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**Difference, Product and Quotient Functions**

Combining Functions Sum, Difference, Product, and Quotient Functions Combining Functions Once simple functions are defined, more complex functions can be constructed from them by combing then in various ways. The first set of combinations are simply the arithmetic operations that we normally use on numbers. The next area we look into is the composition of functions, that is, considering a function of another function. This leads eventually to discussion of inverse functions. Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

The important thing is not to stop questioning Albert Einstein Einstein’s words are typical of the scientist – and of the mathematician. In 1900 the population of the world was approximately 1.6 billion. At that time the Swedish scientist Svante Arrhenius first predicted a greenhouse effect resulting from emissions of carbon dioxide by the industrialized countries. His classic calculation made use of logarithms and predicted that a doubling of the carbon dioxide concentration in the atmosphere would raise the average global temperature by 7F to 11F. Since then, the world population has increased dramatically to approximately 6 billion. With this increase in population, there has been a corresponding increase in emissions of greenhouse gases such as carbon dioxide, methane , and chlorofluorocarbons. Theoretically, these emissions have created the potential to alter the earth’s climate and destroy portions of the ozone layer – or so the theory says. When quantities such as population and pollution increase rapidly, nonlinear functions and equations are used to model their growth. In this chapter, exponential and logarithmic functions are introduced. They occur in a wide variety of applications. Examples include acid rain, the decline of the bluefin tuna, air pollution, global warming, salinity of the ocean, demand for organ transplants, diversity of bird species, the relationship between caloric intake and land ownership in developing countries, hurricanes, earthquakes, and increased skin cancer due to decrease in the ozone layer. Understanding these issues and discovering solutions will require creativity, innovation, and mathematics. This chapter presents some of the mathematical tools necessary for modeling these trends. 10/9/2013 Combining Functions 2 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions Domain f Range f Range of f(x) + g(x) and (f+g)(x) f f(x) g x + f(x)+g(x) = (f+g)(x) g(x) New Functions from Old: Addition By combining existing functions in a variety of ways, we can create new functions from the old ones. We first illustrate functions created through use of the basic arithmetic operations of addition, subtraction, multiplication, and division. We define a new function, f+g, created from f and g. The intuitive formulation is simply the sum of f(x) and g(x), which means both f and g are defined at x. That is, x is the domain of both f and g, i.e. in the common domain of f and g. Function f and g may or may not have the same range. The illustration shows a common range as the intersection of the ranges of f and g. However, the ranges of f and g may be completely disjoint, i.e. may have no points in common. In that case, the range of f+g is the union of the ranges of f and g. Domain g Range g h = f+g 10/9/2013 Combining Functions 3 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions How can we find the domain of f+g ? Domain of f Ranges of f and g f g(y) does not exist !! y g f(y) ? WHY ? So f(y) + g(y) does not exist !! x How Can We Find the Domain of f+g ? The illustration shows a value x in the domain of f but NOT in the domain of g. Hence, while f(x) exists in the range of f, g(x) does NOT EXIST ! This means that we cannot add f(x) and g(x). So, the sum f(x) + g(x) cannot exist. Since we define (f+g)(x) to be f(x) + g(x), then (f+g)(x) cannot exist either. The upshot here is that (f+g)(x) exists only for x in the common domain of f and g. Since f(x) does exist, then f(x) = a, for some value a. Thus the ordered pair (x, a) lies in the set representing f as a function. But there is no corresponding ordered pair (x, b) in the set representing the function g, since g(x) does not exist. As the illustration shows, the same thing happens if x is chosen in the domain of g but not in the domain of f. Then g(x) exists and f(x) does not. Hence (f+g)(x) does not exist for the given value of x. ... and neither does (f+g)(y) Domain of g ... since (f+g)(y) = f(y) + g(y) 10/9/2013 Combining Functions 4 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions How can we find the domain of f+g ? Domain of f Ranges of f and g f f(z) does not exist !! y g f(y) ? WHY ? So f(z) + g(z) does not exist !! x g g(z) ? z f How Can We Find the Domain of f+g ? The illustration shows a value x in the domain of f but NOT in the domain of g. Hence, while f(x) exists in the range of f, g(x) does NOT EXIST ! This means that we cannot add f(x) and g(x). So, the sum f(x) + g(x) cannot exist. Since we define (f+g)(x) to be f(x) + g(x), then (f+g)(x) cannot exist either. The upshot here is that (f+g)(x) exists only for x in the common domain of f and g. Since f(x) does exist, then f(x) = a, for some value a. Thus the ordered pair (x, a) lies in the set representing f as a function. But there is no corresponding ordered pair (x, b) in the set representing the function g, since g(x) does not exist. As the illustration shows, the same thing happens if x is chosen in the domain of g but not in the domain of f. Then g(x) exists and f(x) does not. Hence (f+g)(x) does not exist for the given value of x. ... and neither does (f+g)(z) Domain of g ... since (f+g)(z) = f(z) + g(z) Combining Functions 5 10/9/2013 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions How can we find the domain of f+g ? Domain of f Ranges of f and g (f+g)(x) does exist !! y WHY ? f+g (f+g)(x) Both f(x) and g(x) do exist !! x g g(z) ? How Can We Find the Domain of f+g ? The illustration shows a value x in the domain of f but NOT in the domain of g. Hence, while f(x) exists in the range of f, g(x) does NOT EXIST ! This means that we cannot add f(x) and g(x). So, the sum f(x) + g(x) cannot exist. Since we define (f+g)(x) to be f(x) + g(x), then (f+g)(x) cannot exist either. The upshot here is that (f+g)(x) exists only for x in the common domain of f and g. Since f(x) does exist, then f(x) = a, for some value a. Thus the ordered pair (x, a) lies in the set representing f as a function. But there is no corresponding ordered pair (x, b) in the set representing the function g, since g(x) does not exist. As the illustration shows, the same thing happens if x is chosen in the domain of g but not in the domain of f. Then g(x) exists and f(x) does not. Hence (f+g)(x) does not exist for the given value of x. z f ... so f(x) + g(x) does exist Domain of g ... so (f+g)(x) = f(x) + g(x) Combining Functions 6 10/9/2013 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions How can we find the domain of f+g ? Domain of f Ranges of f and g y (f+g)(x) exists only if x lies in the domain of both f and g f+g (f+g)(x) x How Can We Find the Domain of f+g ? The illustration shows a value x in the domain of f but NOT in the domain of g. Hence, while f(x) exists in the range of f, g(x) does NOT EXIST ! This means that we cannot add f(x) and g(x). So, the sum f(x) + g(x) cannot exist. Since we define (f+g)(x) to be f(x) + g(x), then (f+g)(x) cannot exist either. The upshot here is that (f+g)(x) exists only for x in the common domain of f and g. Since f(x) does exist, then f(x) = a, for some value a. Thus the ordered pair (x, a) lies in the set representing f as a function. But there is no corresponding ordered pair (x, b) in the set representing the function g, since g(x) does not exist. As the illustration shows, the same thing happens if x is chosen in the domain of g but not in the domain of f. Then g(x) exists and f(x) does not. Hence (f+g)(x) does not exist for the given value of x. z Domain of g Dom f+g = Dom f ∩ Dom g Combining Functions 7 10/9/2013 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions Given functions f(x) and g(x) Define functions (f+g)(x) and (f–g)(x) for all x in the domains of BOTH f and g as (f+g)(x) = f(x) + g(x) (f–g)(x) = f(x) – g(x) Combining Functions: Examples Here we introduce a second combined function, f–g, defined as before in the intuitive way: (f–g)(x) = f(x) – g(x) . As before, this new function is defined only where both f and g are defined. That is, (f–g)(x) exists only where x is in the common domain of f and g. Similarly, we define the product function, fg and the quotient function f/g, on the common domain of f and g, with the additional caveat that the quotient exists only where g(x) ≠ 0. Once again we emphasize the fact that the combined functions are defined only for values of x that lie in the domain of both f and g. Remember: Dom f+g = Dom f–g = Dom f ∩ Dom g Combining Functions 8 10/9/2013 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions Given functions f(x) and g(x) Define function (fg)(x) for all x in the domain of BOTH f and g as (fg)(x) = f(x) • g(x) = f(x)g(x) Combining Functions: Examples Here we introduce a second combined function, f–g, defined as before in the intuitive way: (f–g)(x) = f(x) – g(x) . As before, this new function is defined only where both f and g are defined. That is, (f–g)(x) exists only where x is in the common domain of f and g. Similarly, we define the product function, fg and the quotient function f/g, on the common domain of f and g, with the additional caveat that the quotient exists only where g(x) ≠ 0. Once again we emphasize the fact that the combined functions are defined only for values of x that lie in the domain of both f and g. Remember: Dom fg = Dom f ∩ Dom g Combining Functions 9 10/9/2013 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions Given functions f(x) and g(x) Define function (f/g)(x) for all x in the domain of BOTH f and g, with g(x) ≠ 0, as (f/g)(x) = f(x)/g(x) Combining Functions: Examples Here we introduce a second combined function, f–g, defined as before in the intuitive way: (f–g)(x) = f(x) – g(x) . As before, this new function is defined only where both f and g are defined. That is, f(–g)(x) exists only where x is in the common domain of f and g. Similarly, we define the product function, fg and the quotient function f/g, on the common domain of f and g, with the additional caveat that the quotient exists only where g(x) ≠ 0. Once again we emphasize the fact that the combined functions are defined only for values of x that lie in the domain of both f and g. Remember: Dom f/g = Dom f ∩ Dom g … provided g(x) ≠ 0 Combining Functions 10 10/9/2013 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions Examples 1. f(x) = 1 – x and g(x) = x Domain f(x) = ? Domain g(x) = ? { x | x ≤ 1 } { x | 0 ≤ x } (f+g)(x) = f(x) + g(x) 1 – x x + = Dom (f+g) = { x | x ≤ 1 } ∩ { x | 0 ≤ x } Combining Functions: Examples These two examples illustrate the dependence of the combined function on the domains of the original component functions and their domains. No value of x which is not in the domain of both f and g can be used to build the ordered pairs of the combined functions f+g. In Example 1 we see clearly that the intersection of the domains of f and g forms the only pool of values available to f+g. That is, the domain of f+g is contained in the intersection of the domains of f and g. In Example 2, this is less obvious, since the combined function seems to have a form, namely (f+g)(x) = 2x, which could have a domain larger than the domains of either f or g. This is an illusion, since the values of x available for use in f(x) and g(x) are all non-negative numbers. Hence, no negative number can be used as an input to f+g, even though its ultimate form is 2x. Clearly, combining functions can create a new function with a smaller domain, but can never create one with a larger domain. Another way to view the combined function is as the set of ordered pairs whose second components are the sums of the second components in f and g. That is, if and then f+g = { (x, y + z) │ (x, y) f , (x, z) g } . Clearly if either (x, y) is not in f or (x, z) is not in g, then (x, y + z) is not in f+g. = { x | 0 ≤ x ≤ 1 } = [ 0, 1 ] Question: Can addition make the domain bigger ? 10/9/2013 Combining Functions 11 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions Examples 2. f(x) = x + x and x – x g(x) = Domain f(x) = ? Domain g(x) = ? { x | 0 ≤ x } { x | 0 ≤ x } (f+g)(x) = f(x) + g(x) = 2x Domain ? Dom (f+g) = { x | 0 ≤ x } ∩ { x | 0 ≤ x } Combining Functions: Examples These two examples illustrate the dependence of the combined function on the domains of the original component functions and their domains. No value of x which is not in the domain of both f and g can be used to build the ordered pairs of the combined functions f+g. In Example 1 we see clearly that the intersection of the domains of f and g forms the only pool of values available to f+g. That is, the domain of f+g is contained in the intersection of the domains of f and g. In Example 2, this is less obvious, since the combined function seems to have a form, namely (f+g)(x) = 2x, which could have a domain larger than the domains of either f or g. This is an illusion, since the values of x available for use in f(x) and g(x) are all non-negative numbers. Hence, no negative number can be used as an input to f+g, even though its ultimate form is 2x. Clearly, combining functions can create a new function with a smaller domain, but can never create one with a larger domain. Another way to view the combined function is as the set of ordered pairs whose second components are the sums of the second components in f and g. That is, if and then f+g = { (x, y + z) │ (x, y) f , (x, z) g } . Clearly if either (x, y) is not in f or (x, z) is not in g, then (x, y + z) is not in f+g. = [ 0, ) ∞ ≠ R Question: Can addition make the domain bigger ? Combining Functions 12 10/9/2013 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions Examples 3. f(x) = 1 – x and g(x) = x Domain f(x) = ? Domain g(x) = ? { x | x ≤ 1 } { x | 0 ≤ x } ( 1 – x x ) • ( ) = (fg)(x) = f(x) • g(x) Combining Functions: Examples These examples are intended to drive home the point that, when combining functions, the domain of the newly formed function is always a subset of the intersection of the domains of the component functions. That is to say, the domain of the new function can never be larger than the domains of the component functions, but it can be smaller because of the form of the new function. The reason for this is easily seen by considering the set of ordered pairs that make up the component functions. No value of the independent variable (input) not already in those ordered pairs can be used in ordered pairs of the newly formed function. In Example 3, this means that no x greater than 1 can be used by f, and no x less than 0 can be used by g. Thus no x outside [ 0, 1 ] can be used by fg, that is, Dom fg = { x │ x Dom f Dom g } = [ 0, 1 ] . In Example 4, we find an additional restriction on the domain of f/g by noting that the expression for (f/g)(x) makes sense only if g(x) ≠ 0. If we form the new function (f/g)(x) = 2x and if we view the function as simply a “black box” with a formula, ignoring the domain, we might be tempted to think that the new function f/g has the set of all real numbers as its domain. This is a false assumption, since neither f nor g can produce a functional value for any negative number, and additionally (f/g)(x) does not exist for g(x) = 0, that is, for x = 0. Hence Dom f/g = { x │ x Dom f Dom g ; x ≠ 0 } = { x │ 0 < x }. Once again, the domain of the combined function can be smaller than the domains of the component functions but not larger. Dom (fg)(x) = Dom f ∩ Dom g = { x | 0 ≤ x ≤ 1 } = [ 0, 1 ] Question: Can the domain get bigger ? Smaller ? Combining Functions 13 10/9/2013 Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Combining Functions Difference, Product and Quotient Functions Examples 4. f(x) = x + x and x – x g(x) = Domain f(x) Domain g(x) = { x | 0 ≤ x } = { x | 0 ≤ x } (f/g)(x) x + x x – x = Combining Functions: Examples These examples are intended to drive home the point that, when combining functions, the domain of the newly formed function is always a subset of the intersection of the domains of the component functions. That is to say, the domain of the new function can never be larger than the domains of the component functions, but it can be smaller because of the form of the new function. The reason for this is easily seen by considering the set of ordered pairs that make up the component functions. No value of the independent variable (input) not already in those ordered pairs can be used in ordered pairs of the newly formed function. In Example 3, this means that no x greater than 1 can be used by f, and no x less than 0 can be used by g. Thus no x outside [ 0, 1 ] can be used by fg, that is, Dom fg = { x │ x Dom f Dom g } = [ 0, 1 ] . In Example 4, we find an additional restriction on the domain of f/g by noting that the expression for (f/g)(x) makes sense only if g(x) ≠ 0. If we form the new function (f/g)(x) = 2x and if we view the function as simply a “black box” with a formula, ignoring the domain, we might be tempted to think that the new function f/g has the set of all real numbers as its domain. This is a false assumption, since neither f nor g can produce a functional value for any negative number, and additionally (f/g)(x) does not exist for g(x) = 0, that is, for x = 0. Hence Dom f/g = { x │ x Dom f Dom g ; x ≠ 0 } = { x │ 0 < x }. Once again, the domain of the combined function can be smaller than the domains of the component functions but not larger. Domain (f/g)(x) = Dom f ∩ Dom g; g(x) ≠ 0 = { x | 0 < x } Question: Can the domain get bigger ? Smaller ? 10/9/2013 Combining Functions Combining Functions 10/9/2013

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**Difference, Product and Quotient Functions**

Think about it ! 10/9/2013 Combining Functions 15 Combining Functions 10/9/2013

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