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Chapter 9. Exponential and Logarithmic Functions. The Algebra of Functions; Composite Functions. § 9.1. Operations on Functions. It is possible to add, subtract, multiply, and divide functions. The results of these operations will also be functions (assuming we don’t divide by zero).

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## Exponential and Logarithmic Functions

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**Chapter 9**Exponential and Logarithmic Functions**Operations on Functions**It is possible to add, subtract, multiply, and divide functions. The results of these operations will also be functions (assuming we don’t divide by zero).**Operations on Functions**Algebra of Functions Let f and g be functions. New functions from f and g are defined as follows: Sum (f + g)(x) = f(x) + g(x) Difference (f – g)(x) = f(x) – g(x) Product (f·g)(x) = f(x) ·g(x) Quotient**Operations on Functions**Example: • If f(x) = 4x + 3 and g(x) = x2, then find each of the following • (f + g)(x) • 4x + 3 + x2 = x2 + 4x + 3 • (f – g)(x) • 4x + 3 – x2 = -x2 + 4x + 3 • (f·g)(x) • (4x + 3)x2 = 4x3 + 3x2**If f(x) = 4x + 3 and g(x) = x2, then find**x 0 Operations on Functions Example:**Function Composition**We can also combine functions through a function composition. A function composition uses the output from the first function as the input to the second function. Composition of a Function The composition of function f and g is This means the value of x is first substituted into the function g. Then the value that results from the function g is input into the function f.**Function Composition**Notice, that with function composition, we actually activate the functions from right to left in the notation. The function named on the right side of the composition notation is the one we substitute the value for the variable into first.**Function Composition**Example: If f(x) = 4x + 3 and g(x) = x2, then find f(g(x)) = 4(x2) + 3 = 4x2 + 3 g(f(x)) = (4x + 3)2 = 16x2 + 24x + 9 Notice the results are different with a different order.**Function Composition**Example: If H(x) = x3 + 3, name two functions whose composition will result in H(x). Note: There may be more than one way to select the two functions. Answers are not necessarily unique. Let f(x) = x + 3, and g(x) = x3**§ 9.2**Inverse Functions**One-to-One Functions**We have studied functions which are defined to require that each element of the domain (input values) produce a unique element of the range (output values). 1-1 functions also require that each element of the range be produced from only one element of the domain. each x only one y is a function each y only one x is a 1-1 function**One-to-One Functions**Example: • Determine whether each function described is one to-one. • r = {(1, 2), (3, 4), (5, 6), (6, 7)} • It is 1-to-1, since the 2nd coordinate is unique. • g = {(0, 3), (3, 7), (6, 7), (-2, -2)} • It is not 1-to-1, since the second coordinate, 7, is produced from two different domain elements, both 3 and 6.**One-to-One Functions**• We found previously that the graph of a function must satisfy the Vertical Line Test. • Every vertical line must intersect the graph of the function at most one time. • Similarly, a 1-to-1 function must satisfy the Horizontal Line Test. • Horizontal Line Test • Every horizontal line must intersect the graph of the 1-to-1 function at most one time.**y**x One-to-One Functions Example: Is this the graph of a 1-to-1 function? No, because we can find a horizontal line that intersects the graph in more than one point.**y**x One-to-One Functions Example: Is this the graph of a 1-to-1 function? Yes, because every horizontal line intersects the graph in only one point.**One-to-One Functions**Note that while we previously discovered that a horizontal line represents a function, it is NOT a 1-to-1 function. It will not pass the Horizontal Line Test.**Inverse Functions**For each 1-to-1 function, we can find an inverse function by switching the coordinates of the ordered pairs of the original function. For a particular function f, the notation for the inverse function is f –1, and is read “f inverse”. Note that since we switch the coordinates in the ordered pairs to find the inverse function, the domain of f is the range of f -1, and the range of f is the domain of f -1. So, for the 1-to-1 function f consisting of the ordered pairs (x, y), the inverse function f–1 consists of the ordered pairs (y, x).**Inverse Functions**Example: Find the inverse of the 1-to-1 function g = {(3, 4), (6, 2), (2, -3), (4, -5)}. g-1 = {(4, 3), (2, 6), (-3, 2), (-5, 4)}**Inverse Functions**• If a 1-to-1 function is defined as a set of ordered pairs, it is relatively easy to find the inverse function (just switch the 1st and 2nd coordinates) • Finding the Inverse of a One-to-One Function • 1.) Replace f(x) with y. • 2.) Interchange x and y. • 3.) Solve the equation for y. • 4.) Replace y with the notation f–1(x).**4)**Inverse Functions Example: • Find an equation of the inverse of f(x) = 6x – 1. • Then graph both f and f–1 on the same set of axes. • y = 6x – 1 • x = 6y – 1 • 6y = x + 1 Continued.**y**x Inverse Functions Example continued: f(x) = 6x – 1 Notice that f and f–1 are mirror images of each other around the line y = x. A function and its inverse are symmetric about the line y = x.**Inverse Functions**If f is a 1-to-1 function, then the inverse of f is the function f–1 such that and**If f(x) = x3 – 5, show that f–1(x) =**is the inverse function of f(x) = x3 – 5. Inverse Functions Example: Since both of the compositions give us x,**§ 9.3**Exponential Functions**Exponential Expressions**We have previously worked with exponential expressions, where the exponent was a rational number The expression bx can actually be defined for all real numbers, x, including irrational numbers. However, the proof of this would have to wait until a higher level math course.**Exponential Functions**Exponential Functions A function of the form f(x) = bx is called an exponential function if b > 0, b is not 1, and x is a real number.**Exponential Functions**• We can graph exponential functions of the form f(x) = 3x, g(x) = 5x or h(x) = (½)xby substituting in values for x, and finding the corresponding function values to get ordered pairs. • We would find all graphs satisfy the following properties: • 1-to-1 function • y-intercept (0, 1) • no x-intercept • domain is (-, ) • range is (0, )**y**x Graphs of Exponential Functions We would find a pattern in the graphs of all the exponential functions of the type bx, where b > 1.**y**x Graphs of Exponential Functions We would find a pattern in the graphs of all the exponential functions of the type bx, where 0 < b < 1.**y**(h, 1) x Graphs of Exponential Functions We would find a pattern in the graphs of all the exponential functions of the type bx-h, where b > 1. The graph has the same shape as the graph for bx, except it is shifted to the right h units.**y**(-h, 1) x Graphs of Exponential Functions We would find a pattern in the graphs of all the exponential functions of the type bx+h, where b > 1. The graph has the same shape as the graph for bx, except it is shifted to the left h units.**Uniqueness of bx**Uniqueness of bx Let b > 0 and b 1. Then bx = by is equivalent to x = y. Example: Solve 6x = 36 6x = 62 x = 2**Solving Exponential Functions**Example: Solve 92x+1 = 81 92x+1 = 92 2x + 1 = 2 2x = 1 x = ½**Solve**Solving Exponential Functions Example: 3-3 = 32x –3 = 2x**Solving Exponential Functions**Example: Solve 43x-6 = 322x (22)3x-6 = (25)2x (22)3x-6 = 210x 26x-12 = 210x 6x – 12 = 10x –12 = 4x x = –3**Applications of Exponential Functions**Many applications use exponential functions of various types. Compound interest formulas are exponential functions used to determine the amount of money accumulated or borrowed. Exponential functions with negative exponents can be used to describe situations of decay, while those with positive exponents can be used to describe situations of growth.**Applications of Exponential Functions**Example: Find the total amount invested in a savings account if $5000 was invested and earned 6% compounded monthly for 18 years. Round your answer to two decimal places. The formula that is used for calculating compound interest is where P is the initial principal invested, r is the interest rate, n is the number of times interest is compounded each year, t is the time of the investment (in years) and A is the amount of money in the account. Continued.**$**Applications of Exponential Functions Example continued:**Applications of Exponential Functions**Example: An accidental spill of 100 grams of radioactive material in a local stream has led to the presence of radioactive debris decaying at a rate of 5% each day. Find how much debris still remains after 30 days. The formula that would be used for this problem is where A is the amount of radioactive material to start, r is the rate of decay, t is the number of days and y is the amount of radioactive material after the time period. Continued.**(exact answer)**Applications of Exponential Functions Example continued:**§ 9.4**Logarithmic Functions**If we graph an exponential function where the base > 1, we**get an increasing function, as shown below. y x Graph of a Exponential Function**y**x Graph of a Logarithmic Function We can graph the inverse of the function, as shown below. This inverse function is referred to as a logarithmic function.**Logarithmic Functions**Logarithmic Definition If b > 0 and b≠ 1, then y = logbx means x = by for every x > 0 and every real number y.**c) log3 = ½**Writing Exponential Functions Example: • Write each of the following as an exponential equation. • log4 16 = 2 • 4² = 16 • log8 ⅛ = –1 • 8–1 = ⅛**c) 41/3 =**log4 = ⅓ Writing Logarithmic Functions Example: Write each of the following as a logarithmic equation. a) 54 = 625 log5 625 = 4 b) 2–3 = ⅛ log2 ⅛ = –3**b) log5**• Since 5–2 = , then log5 = –2 Values of Logarithmic Expressions Example: Find the value of each of the following logarithmic expressions. • a) log2 32 • Since 25 = 32, then log2 32 = 5 • c) log4 2 • Since 4½ = 2, then log4 2= ½**Solving Logarithmic Equations**Example: Solve log3 1 = x for x. • First we rewrite the equation as an exponential equation. • 3x = 1 • Since 30 = 1, then x = 0.**Solving Logarithmic Equations**Example: Solve logx 81 = 4 for x. • First we rewrite the equation as an exponential equation. • x4 = 81 • Since 34 = 81, then x = 3.

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