represent exponential growth or decay? • If h(x) = abx and h(0) = 5, and h(1) = 15, what are the constants a and b? • According to the rule of 72, how long does it take to double an investment that grows at a rate of 6% per year? • A colony of bacteria is dying . Every 5 days, they loose 1/3 of their population. Today they have 2.6million bacteria. How many will they have in a week from now? This function is neither exp’l growth nor decay! (a < 0) a = 5, b = 3, i.e., h(x) = 5∙3x 72/6 ~ 12 years P(7) = 2,600,000(2/3)7/5 ~ 1,474,000 bacteria

  9. Homework L5.3 pp 183-186 #7, 10, 12-14, 17, 22, 26 a & b

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