Copyright © 2013, 2009, 2005 Pearson Education, Inc. Graphs and Functions Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1
2.8 Function Operations and Composition Arithmetic Operations on Functions The Difference Quotient Composition of Functions and Domain
Operations on Functions and Domains Given two functions and g, then for all values of x for which both (x) and g(x) are defined, the functions + g, – g, g, and are defined as follows. Sum Difference Product Quotient
Domains For functions and g, the domains of + g, – g, and g include all real numbers in the intersection of the domains of and g, while the domain of includes those real numbers in the intersection of the domains of and g for which g(x) ≠ 0.
Note The condition g(x) ≠ 0 in the definition of the quotient means that the domain of (x) is restricted to all values of x for which g(x) is not 0. The condition does not mean that g(x) is a function that is never 0.
Let (x) = x2 + 1 and g(x) = 3x + 5. Find each of the following. USING OPERATIONS ON FUNCTIONS Example 1 Let (x) = x2 + 1 and g(x) = 3x + 5. Find each of the following. (a) Solution First determine (1) = 2 and g(1) = 8. Then use the definition.
Let (x) = x2 + 1 and g(x) = 3x + 5. Find each of the following. USING OPERATIONS ON FUNCTIONS Example 1 Let (x) = x2 + 1 and g(x) = 3x + 5. Find each of the following. (b) Solution First determine that (– 3) = 10 and g(– 3) = – 4. Then use the definition.
Let (x) = x2 + 1 and g(x) = 3x + 5. Find each of the following. USING OPERATIONS ON FUNCTIONS Example 1 Let (x) = x2 + 1 and g(x) = 3x + 5. Find each of the following. (c) Solution
Let (x) = x2 + 1 and g(x) = 3x + 5. Find each of the following. USING OPERATIONS ON FUNCTIONS Example 1 Let (x) = x2 + 1 and g(x) = 3x + 5. Find each of the following. (d) Solution
Let Find each function. (a) Solution Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let Find each function. (a) Solution
Let Find each function. (b) Solution Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let Find each function. (b) Solution
Let Find each function. (c) Solution Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let Find each function. (c) Solution
Let Find each function. (d) Solution Example 2 USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let Find each function. (d) Solution
(e) Give the domains of the functions in parts (a)-(d). USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let Find each function. (e) Give the domains of the functions in parts (a)-(d). Solution To find the domains of the functions, we first find the domains of and g. The domain of is the set of all real numbers (– , ). Because g is defined by a square root radical, the radicand must be non-negative (that is, greater than or equal to 0).
(e) Give the domains of the functions in parts (a)-(d). USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let Find each function. (e) Give the domains of the functions in parts (a)-(d). Solution Thus, the domain of g is
(e) Give the domains of the functions in parts (a)-(d). USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let Find each function. (e) Give the domains of the functions in parts (a)-(d). Solution The domains of + g, – g, g are the intersection of the domains of and g, which is
(e) Give the domains of the functions in parts (a)-(d). USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let Find each function. (e) Give the domains of the functions in parts (a)-(d). Solution The domain of includes those real numbers in the intersection of the domains for which That is, the domain of is
EVALUATING COMBINATIONS OF FUNCTIONS Example 3 If possible, use the given representations of functions and g to evaluate
(a) EVALUATING COMBINATIONS OF FUNCTIONS Example 3 For ( – g)(–2), although (–2) = – 3, g(–2) is undefined because –2 is not in the domain of g.
(a) EVALUATING COMBINATIONS OF FUNCTIONS Example 3 The domains of and g include 1, so The graph of g includes the origin, so Thus, is undefined.
EVALUATING COMBINATIONS OF FUNCTIONS Example 3 If possible, use the given representations of functions and g to evaluate (b) x (x) g(x) – 2 – 3 undefined 1 3 4 9 2 In the table, g(– 2) is undefined. Thus, ( – g)(– 2) is undefined.
EVALUATING COMBINATIONS OF FUNCTIONS Example 3 If possible, use the given representations of functions and g to evaluate (b) x (x) g(x) – 2 – 3 undefined 1 3 4 9 2
EVALUATING COMBINATIONS OF FUNCTIONS Example 3 If possible, use the given representations of functions and g to evaluate (c) Using we can find (f + g)(4) and (fg)(1). Since –2 is not in the domain of g, (f – g)(–2) is not defined.
Solution We use a three-step process. FINDING THE DIFFERENCE QUOTIENT Example 4 Let (x) = 2x2 – 3x. Find and simplify the expression for the difference quotient, Solution We use a three-step process. Step 1 Find the first term in the numerator, (x + h). Replace x in (x) with x + h.
Remember this term when squaring x + h FINDING THE DIFFERENCE QUOTIENT Example 4 Let (x) = 2x2 – 3x. Find and simplify the expression for the difference quotient, Solution Step 2 Find the entire numerator Substitute Remember this term when squaring x + h Square x + h
FINDING THE DIFFERENCE QUOTIENT Example 4 Let (x) = 2x2 – 3x. Find and simplify the expression for the difference quotient, Solution Step 2 Distributive property Combine like terms.
Step 3 Find the difference quotient by dividing by h. FINDING THE DIFFERENCE QUOTIENT Example 4 Let (x) = 2x2 – 3x. Find and simplify the expression for the difference quotient, Solution Step 3 Find the difference quotient by dividing by h. Substitute. Factor out h. Divide.
Caution In Example 4, notice that the expression (x + h) is not equivalent to (x) + (h). These expressions differ by 4xh. In general, (x + h) is not equivalent to (x) + (h).
Composition of Functions and Domain If and g are functions, then the composite function, or composition, of g and is defined by The domain of is the set of all numbers x in the domain of such that (x) is in the domain of g.
Solution First find g(2): EVALUATING COMPOSITE FUNCTIONS Example 5 Let (x) = 2x – 1 and g(x) (a) Solution First find g(2): Now find
Let (x) = 2x – 1 and g(x) (b) Solution EVALUATING COMPOSITE FUNCTIONS Example 5 Let (x) = 2x – 1 and g(x) (b) Solution
Find each of the following. DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS Example 6 Given that Find each of the following. (a) Solution The domain and range of g are both the set of real numbers. The domain of f is the set of all nonnegative real numbers. Thus, g(x), which is defined as 4x + 2, must be greater than or equal to zero.
Find each of the following. DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS Example 6 Given that Find each of the following. (a) Solution Therefore, the domain of
Find each of the following. DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS Example 6 Given that Find each of the following. (b) Solution The domain and range of f are both the set of all nonnegative real numbers. The domain of g is the set of all real numbers. Therefore, the domain of
DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS Example 7 (a) Solution Multiply the numerator and denominator by x.
DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS Example 7 (a) Solution The domain of g is all real numbers except 0, which makes g(x) undefined. The domain of is all real numbers except 3. The expression for g(x), therefore, cannot equal 3. We determine the value that makes g(x) = 3 and exclude it from the domain of
DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS Example 7 (a) Solution The solution must be excluded. Multiply by x. Divide by 3.
DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS Example 7 (a) Solution Therefore the domain of is the set of all real numbers except 0 and 1/3, written in interval notation as
DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS Example 7 (b) Solution Note that this is meaningless if x = 3
DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS Example 7 (b) Solution The domain of is all real numbers except 3, and the domain of g is all real numbers except 0. The expression for (x), which is , is never zero, since the numerator is the nonzero number 6. Therefore, the domain of is the set of all real numbers except 3, written
Let (x) = 4x + 1 and g(x) = 2x2 + 5x. SHOWING THAT IS NOT EQUIVALENT TO Example 8 Let (x) = 4x + 1 and g(x) = 2x2 + 5x. Solution Square 4x + 1; distributive property.
Let (x) = 4x + 1 and g(x) = 2x2 + 5x. SHOWING THAT IS NOT EQUIVALENT TO Example 8 Let (x) = 4x + 1 and g(x) = 2x2 + 5x. Solution Distributive property. Combine like terms.
Let (x) = 4x + 1 and g(x) = 2x2 + 5x. SHOWING THAT IS NOT EQUIVALENT TO Example 8 Let (x) = 4x + 1 and g(x) = 2x2 + 5x. Solution Distributive property
Find functions and g such that FINDING FUNCTIONS THAT FORM A GIVEN COMPOSITE Example 9 Find functions and g such that Solution Note the repeated quantity x2 – 5. If we choose g(x) = x2 – 5 and (x) = x3 – 4x + 3, then There are other pairs of functions and g that also work.