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Objective To perform operations on functions and to determine the domains of the resulting functions.

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Presentation on theme: "Objective To perform operations on functions and to determine the domains of the resulting functions."— Presentation transcript:

1 Objective To perform operations on functions and to determine the domains of the resulting functions.

2 In real life, we need consider the profit function in terms of the item we made and sold. In fact, the profit function is the difference between the sales income and the cost in terms of the item we made and sold. If let x denote the number of item we made and sold, then the profit function can be expressed as P(x) = S(x) – C(x) In real life example, we can see that we do have the demand to consider the difference between the two functions

3 The sum f + g This just says that to find the sum of two functions, add them together. You should simplify by finding like terms. Combine like terms & put in power descending order

4 The difference f - g To find the difference between two functions, subtract the first from the second. CAUTION: Make sure you distribute the – to each term of the second function. You should simplify by combining like terms. Distribute negative

5 The product f g To find the product of two functions, put parenthesis around them and multiply each term from the first function to each term of the second function. Distributive Property Good idea to put in power descending order but not required.

6 The quotient f /g To find the quotient of two functions, put the first one over the second. Nothing more you could do here. (If you can reduce these you should).

7 Note that the domain of the resulting functions of the first 4 operations is determined at the set-up time, not the simplified time. Example 1 Let. Find the resulting function and domain.

8 So the first 4 operations on functions are pretty straight forward. The rules for the domain of functions would apply to these combinations of functions as well. The domain of the sum, difference or product would be the numbers x in the domains of both f and g. For the quotient, you would also need to exclude any numbers x that would make the resulting denominator 0. In one word, the domain of the resulting functions of the first 4 operations is determined at the set-up time, not the simplified time.

9 COMPOSITION OF FUNCTIONS “SUBSTITUTING ONE FUNCTION INTO ANOTHER”

10 The Composition Function This is read “f composition g” and means to copy the f function down but where ever you see an x, substitute in the g function. Square first and then distribute the 2

11 This is read “g composition f” and means to copy the g function down but where ever you see an x, substitute in the f function. You could multiply this out but since it’s to the 3 rd power we won’t

12 This is read “f composition f” and means to copy the f function down but where ever you see an x, substitute in the f function. (So sub the function into itself).

13 The DOMAIN of the Composition Function The domain of f composition g is the set of all numbers x in the domain of g such that g(x) is in the domain of f. The domain of g is x  1 We also have to worry about any “illegals” in this composition function, specifically dividing by 0. This would mean that x  1 so the domain of the composition would be combining the two restrictions.

14 The DOMAIN and RANGE of Composite Functions We could first look at the domain and range of f(x) and g(x). Hence we must exclude 6 from the domain of f(x) For g(x) to cope with the output from f(x) we must ensure that the output of f(x) does not include 1

15 The DOMAIN and RANGE of Composite Functions Hence we must exclude 6 from the domain of f(x) Or we could find g o f (x) and determine the domain and range of the resulting expression.

16 The DOMAIN and RANGE of Composite Functions We could first look at the domain and range of f(x) and g(x). Hence we must exclude 1 from the domain of g(x) For f(x) to cope with the output from g(x) we must ensure that the output does not include 0

17 The DOMAIN and RANGE of Composite Functions Or we could find f o g (x) and determine the domain and range of the resulting expression. Domain:Range:

18 The DOMAIN and RANGE of Composite Functions We could first look at the domain and range of f(x) and g(x).

19 The DOMAIN and RANGE of Composite Functions Or we could find g o f (x) and determine the domain and range of the resulting expression. However, the approach that finding domain and range directly from the expression of the composition must be used with CAUTION. Domain:Range: Note that the domain of the resulting functions is determined at the set-up time, not the simplified time. Not:

20 The DOMAIN and RANGE of Composite Functions We could first look at the domain and range of f(x) and g(x). f o g (x) is a function for the natural domain of g(x) f(x) can cope with all the numbers in the range of g(x) because the range of g(x) is contained within the domain of f(x)

21 The DOMAIN and RANGE of Composite Functions We could first look at the domain and range of f(x) and g(x). g(x) cannot cope with all the numbers in the range of f(x). Need to restrict the domain f(x) to give an output that g(x) can cope with.

22 The DOMAIN and RANGE of Composite Functions We could first look at the domain and range of f(x) and g(x). g o f (x) is not a function for the domain of g(x) unless we restrict the domain of f(x) g(x) cannot cope with all the numbers in the range of f(x). Need to restrict the domain f(x) to give an output that g(x) can cope with.

23 Example 2. Let Find a rule forAnd give the domain of the composite function. The domain ofeven though the expression is also defined for x < 2.

24 Example 3. Let Find rules for and give the domain of each composite function.

25 A Challenge Example of Composite Functions Try it !!

26 Method 1 Complete the square with (x – 5)

27 Method 1

28 Method 2

29 Of the Method 1 and Method 2, the Method 2 is BETTER!!!

30 Assignment P.123 #18; P. 128 #1 – 25 (odd); WS 4.1

31 The DOMAIN and RANGE of Composite Functions Or we could find g o f (x) and determine the domain and range of the resulting expression. However this approach must be used with CAUTION. Domain:Range:


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