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Ch. 2 - Functions and Graphs
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBlitzer 8th EditionCh. 2 - Functions and GraphsProblema 49a
Capitolo 3, Problema 49a

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

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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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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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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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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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Imaginary Roots with the Square Root Property
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