Exponential Functions MATH 160, Precalculus J. Robert Buchanan Department of Mathematics Fall 2011
Objectives In this lesson we will learn to: recognize and evaluate exponential functions with base a, graph exponential functions and use the One-to-One Property, recognize and evaluate exponential functions with base e, use exponential functions to model and solve real-world problems.
Exponential Functions Definition The exponential function f with base a is denoted by f (x) = a x where a > 0, a 1, and x is any real number.
Exponential Functions Definition The exponential function f with base a is denoted by f (x) = a x where a > 0, a 1, and x is any real number. Example Use a calculator if necessary to evaluate each of the following expressions. 3 1.2 = 1.2 3 = 2 2.7 = 0.8 0.2 = 1.01 12.7 =
Exponential Functions Definition The exponential function f with base a is denoted by f (x) = a x where a > 0, a 1, and x is any real number. Example Use a calculator if necessary to evaluate each of the following expressions. 3 1.2 = 3.73719 1.2 3 = 2 2.7 = 0.8 0.2 = 1.01 12.7 =
Exponential Functions Definition The exponential function f with base a is denoted by f (x) = a x where a > 0, a 1, and x is any real number. Example Use a calculator if necessary to evaluate each of the following expressions. 3 1.2 = 3.73719 1.2 3 = 1.728 2 2.7 = 0.153893 0.8 0.2 = 0.956352 1.01 12.7 = 1.1347
Graphs, a > 1 When a > 1 the graph of an exponential function resembles the following. x 2 x 1 4 16 3 1 8 2 1 4 1 1 2 0 1 1 2 2 4 3 8 4 16 15 10 5 x
Effect of the Base If we increase the value of the base a, the graph becomes steeper. 30 25 20 15 10 5 x
Graphs, 0 < a < 1 When 0 < a < 1 the graph of an exponential function resembles the following. x ( 1 3 4 81 3 27 2 9 1 3 0 1 1 1 3 2 1 9 3 1 27 4 1 81 ) x 4 35 30 25 20 15 10 5 x
Asymptote Regardless of the base a, the function f (x) = a x approaches, but never touches the line y = 0 (the x-axis). In situations like this we may say, y = 0 is an asymptote of the graph of f (x) = a x, or y = 0 is a horizontal asymptote of the graph of f (x) = a x, or the graph of f (x) = a x approaches the line y = 0 asymptotically.
General Concepts of Exponential Functions For a > 1: For 0 < a < 1: a x > 0 a x increases on (, ) and is called an exponential growth function. a 0 = 1, so the point (0, 1) is on the graph. a x approaches the x-axis for negative values of x (The x-axis is a horizontal asymptote). a x > 0 a x decreases on (, ) and is called an exponential decay function. a 0 = 1, so the point (0, 1) is on the graph. a x approaches the x-axis for positive values of x (The x-axis is a horizontal asymptote).
One-to-One Property Since exponential functions are either always increasing or always decreasing, they all pass the Horizontal Line Test and are therefore one-to-one functions. One-to-One Property For a > 0 and a 1, a x = a y if and only if x = y.
One-to-One Property Since exponential functions are either always increasing or always decreasing, they all pass the Horizontal Line Test and are therefore one-to-one functions. One-to-One Property For a > 0 and a 1, a x = a y if and only if x = y. Example Use the One-to-One Property to solve the following equation. 2 x 3 = 16
One-to-One Property Since exponential functions are either always increasing or always decreasing, they all pass the Horizontal Line Test and are therefore one-to-one functions. One-to-One Property For a > 0 and a 1, a x = a y if and only if x = y. Example Use the One-to-One Property to solve the following equation. 2 x 3 = 16 2 x 3 = 2 4
One-to-One Property Since exponential functions are either always increasing or always decreasing, they all pass the Horizontal Line Test and are therefore one-to-one functions. One-to-One Property For a > 0 and a 1, a x = a y if and only if x = y. Example Use the One-to-One Property to solve the following equation. 2 x 3 = 16 2 x 3 = 2 4 x 3 = 4 x = 7
Natural Base e In many applications (particularly to the physical and social sciences) we will prefer to use the base e 2.718281828.... This is called the natural base and the function f (x) = e x is called the natural exponential function.
Natural Base e In many applications (particularly to the physical and social sciences) we will prefer to use the base e 2.718281828.... This is called the natural base and the function f (x) = e x is called the natural exponential function. Example Use a calculator if necessary to evaluate each of the following expressions. e 1.2 = e 3 = e 2.7 = e 0.2 = e π =
Natural Base e In many applications (particularly to the physical and social sciences) we will prefer to use the base e 2.718281828.... This is called the natural base and the function f (x) = e x is called the natural exponential function. Example Use a calculator if necessary to evaluate each of the following expressions. e 1.2 = 3.32012 e 3 = e 2.7 = e 0.2 = e π =
Natural Base e In many applications (particularly to the physical and social sciences) we will prefer to use the base e 2.718281828.... This is called the natural base and the function f (x) = e x is called the natural exponential function. Example Use a calculator if necessary to evaluate each of the following expressions. e 1.2 = 3.32012 e 3 = 20.0855 e 2.7 = 0.0672055 e 0.2 = 1.2214 e π = 23.1407
Application: Compound Interest Definition After t years, the balance A in an account with principal P and annual interest rate r (expressed as a decimal) is given by the following formulas. ( 1 For n compounding periods per year: A = P 1 + r ) n t n 2 For continuous compounding: A = Pe r t
Example Determine the balance A at the end of 20 years if $1500 is invested at 6.5% interest and the interest is compounded quarterly monthly weekly continuously
Example Determine the balance A at the end of 20 years if $1500 is invested at 6.5% interest and the interest is compounded quarterly ( A = 1500 1 + 0.065 ) (4)(20) 5446.73 4 monthly weekly continuously
Example Determine the balance A at the end of 20 years if $1500 is invested at 6.5% interest and the interest is compounded quarterly ( A = 1500 1 + 0.065 ) (4)(20) 5446.73 4 monthly ( A = 1500 1 + 0.065 ) (12)(20) 5484.67 12 weekly continuously
Example Determine the balance A at the end of 20 years if $1500 is invested at 6.5% interest and the interest is compounded quarterly ( A = 1500 1 + 0.065 ) (4)(20) 5446.73 4 monthly weekly ( A = 1500 1 + 0.065 ) (12)(20) 5484.67 12 ( A = 1500 1 + 0.065 ) (52)(20) 5499.48 52 continuously
Example Determine the balance A at the end of 20 years if $1500 is invested at 6.5% interest and the interest is compounded quarterly ( A = 1500 1 + 0.065 ) (4)(20) 5446.73 4 monthly weekly ( A = 1500 1 + 0.065 ) (12)(20) 5484.67 12 ( A = 1500 1 + 0.065 ) (52)(20) 5499.48 52 continuously A = 1500e (0.065)(20) 5503.93
Application: Exponential Growth The number of fruit flies in an experimental population after t hours is given by Q(t) = 20e 0.03t for t 0. 1 Find the initial number of fruit flies in the population. 2 How large is the population of fruit flies after 72 hours?
Application: Exponential Growth The number of fruit flies in an experimental population after t hours is given by Q(t) = 20e 0.03t for t 0. 1 Find the initial number of fruit flies in the population. Q(0) = 20e (0.03)(0) = 20 2 How large is the population of fruit flies after 72 hours?
Application: Exponential Growth The number of fruit flies in an experimental population after t hours is given by Q(t) = 20e 0.03t for t 0. 1 Find the initial number of fruit flies in the population. Q(0) = 20e (0.03)(0) = 20 2 How large is the population of fruit flies after 72 hours? Q(72) = 20e (0.03)(72) 173
Application: Radioactive Decay Let Q represent the mass of carbon-14 ( 14 C) in grams. The quantity of carbon-14 present after t years is given by Q(t) = 10 ( ) 1 t/5715 2 1 Find the initial quantity of carbon-14 present. 2 Determine the quantity present after 2500 years.
Application: Radioactive Decay Let Q represent the mass of carbon-14 ( 14 C) in grams. The quantity of carbon-14 present after t years is given by Q(t) = 10 ( ) 1 t/5715 2 1 Find the initial quantity of carbon-14 present. Q(0) = 10 ( ) 1 0/5715 = 10 grams 2 2 Determine the quantity present after 2500 years.
Application: Radioactive Decay Let Q represent the mass of carbon-14 ( 14 C) in grams. The quantity of carbon-14 present after t years is given by Q(t) = 10 ( ) 1 t/5715 2 1 Find the initial quantity of carbon-14 present. Q(0) = 10 ( ) 1 0/5715 = 10 grams 2 2 Determine the quantity present after 2500 years. Q(2500) = 10 ( ) 1 2500/5715 7.38441 grams 2
Half-Life 10 8 6 4 2 0 2000 4000 6000 8000 10 000 Q(t) = 10 ( ) 1 t/5715 2
Homework Read Section 3.1. Exercises: 1, 5, 9, 13,..., 69, 73