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Ch. 2 - Functions and Graphs
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 2 - Functions and GraphsProblem 49a
Chapter 3, Problem 49a

Find ƒ+g, f−g, fg, and f/g. Determine the domain for each function. f(x) = √(x -2), g(x) = √(2-x)

Verified step by step guidance
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Step 1: Understand the problem. We are tasked with finding the sum (ƒ+g), difference (ƒ−g), product (ƒg), and quotient (ƒ/g) of the two functions ƒ(x) = √(x - 2) and g(x) = √(2 - x). Additionally, we need to determine the domain for each resulting function.
Step 2: Find ƒ+g. The sum of the two functions is given by (ƒ+g)(x) = ƒ(x) + g(x). Substituting the given functions, we have (ƒ+g)(x) = √(x - 2) + √(2 - x). To determine the domain, both square roots must be defined, meaning the expressions inside the square roots must be non-negative. Solve x - 2 ≥ 0 and 2 - x ≥ 0 to find the domain.
Step 3: Find ƒ−g. The difference of the two functions is given by (ƒ−g)(x) = ƒ(x) - g(x). Substituting the given functions, we have (ƒ−g)(x) = √(x - 2) - √(2 - x). The domain is the same as in Step 2, as it depends on the square root expressions being defined.
Step 4: Find ƒg. The product of the two functions is given by (ƒg)(x) = ƒ(x) * g(x). Substituting the given functions, we have (ƒg)(x) = √(x - 2) * √(2 - x). The domain is again determined by ensuring both square roots are defined, as in Step 2.
Step 5: Find ƒ/g. The quotient of the two functions is given by (ƒ/g)(x) = ƒ(x) / g(x). Substituting the given functions, we have (ƒ/g)(x) = √(x - 2) / √(2 - x). In addition to the domain restrictions from Step 2, we must also ensure that the denominator √(2 - x) ≠ 0. Solve 2 - x ≠ 0 to refine the domain.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Function Operations

Function operations involve combining two functions through addition, subtraction, multiplication, and division. For functions f and g, these operations are defined as (f + g)(x) = f(x) + g(x), (f - g)(x) = f(x) - g(x), (fg)(x) = f(x) * g(x), and (f/g)(x) = f(x) / g(x), provided that g(x) is not zero.
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Multiplying & Dividing Functions

Domain of a Function

The domain of a function is the set of all possible input values (x-values) for which the function is defined. For functions involving square roots, the expression inside the square root must be non-negative. Therefore, determining the domain requires solving inequalities to find the valid x-values for each function.
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Domain Restrictions of Composed Functions

Square Root Functions

Square root functions, such as f(x) = √(x - 2) and g(x) = √(2 - x), are defined only for values that make the expression under the square root non-negative. This means that for f(x), x must be greater than or equal to 2, while for g(x), x must be less than or equal to 2. Understanding these constraints is crucial for determining the domain of the combined functions.
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Related Practice
Textbook Question

In Exercises 31–50, find ƒ+g, f−g, fg, and f/g. Determine the domain for each function. f(x) = √(x -2), g(x) = √(2-x)

Textbook Question

Graph each equation in a rectangular coordinate system. y = -2

Textbook Question

In Exercises 39-52, a. Find an equation for ƒ¯¹(x). b. Graph ƒ and ƒ¯¹(x) in the same rectangular coordinate system. c. Use interval notation to give the domain and the range off and ƒ¯¹. f(x) = √(x-1)

Textbook Question

In Exercises 41–52, give the center and radius of the circle described by the equation and graph each equation. Use the graph to identify the relation's domain and range. x² + (y − 1)² = 1

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Textbook Question

In Exercises 46–49, give the slope and y-intercept of each line whose equation is given. Then graph the line. 2x + 3y + 6 = 0

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Textbook Question

In Exercises 31–50, find ƒ+g, f−g, fg, and f/g. Determine the domain for each function. f(x) = √(x -2), g(x) = √(2-x)