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**LIAL HORNSBY SCHNEIDER**

COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER

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**2.8 Function Operations and Composition**

Arithmetic Operations on Functions The Difference Quotient Composition of Functions and Domain

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**Operations of Functions**

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 /g are defined as follows. Sum Difference Product Quotient

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Note The condition g(x) ≠ 0 in the definition of the quotient means that the domain of (/g)(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.

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**Let (x) = x2 + 1 and g(x) = 3x + 5. Find the following.**

USING OPERATIONS ON FUNCTIONS Example 1 Let (x) = x2 + 1 and g(x) = 3x Find the following. a. Solution Since (1) = 2 and g(1) = 8, use the definition to get

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**Let (x) = x2 + 1 and g(x) = 3x + 5. Find the following.**

USING OPERATIONS ON FUNCTIONS Example 1 Let (x) = x2 + 1 and g(x) = 3x Find the following. b. Solution Since (– 3) = 10 and g(– 3) = – 4, use the definition to get

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**Let (x) = x2 + 1 and g(x) = 3x + 5. Find the following.**

USING OPERATIONS ON FUNCTIONS Example 1 Let (x) = x2 + 1 and g(x) = 3x Find the following. c. Solution Since (5) = 26 and g(5) = 20, use the definition to get

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**Let (x) = x2 + 1 and g(x) = 3x + 5. Find the following.**

USING OPERATIONS ON FUNCTIONS Example 1 Let (x) = x2 + 1 and g(x) = 3x Find the following. d. Solution Since (0) = 1 and g(0) = 5, use the definition to get

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y Domains For functions and g, the domains of + g, – g, and g include all real numbers in the intersections of the domains of and g, while the domain of /g includes those real numbers in the intersection of the domains of and g for which g(x) ≠ 0.

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**USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS**

Example 2 Let a. Solution

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**USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS**

Example 2 Let b. Solution

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**USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS**

Example 2 Let c. Solution

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**USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS**

Example 2 Let d. Solution

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**e. Give the domains of the functions.**

USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let e. Give the domains of the functions. 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 (– , ).

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**e. Give the domains of the functions.**

USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let e. Give the domains of the functions. Solution Since , the domain of g includes just the real numbers that make 2x – 1 nonnegative. Solve 2x – 1 0 to get x ½ . The domain of g is

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**e. Give the domains of the functions.**

USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let e. Give the domains of the functions. Solution The domains of + g, – g, g are the intersection of the domains of and g, which is

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**e. Give the domains of the functions.**

USING OPERATIONS OF FUNCTIONS AND DETERMINING DOMAINS Example 2 Let e. Give the domains of the functions. Solution The domains of includes those real numbers in the intersection for which that is, the domain of is

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**EVALUATING COMBINATIONS OF FUNCTIONS**

Example 3 If possible, use the given representations of functions and g to evaluate …

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**a. EVALUATING COMBINATIONS OF FUNCTIONS Example 3**

y 9 a. 5 For ( – g)(–2),although (–2) = – 3, g(–2) is undefined because –2 is not in the domain of g. x – 4 – 2 2 4

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**a. EVALUATING COMBINATIONS OF FUNCTIONS Example 3**

y 9 a. 5 The domains of and g include 1, so x – 4 – 2 2 4

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**a. EVALUATING COMBINATIONS OF FUNCTIONS Example 3**

y 9 a. 5 The graph of g includes the origin, so x – 4 – 2 2 4 Thus, is undefined.

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**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.

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**EVALUATING COMBINATIONS OF FUNCTIONS**

Example 3 If possible, use the given representations of functions and g to evaluate b. x (x) h(x) – 2 – 3 undefined 1 3 4 9 2

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**EVALUATING COMBINATIONS OF FUNCTIONS**

Example 3 If possible, use the given representations of functions and g to evaluate b. x (x) h(x) – 2 – 3 undefined 1 3 4 9 2 and

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**EVALUATING COMBINATIONS OF FUNCTIONS**

Example 3 If possible, use the given representations of functions and g to evaluate c.

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**EVALUATING COMBINATIONS OF FUNCTIONS**

Example 3 c.

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**FINDING THE DIFFERENCE QUOTIENT**

Example 4 Let (x) = 2x2 – 3x. Find the difference quotient and simplify the expression. Solution Step 1 Find the first term in the numerator, (x + h). Replace the x in (x) with x + h.

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**Remember this term when squaring x + h**

FINDING THE DIFFERENCE QUOTIENT Example 4 Let (x) = 2x – 3x. Find the difference quotient and simplify the expression. Solution Step 2 Find the entire numerator Substitute Remember this term when squaring x + h Square x + h

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**Step 2 Find the entire numerator**

FINDING THE DIFFERENCE QUOTIENT Example 4 Let (x) = 2x – 3x. Find the difference quotient and simplify the expression. Solution Step 2 Find the entire numerator Distributive property Combine terms.

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**Step 3 Find the quotient by dividing by h.**

FINDING THE DIFFERENCE QUOTIENT Example 4 Let (x) = 2x – 3x. Find the difference quotient and simplify the expression. Solution Step 3 Find the quotient by dividing by h. Substitute. Factor out h. Divide.

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**Caution Notice that (x + h) is not the same as (x) + (h)**

Caution Notice that (x + h) is not the same as (x) + (h). For (x) = 2x2 – 3x in Example 4. but These expressions differ by 4xh.

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**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.

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**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

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**Don’t confuse composition with multiplication**

EVALUATING COMPOSITE FUNCTIONS Example 5 Let (x) = 2x – 1 and g(x) b. Solution Don’t confuse composition with multiplication

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**a. Solution DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS**

Example 7 a. Solution Multiply the numerator and denominator by x.

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**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

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**a. Solution DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS**

Example 7 a. Solution The solution must be excluded. Multiply by x. Divide by 3.

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**DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS**

Example 7 a. Solution Divide by 3. Therefore the domain of is the set of all real numbers except 0 and ⅓, written in interval notation as

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**b. Solution DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS**

Example 7 b. Solution Note that this is meaningless if x = 3

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**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.

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**Solution Therefore, the domain of **

DETERMINING COMPOSITE FUNCTIONS AND THEIR DOMAINS Example 7 b. Solution Therefore, the domain of is the set of all real numbers except 3, written

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**Let (x) = 4x + 1 and g(x) = 2x2 + 5x.**

SHOWING THAT Example 8 Let (x) = 4x + 1 and g(x) = 2x2 + 5x. Solution Square 4x + 1; distributive property.

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**Let (x) = 4x + 1 and g(x) = 2x2 + 5x.**

SHOWING THAT Example 8 Let (x) = 4x + 1 and g(x) = 2x2 + 5x. Solution Distributive property. Combine terms.

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**Let (x) = 4x + 1 and g(x) = 2x2 + 5x.**

SHOWING THAT Example 8 Let (x) = 4x + 1 and g(x) = 2x2 + 5x. Solution Distributive property

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**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.

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