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Two functions fand g are defined in the figure below. (a) Domain of f (b) 3 Range of f 4 6 7 9 Domain of g g Find the domain and range of the composition gof. Write your answers in set notation. Domain of g of:] Range of g of : ] Range of g --... x 5 ?

Two functions fand g are defined in the figure below. (a) Domain of f (b) 3 Range of f 4 6 7 9 Domain of g g Find the domain and range of the composition gof. Write your answers in set notation. Domain of g of:] Range of g of : ] Range of g --... x 5 ?

College Algebra
College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter3: Functions
Section3.7: Inverse Functions
Problem 1SE: Describe why the horizontal line test is an effective way to determine whether a function is...
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### Composition of Functions: Domain and Range In the figure below, two functions \( f \) and \( g \) are defined. We need to determine the domain and range of the composition \( g \circ f \). #### Function Diagrams **Function \( f \):** - **Domain of \( f \)**: {1, 2, 3, 4, 7} - **Range of \( f \)**: {0, 1, 3, 6} Mappings for \( f \): - \( f(1) = 0 \) - \( f(2) = 1 \) - \( f(3) = 3 \) - \( f(4) = 6 \) - \( f(7) = 6 \) **Function \( g \):** - **Domain of \( g \)**: {0, 1, 4, 6} - **Range of \( g \)**: {7, 3, 9} Mappings for \( g \): - \( g(0) = 7 \) - \( g(1) = 7 \) - \( g(4) = 3 \) - \( g(6) = 9 \) #### Composition \( g \circ f \) To find the domain and range of \( g \circ f \), we need to follow the values mapped by \( f \) and then apply \( g \). **Domain of \( g \circ f \):** - Since \( g \circ f(x) \) is defined only when \( f(x) \) is in the domain of \( g \), - Domain values of \( f \) (1, 2, 3, 4, 7) need to map to the domain values of \( g \) (0, 1, 4, 6) through \( f \). **Range of \( g \circ f \):** - Follow the mappings: - \( f(1) = 0 \) and \( g(0) = 7 \) - \( f(2) = 1 \) and \( g(1) = 7 \) - \( f(3) = 3 \), and 3 is not in the domain of \( g \), so this does not contribute to the range. - \( f(4) =
Transcribed Image Text:### Composition of Functions: Domain and Range In the figure below, two functions \( f \) and \( g \) are defined. We need to determine the domain and range of the composition \( g \circ f \). #### Function Diagrams **Function \( f \):** - **Domain of \( f \)**: {1, 2, 3, 4, 7} - **Range of \( f \)**: {0, 1, 3, 6} Mappings for \( f \): - \( f(1) = 0 \) - \( f(2) = 1 \) - \( f(3) = 3 \) - \( f(4) = 6 \) - \( f(7) = 6 \) **Function \( g \):** - **Domain of \( g \)**: {0, 1, 4, 6} - **Range of \( g \)**: {7, 3, 9} Mappings for \( g \): - \( g(0) = 7 \) - \( g(1) = 7 \) - \( g(4) = 3 \) - \( g(6) = 9 \) #### Composition \( g \circ f \) To find the domain and range of \( g \circ f \), we need to follow the values mapped by \( f \) and then apply \( g \). **Domain of \( g \circ f \):** - Since \( g \circ f(x) \) is defined only when \( f(x) \) is in the domain of \( g \), - Domain values of \( f \) (1, 2, 3, 4, 7) need to map to the domain values of \( g \) (0, 1, 4, 6) through \( f \). **Range of \( g \circ f \):** - Follow the mappings: - \( f(1) = 0 \) and \( g(0) = 7 \) - \( f(2) = 1 \) and \( g(1) = 7 \) - \( f(3) = 3 \), and 3 is not in the domain of \( g \), so this does not contribute to the range. - \( f(4) =
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