Copyright 2006 Pearson Education Inc Publishing as Pearson
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
5 5. 1 5. 2 5. 3 5. 4 5. 5 5. 6 5. 7 Exponential and Logarithmic Functions Combining Functions Inverse Functions and Their Representations Exponential Functions and Models Logarithmic Functions and Models Properties of Logarithms Exponential and Logarithmic Equations Constructing Nonlinear Models Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
5. 1 Combining Functions ♦ Perform arithmetic operations on functions ♦ Perform composition of functions Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Five Ways of Combining Two Functions f and g • • Addition • Subtraction • Multiplication • Division • Composition • • Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4
Definitions If f(x) and g(x) both exist, the sum, difference, product, quotient and composition of two functions f and g are defined by • Addition • Subtraction • Multiplication • Division • Composition Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 5
Examples of Evaluating Combinations of Functions – Using Symbolic Representations Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 6
Example of Addition of Functions: Let f(x) = x 2 + 2 x and g(x) = 3 x - 1 • Find the symbolic representation for the function f + g and use this to evaluate (f + g)(2) (f + g)(x) = (x 2 + 2 x) + (3 x 1) (f + g)(x) = x 2 + 5 x 1 (f + g)(2) = 22 + 5(2) 5( 1 = 13 or (f + g)(2) = f(2) + g(2) = 22 + 2(2) + 3(2) 1 = 13 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 7
Example of Subtraction of Functions: Let f(x) = x 2 + 2 x and g(x) = 3 x 1 • Find the symbolic representation for the function f g and use this to evaluate (f g)(2) • (f g)(x) = (x 2 + 2 x) (3 x 1) (f g)(x) = x 2 x + 1 • So (f g)(2) = 22 2 + 1 = 3 • Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 8
Example of Multiplication of Functions: Let f(x) = x 2 + 2 x and g(x) = 3 x 1 • Find the symbolic representation for the function fg and use this to evaluate (fg)(2) • (fg)(x) = (x 2 + 2 x)(3 x 1) • (fg)(x) = 3 x 3 + 6 x 2 2 x • (fg)(x) = 3 x 3 + 5 x 2 2 x • So (fg)(2) = 3(2)3 +5(2)2 2(2) = 40 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 9
Example of Division of Functions: Let f(x) = x 2 + 2 x and g(x) = 3 x 1 • Find the symbolic representation for the function and use this to evaluate • • So Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 10
Example of Composition of Functions: Let f(x) = x 2 + 2 x and g(x) = 3 x - 1 • Find the symbolic representation for the function f g and use this to evaluate (f g)(2) • (f g)(x) = f(g(x)) = f(3 x – 1) = (3 x – 1)2 + 2(3 x – 1) (f g)(x) = (3 x – 1) ( 3 x – 1) + 6 x – 2 (f g)(x) = 9 x 2 – 3 x + 1 + 6 x – 2 (f g)(x) = 9 x 2 – 1 • So (f g)(2) = 9(2)2 – 1 = 35 • • • Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 11
Evaluating Combinations of Functions Numerically • • Given numerical representations for f and g in the table Evaluate combinations of f and g as specified. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 12
(f + g)(5) = f(5) + g(5) = 8 + 6 = 14 (fg)(5) = f(5) g(5) = 8 6 = 48 (f g)(5) = f(g(5)) = f(6) = 7 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 13
Answers: Given Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 14
Evaluating Combinations of Functions Graphically • Use graph of f and g below to evaluate • (f + g) (1) • (f – g) (1) (f/g) (1) • (f g) (1) • • y = f(x) y = g(x) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 15
Answers: • (f + g) (1) = f(1) + g(1) = 3 + 0 = 3 • (f – g) (1) = f(1) – g(1) = 3 – 0 = 3 y = f(x) y = g(x) • (fg) (1) = f(1) g(1) = 3 0 = 0 (f/g) (1) is undefined, because division by 0 is undefined. • (f g) (1) = f(g(1)) = f(0) = 2 • Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 16
5. 2 ♦ ♦ Inverse Functions and Their Representations Calculate inverse operations Identify one-to-one functions Find inverse functions symbolically Use other representations to find inverse functions Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Inverse Operations • Actions: • • • Put on socks and put on shoes Put a gift inside a box and wrap the box Multiply x by 3 and add 2 Take cube root of x and subtract 1 Inverse Actions • • Take off shoes and take off socks Unwrap the box and take the gift out of the box Subtract 2 from x and divide by 3 Add 1 to x and cube the result Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 18
Reminder of the definition of a function • • • y = f(x) means that given an input x, there is just one corresponding output y. Graphically, this means that the graph passes the vertical line test. Numerically, this means that in a table of values for y = f(x) there are no x-values repeated. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 19
Example • Given y 2 = x, is y = f(x)? That is, is y a function of x? • No, because if x = 4, y could be 2 or – 2. • x y 4 – 2 1 – 1 0 0 1 1 4 2 • Note that the graph fails the vertical line test. Note that there is a value of x in the table for which there are two different values of y (that is, x-values are repeated. ) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 20
Idea Behind a One-to-One Function • Given a function y = f(x), f is 1 -1 (pronounced “one-to-one”) means that • given an output y there was just one input x which produced that output. • • Graphically, this means that the graph passes the horizontal line test. (Every horizontal line intersects the graph at most once. ) Numerically, this means there are no y-values repeated in a table of values. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 21
Example • Given y = f(x) = |x|, is f 1 -1? • No, because if y = 2, x could be 2 or – 2. • Note that the graph fails the horizontal line test. (-2, 2) x y – 2 2 – 1 1 0 0 1 1 2 2 • (2, 2) Note that there is a value of y in the table for which there are two different values of x (that is, y-values are repeated. ) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 22
Formal Definition of One-to-One Function • • A function f is a one-to-one function if, for elements c and d in the domain of f, c ≠ d implies f(c) ≠ f(d) Example: Given y = f(x) = |x|, f is not 1 -1 because – 2 ≠ 2 yet | – 2 | = | 2 | Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 23
Given a 1 -1 function f • • f -1 is a symbol for the inverse of the function f, not to be confused with the reciprocal. If f -1(x) does NOT mean 1/ f(x), what does it mean? • • y = f -1(x) means that x = f(y) Note that y = f -1(x) is pronounced “y equals f inverse of x. ” Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 24
Example of An Inverse Function • • • Let F be Fahrenheit temperature and let C be Centigrade temperature. F = f(C) = (9/5)C + 32 C = f -1(F) = ? ? ? • • The function f multiplies an input C by 9/5 and adds 32. To undo multiplying by 9/5 and adding 32, one should • subtract 32 and divide by 9/5 So C = f -1(F) = (F – 32)/(9/5) So C = f -1(F) = (5/9)(F – 32) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 25
Example of An Inverse Function (con’t) • • • F = f(C) = (9/5)C + 32 C = f -1(F) = (5/9)(F – 32) Evaluate f(0) and interpret. • • • Evaluate f -1(32) and interpret. • • • f(0) = (9/5)(0) + 32 = 32 When the Centigrade temperature is 0, the Fahrenheit temperature is 32. f -1(32) = (5/9)(32 - 32) = 0 When the Fahrenheit temperature is 32, the Centigrade temperature is 0. Note that f(0) = 32 and f -1(32) = 0 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 26
Example 2 • To undo multiplying by 3 and adding 2, one must • • subtract 2 and divide by 3 • So if f(x) = 3 x + 2, then f -1(x) = (x – 2)/3 • Let’s check • • f(1) = 3(1) + 2 = 5 and f -1(5) = (5 – 2)/3 = 1 So f(1) = 5 and f -1(5) = 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 27
Example 3 • • Let f(x) compute the distance traveled in miles after x hours by a car with a velocity of 60 miles per hour. Explain what f -1 computes. We are given that distance = f(time) so time = f -1(distance). f -1 computes the time it takes a car with a velocity of 60 mph to travel x miles. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 28
Example 4 • • Describe verbally the inverse of the statement. Then express both the statement and its inverse symbolically. • Take the cube root of x and add 1. To undo taking a cube root and adding 1, one must subtract 1 and cube the result. • Expressing the given statement symbolically • Expressing the inverse symbolically Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 29
Let’s check So f(8) = 3 and f -1(3) = 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 30
Note that: Since f(8) = 3 and f -1(3) = 8 • (f -1 o f)(8) = f -1(f(8)) = f -1(3) = 8 • (f o f -1 )(3) = f(f -1 (3)) = f (8) = 3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 31
Formal Definition of an Inverse Function Let f be a 1 -1 function. Then f -1 is the inverse function of f, if • (f -1 o f)(x) = f -1(f(x)) = x for every x in the domain of f • (f o f -1 )(x) = f(f -1(x)) = x for every x in the domain of f -1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 32
Using composition of functions verify that if 3 = f (x) x + 1 then Step 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 33
Using composition of functions verify that if then Step 2 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 34
Procedure for finding Symbolic Representation of f -1 • • Make sure that f is a 1 -1 function. If not, f -1 does not exist. Solve the equation y = f(x) for x, resulting in the equation x = f -1(y) Interchange x and y to obtain y = f -1(x) Example. • • • f(x) = 3 x + 2 y = 3 x + 2 Solving for x gives: 3 x = y – 2 x = (y – 2)/3 Interchanging x and y gives: y = (x – 2)/3 So f -1(x) = (x – 2)/3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 35
Recall • To undo multiplying by 3 and adding 2, one must • • • subtract 2 and divide by 3 So if f(x) = 3 x + 2, then f -1(x) = (x – 2)/3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 36
Why does f need to be 1 -1 to have an inverse function? Suppose f(x) = x 2 • f(3) = 9 and f( 3) = 9 • If we defined f -1(9), it could have two different values, namely 3 and – 3 and thus f -1 would not be a function. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 37
Evaluating an inverse function numerically x 1 2 3 4 5 f(x) – 5 – 3 0 3 5 • The function is 1 -1 so f -1 exists. • f -1(– 5) = 1 • f -1(– 3) = 2 • f -1(0) = 3 • f -1(3) = 4 • f -1(5) = 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 38
Evaluating an inverse function graphically • The graph of f below passes the horizontal line test so f is 1 -1. Evaluate f -1(4). • Since the point (2, 4) is on the graph of f, the point (4, 2) will be on the graph of f -1 and thus f -1(4) = 2 f(2)=4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 39
Graphs of Functions and Their Inverses • The graph of f -1 is a reflection of the graph of f across the line y = x Note that the domain of f equals the range of f -1 and the range of f equals the domain of f -1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 40
5. 3 ♦ ♦ Exponential Functions and Models Distinguish between linear and exponential growth Model data with exponential functions Calculate compound interest Use the natural exponential functions in applications Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Population Growth by a Constant Number vs by a Constant Percentage Suppose a population is 10, 000 in January 2004. Suppose the population increases by… • • • 500 people per year What is the population in Jan 2005? • 10, 000 + 500 = 10, 500 What is the population in Jan 2006? • 10, 500 + 500 = 11, 000 • • • 5% per year What is the population in Jan 2005? • 10, 000 +. 05(10, 000) = 10, 000 + 500 = 10, 500 What is the population in Jan 2006? • 10, 500 +. 05(10, 500) = 10, 500 + 525 = 11, 025 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 42
Suppose a population is 10, 000 in Jan 2004. Suppose the population increases by 500 per year. What is the population in …. • • Jan 2005? • 10, 000 + 500 = 10, 500 Jan 2006? • 10, 000 + 2(500) = 11, 000 Jan 2007? • 10, 000 + 3(500) = 11, 500 Jan 2008? • 10, 000 + 4(500) = 12, 000 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 43
Suppose a population is 10, 000 in Jan 2004 and increases by 500 per year. • • • Let t be the number of years after 2004. Let P(t) be the population in year t. What is the symbolic representation for P(t)? We know… Population in 2004 = P(0) = 10, 000 + 0(500) Population in 2005 = P(1) = 10, 000 + 1(500) Population in 2006 = P(2) = 10, 000 + 2(500) Population in 2007 = P(3) = 10, 000 + 3(500) Population t years after 2004 = P(t) = 10, 000 + t(500) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 44
Population is 10, 000 in 2004; increases by 500 per yr P(t) = 10, 000 + t(500) • • P is a linear function of t. What is the slope? • • 500 people/year What is the y-intercept? • number of people at time 0 (the year 2004) = 10, 000 When P increases by a constant number of people per year, P is a linear function of t. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 45
Suppose a population is 10, 000 in Jan 2004. More realistically, suppose the population increases by 5% per year. What is the population in …. • • • Jan 2005? • 10, 000 +. 05(10, 000) = 10, 000 + 500 = 10, 500 Jan 2006? • 10, 500 +. 05(10, 500) = 10, 500 + 525 = 11, 025 Jan 2007? • 11, 025 +. 05(11, 025) = 11, 025 + 551. 25 = 11, 576. 25 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 46
Suppose a population is 10, 000 in Jan 2004 and increases by 5% per year. • • • Let t be the number of years after 2004. Let P(t) be the population in year t. What is the symbolic representation for P(t)? We know… Population in 2004 = P(0) = 10, 000 Population in 2005 = P(1) = 10, 000 +. 05 (10, 000) = 1. 051(10, 000) =10, 500 Population in 2006 = P(2) = 10, 500 +. 05 (10, 500) = 1. 05 (1. 05)(10, 000) = 1. 052(10, 000) = 11, 025 Population t years after 2004 = P(t) = 10, 000(1. 05)t Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 47
Population is 10, 000 in 2004; increases by 5% per yr P(t) = 10, 000 (1. 05)t • • • P is an EXPONENTIAL function of t. More specifically, an exponential growth function. What is the base of the exponential function? • 1. 05 What is the y-intercept? • number of people at time 0 (the year 2004) = 10, 000 When P increases by a constant percentage per year, P is an exponential function of t. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 48
Linear vs. Exponential Growth • • A Linear Function adds a fixed amount to the previous value of y for each unit increase in x For example, in f(x) = 10, 000 + 500 x 500 is added to y for each increase of 1 in x. • • An Exponential Function multiplies a fixed amount to the previous value of y for each unit increase in x. For example, in f(x) = 10, 000 (1. 05)x y is multiplied by 1. 05 for each increase of 1 in x. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 49
Definition of Exponential Function • A function represented by f(x) = Cax, a > 0, a not 1, and C > 0 is an exponential function with base a and coefficient C. • • If a > 1, then f is an exponential growth function If 0 < a < 1, then f is an exponential decay function Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 50
Caution • • Don’t confuse f(x) = 2 x with f(x) = x 2 f(x) = 2 x is an exponential function. f(x) = x 2 is a polynomial function, specifically a quadratic function. The functions and consequently their graphs are very different. f(x) = x 2 f(x) = 2 x Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 51
Comparison of Exponential and Linear Functions Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 52
Linear Function y = 10000 + 500 x x 0 1 2 3 4 5 6 ? x ? y y 10000 10500 11000 11500 12000 12500 13000 1 1 1 500 500 500 ? y ? x 500/1=500 500/1=500 Linear Function Slope is constant. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 53
Exponential Function Y = 10000 (1. 05)x Note that this constant is the base of the exponential function. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Exponential Function Ratios of consecutive y -values (corresponding to unit increases in x) are constant, in this case 1. 05. 54
Which function is linear and which is exponential? x y -3 3/8 -2 3/4 -1 3/2 0 3 1 6 2 12 3 24 x y -3 9 -2 7 -1 5 0 3 1 1 2 -1 3 -3 For the linear function, tell the slope and y -intercept. For the exponential function, tell the base and the y-intercept. Write the equation of each. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 55
Which function is linear and which is exponential? continued x -3 -2 -1 0 1 2 3 y 3/8 3/4 3/2 3 6 12 24 (3/4)/(3/8) = 2 (3/2)/(3/4) = 2 3/(3/2) = 2 6/3 = 2 12/6 = 2 24/12 = 2 y is an exponential function of x because the ratio of consecutive values of y is constant, namely 2. Thus the base is 2. The y-intercept is 3. Thus the equation is y = 3· 2 x Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 56
Which function is linear and which is exponential? continued x -3 -2 -1 0 1 2 3 y 9 7 5 3 1 -1 -3 y is a linear function of x because the slope is constant, namely – 2/1 = – 2. The y-intercept is 3. Thus the equation is y = – 2 x + 3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 57
Exponential Growth vs Decay • Example of exponential growth function f(x) = 3 • 2 x • Example of exponential decay function Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 58
Recall • In the exponential function f(x) = Cax • • If a > 1, then f is an exponential growth function If 0 < a < 1, then f is an exponential decay function Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 59
Exponential Growth Function f(x) = Cax where a > 1 • • Example f(x) = 3 • 2 x • Properties of an exponential growth function • • • Domain: (-∞, ∞) Range: (0, ∞) f increases on (-∞, ∞) The negative x-axis is a horizontal asymptote. y-intercept is (0, 3). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 60
Exponential Decay Function f(x) = Cax where 0 < a < 1 • Example • Properties of an exponential decay function • • • Domain: (-∞, ∞) Range: (0, ∞) f decreases on ( -∞ , ∞ ) The positive x-axis is a horizontal asymptote. y-intercept is (0, 3). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 61
Example of exponential decay - Carbon -14 dating • • • The time it takes for half of the atoms to decay into a different element is called the half-life of an element undergoing radioactive decay. The half-life of carbon-14 is 5700 years. Suppose C grams of carbon-14 are present at t = 0. Then after 5700 years there will be C/2 grams present. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 62
Recall the half-life of carbon-14 is 5700 years. • • Let t be the number of years. Let A =f(t) be the amount of carbon-14 present at time t. Let C be the amount of carbon-14 present at t = 0. Then f(0) = C and f(5700) = C/2. Thus two points of f are (0, C) and (5700, C/2) Using the point (5700, C/2) and substituting 5700 for t and C/2 for A in A = f(t) = Cat yields: C/2 = C a 5700 Dividing both sides by C yields: 1/2 = a 5700 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 63
Recall the half-life of carbon-14 is 5700 years. Half-life Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 64
Generalizing this • If a radioactive sample containing C units has a half-life of k years, then the amount A remaining after x years is given by Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 65
Example of Radioactive Decay • Radioactive strontium-90 has a half-life of about 28 years and sometimes contaminates the soil. After 50 years, what percentage of a sample of radioactive strontium would remain? Note calculator keystrokes: Since C is present initially and after 50 years. 29 C remains, then 29% remains. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 66
Example of Exponential Growth Compound Interest • • Suppose $10, 000 is deposited into an account which pays 5% interest compounded annually. Then the amount A in the account after t years is: A(t) = 10, 000 (1. 05)t Note the similarity with: Suppose a population is 10, 000 in 2004 and increases by 5% per year. Then the population P, t years after 2004 is: P(t) = 10, 000 (1. 05)t Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 67
Frequencies of Compounding (Adding Interest) • • • annually (1 time per year) semiannually (2 times per year) quarterly (4 times per year) monthly (12 times per year) daily (365 times per year) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 68
Compound Interest Formula • If P dollars is deposited in an account paying an annual rate of interest r, compounded (paid) n times per year, then after t years the account will contain A dollars, where Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 69
Suppose $1000 is deposited into an account yielding 5% interest compounded at the following frequencies. How much money after t years? • Annually • Semiannually • Quarterly • Monthly Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 70
The Natural Exponential Function • The function f, represented by f(x) = ex is the natural exponential function where e 2. 71828 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 71
Continuously Compounded Interest • If a principal of P dollars is deposited in an account paying an annual rate of interest r (expressed in decimal form), compounded continuously, then after t years the account will contain A dollars, where A = Pert Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 72
Example • Suppose $100 is invested in an account with an interest rate of 8% compounded continuously. How much money will there be in the account after 15 years? A = Pert A = $100 e. 08(15) A = $332. 01 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 73
5. 4 ♦ ♦ Logarithmic Functions and Models Evaluate the common logarithm function Solve basic exponential and logarithmic equations Evaluate logarithms with other bases Solve general exponential and logarithmic equations Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Common Logarithm • • The common logarithm of a positive number x, denoted log x, is defined by logx = k if and only if x = 10 k where k is a real number. The function given by f(x) = log x is called the common logarithm function. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 75
Evaluate each of the following. • log 10 • 1 because 101 = 10 • log 100 • 2 because 102 = 100 • log 1000 • 3 because 103 = 1000 • log 10000 • 4 because 104 = 10000 • log (1/10) • – 1 because 10 -1 = 1/10 log (1/100) • – 2 because 10 -2 = 1/100 log (1/1000) • – 3 because 10 -3 = 1/1000 log 1 • 0 because 100 = 1 • • • Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 76
Graph of f(x) = log x x f(x) . 01 -2 . 1 -1 1 0 10 1 100 2 Note that the graph of y = log x is the graph of y = 10 x reflected through the line y = x. This suggests that these are inverse functions. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 77
The Inverse of y = log x • • Note that the graph of f(x) = log x passes the horizontal line test so it is a 1 -1 function and has an inverse function. Find the inverse of y = log x Using the definition of common logarithm to solve for x gives x = 10 y Interchanging x and y gives y = 10 x So yes, the inverse of y = log x is y = 10 x Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 78
Inverse Properties of the Common Logarithm • Recall that f -1(x) = 10 x given f(x) = log x • Since (f f -1 )(x) = x for every x in the domain of f -1 • • log(10 log( x) = x for all real numbers x. Since (f -1 f)(x) = x for every x in the domain of f • 10 logx = x for any positive number x Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 79
Solving Exponential Equations Using The Inverse Property log(10 log( x) = x • • Solve the equation 10 x = 35 Take the common log of both sides • • Using the inverse property log(10 log( x) = x this simplifies to • • log 10 x = log 35 Using the calculator to estimate log 35 we have • x 1. 54 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 80
Solving Logarithmic Equations Using The Inverse Property 10 logx = x • • Solve the equation log x = 4. 2 Exponentiate each side using base 10 • • Using the inverse property 10 logx = x this simplifies to • • 10 logx = 104. 2 Using the calculator to estimate 104. 2 we have • x 15848. 93 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 81
Definition of Logarithm With Base a • • The logarithm with base a of a positive number x, denoted by logax is defined by logax = k if and only if x = ak where a > 0, a ≠ 1, and k is a real number. The function given by f(x) = logax is called the logarithmic function with base a. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 82
Practice with the Definition Practice Questions: • Logbc = d means____ • p = logy m means ____ • True or false: • • True or false: • • log 28 = 3 log 525 = 2 log 255 = 1/2 log 48 = 2 What is the value of log 48? • • Answers: bd = c yp = m True because 23=8 True because 52=25 True because 251/2=5 False because 42=16 not 8 It is 3/2 because 43/2 = 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 83
Practice Evaluating Logarithms • • • Evaluate log 636 log 366 log 232 log 322 log 6(1/36) log 2 (1/32) log 100 log (1/10) log 1 • • • Answers: 2 because 62 = 36 1/2 because 36(1/2) = 6 5 because 25 = 32 1/5 because 32(1/5) = 2 – 2 because 6– 2 = 1/36 – 5 because 2– 5 = 1/32 2 because 102 = 100 – 1 because 10– 1 = 1/10 0 because 100 = 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 84
Calculators and logarithms • • The TI-83 evaluates base 10 logarithms and base e logarithms. Base 10 logs are called common logs. • • • log x means log 10 x. Notice the log button on the calculator. Base e logs are called natural logs. • • ln x means logex. Notice the ln button on the calculator. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 85
Evaluate each of the following without calculator. Then check with calculator. • • lne ln(e 2) ln 1. • • lne = logee = 1 since e 1= e ln(e 2) = loge (e 2) = 2 since 2 is the exponent that goes on e to produce e 2. ln 1 = loge 1 = 0 since e 0= 1 1/2 since 1/2 is the exponent that goes on e to produce e 1/2 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 86
The Inverse of y = logax • • Note that the graph of f(x) = logax passes the horizontal line test so it is a 1 -1 function and has an inverse function. Find the inverse of y = logax Using the definition of common logarithm to solve for x gives x = ay Interchanging x and y gives y = ax So the inverse of y = logax is y = ax Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 87
Inverse Properties of Logarithms With Base a • • • Recall that f -1(x) = ax given f(x) = logax Since (f f -1 )(x) = x for every x in the domain of f -1 x • loga(a ) = x for all real numbers x. Since (f -1 f)(x) = x for every x in the domain of f logax = x for any positive number x • a Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 88
Solving Exponential Equations Using The Inverse Property loga(ax) = x • • Solve the equation 4 x = 1/64 Take the log of both sides to the base 4 • • Using the inverse property loga (ax) =x this simplifies to • • x = log 4(1/64) Since 1/64 can be rewritten as 4– 3 • • log 4 (4 x) = log 4(1/64) x = log 4(4– 3) Using the inverse property loga (ax) = x this simplifies to • x = – 3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 89
Solving Exponential Equations Using The Inverse Property loga(ax) = x • • Solve the equation ex = 15 Take the log of both sides to the base e • • Using the inverse property loga(ax) = x this simplifies to • • ln(ex) = ln(15) x = ln 15 Using the calculator to estimate ln 15 • x 2. 71 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 90
Solving Logarithmic Equations Using The Inverse Property alogax = x • • Solve the equation lnx = 1. 5 Exponentiate both sides using base e • • Using the inverse property alogax = x this simplifies to • • elnx = e 1. 5 Using the calculator to estimate e 1. 5 • x 4. 48 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 91
Recall from section 5. 3 Exponential Decay Function Exponential Growth Function Graph of f(x) = ax where a >1 Graph of f(x) = ax where 0 < a <1 Using the fact that the graph of a function and its inverse are symmetric with respect to the line y = x, graph f-1(x) = logax for the two types of exponential functions listed above. Looking at the two resulting graphs, what is the domain of a logarithmic function? What is the range of a logarithmic function? Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 92
Exponential Decay Function Exponential Growth Function Graph of f(x) = ax where a >1 Graph of f(x) = ax where 0 < a < 1 Superimpose graphs of the inverses of the functions above similar to Figure 5. 58 on page 422 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 93
5. 5 Properties of Logarithms ♦ Apply basic properties of logarithms ♦ Use the change of base formula Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Property 1 • loga 1 = 0 and logaa = 1 • True because a 0 = 1 and a 1 = a Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 95
Property 2 • logam + logan = loga(mn) • The sum of logs is the log of the product. • Example: Let a = 2, m = 4 and n = 8 • • logam + logan = log 24 + log 28 = 2 + 3 loga(mn) = log 2(4 · 8) = log 2(32) = 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 96
Property 3 • • • The difference of logs is the log of the quotient. Example: Let a = 2, m = 4 and n = 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 97
Property 4 • • Example: Let a = 2, m = 4 and r = 3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 98
Example • Expand the expression. Write without exponents. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 99
Example • Write as the logarithm of a single expression Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 100
Change of Base Formula • Example: Use the change of base formula to evaluate log 38 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 101
5. 6 Exponential and Logarithmic Equations ♦ Solve exponential equations ♦ Solve logarithmic equations Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Modeling Compound Interest • • How long does it take money to grow from $100 to $200 if invested into an account which compounds quarterly at an annual rate of 5%? Must solve for t in the following equation Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 103
Solve for t Divide each side by 100 Take common logarithm of each side Property 4: log(mr) = rlogm Divide each side by 4 log 1. 0125 Approximate using calculator Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 104
Alternatively Divide each side by 100 Take natural logarithm of each side Property 4: ln(mr) = r lnm Divide each side by 4 ln 1. 0125 Approximate using calculator Note that the only difference in this and the previous solution is that we took natural logs of both sides of the equation instead of common logs. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 105
Solving the Previous Equation Graphically Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 106
Solve 3(1. 2)x + 2 = 15 for x symbolically Divide each side by 3 Take common logarithm of each side (Could use natural logarithm) Property 4: log(mr) = r logm Divide each side by log 1. 2 Approximate using calculator Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 107
Solve ex+2 = 52 x for x symbolically Take natural logarithm of each side Property 4: ln(mr) = r lnm lne = 1 Subtract 2 xln 5 and 2 from each side Factor x from left-hand side Divide each side by 1 – 2 ln 5 Approximate using calculator Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 108
Solving a Logarithmic Equation Symbolically • In developing countries there is a relationship • • between the amount of land a person owns and the average daily calories consumed. This relationship is modeled by the formula C(x) = 280 ln(x+1) + 1925 where x is the amount of land owned in acres and Source: D. Gregg: The World Food Problem Determine the number of acres owned by someone whose average intake is 2400 calories per day. Must solve for x in the equation 280 ln(x+1) + 1925 = 2400 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 109
Solve 280 ln(x+1) + 1925 = 2400 Symbolically Subtract 1925 from each side Divide each side by 280 Exponentiate each side base e Inverse property elnk = k Subtract 1 from each side Approximate using calculator Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 110
Solving Logarithmic Equations Symbolically Definition of logarithm logax = k if and only if x = ak Add x to both sides of equation Subtract 2 from both sides of equation Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 111
5. 7 Constructing Nonlinear Models ♦ Find an exponential model ♦ Find a logarithmic model ♦ Find a logistic model Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
3 types of nonlinear data Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 113
Exponential Model Example The National Institute for Automotive Service Excellence (ASE) reported that the number of females working in automotive repair is increasing. Year 1988 1989 1990 1991 1992 1993 1994 1995 Total 556 614 654 737 849 1086 1329 1592 a) What type of function might model these data? b) Use least-squares regression to find an exponential function given by f(x) = abx that model the data. c) Use f to estimate the number of certified female technicians in 2005. Round the result to the nearest hundred. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 114
Solution Year 1988 1989 1990 1991 1992 1993 1994 1995 Total 556 614 654 737 849 1086 1329 1592 a) Let y = number of female technicians Let x = 0 correspond to 1988, x = 1 to 1989 and so on, until x = 7 corresponds to 1995. A scatterplot of the data is shown. The data is rapidly increasing, and an exponential function might model these data. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 115
Solution continued b) The formula f(x) = 507. 1(1. 166)x is found using the least-squares regression, as shown in the following figures. c) Since x = 17 corresponds to year 2005, f(17) = 507. 1(1. 166)17 6902. According to the model, the number of certified female automotive technicians in 2005 could be about 6900. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 116
Logarithmic Model Example The table below lists the interest rates for certificates of deposit during January 1997. Use the data to complete the following. Time Yield 6 mo 1 yr 4. 75% 5. 03% 2. 5 yr 5. 25% 5. 54% a) Make a scatterplot of the data. What type of function might model these data? b) Use least-squares regression to obtain a formula, f(x) = a + b ln x, that models these data. c) Graph f and the data in the same viewing rectangle. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 117
Solution Time Yield 6 mo 1 yr 4. 75% 5. 03% 2. 5 yr 5. 25% 5. 54% a) Enter the data points into your calculator. The data increase but are gradually leveling off. A logarithmic modeling function may be appropriate. b) The least-square regression has been used to find logarithmic function. f(x) = 5 + 0. 33 lnx. c) A graph and data are shown. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 118
Logistic Model Example One of the earliest studies about population growth was done using yeast plants in 1913. A small amount of yeast was placed in a container with a fixed amount of nourishment. The units of yeast were recorded every 2 hours. Time 0 2 4 6 8 10 12 14 16 18 Yeast 9. 6 29. 0 71. 1 174. 6 350. 7 513. 3 594. 8 640. 8 655. 9 661. 8 a) Make a scatterplot of the data. Describe the growth. b) Use least-squares regression to find a logistic function that models the data. c) Graph f and the data in the same viewing rectangle. d) Approximate graphically the time when the amount of yeast was 200 units. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 119
Solution Time 0 2 4 6 8 10 12 14 16 18 Yeast 9. 6 29. 0 71. 1 174. 6 350. 7 513. 3 594. 8 640. 8 655. 9 661. 8 a) The yeast increase slowly at first, then they grow more rapidly, until the amount of yeast gradually levels off. b) Least-squares regression to find a logistic function. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 120
Solution continued c) Data and graph in the same viewing rectangle. d) The graphs of y 1 = f(x) and y(2) = 200 intersect near (6. 29, 200). The amount of yeast reached 200 units after about 6. 29 hours. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 121
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