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Define a function satisfying the given condition by defining its value on each element of the domain. Note that there are many correct answers for these questions. 1. f: {5, 15, 17, 18} → {1, 6, 7, 12, 13, 17} which is injective. 2. f: {5, 15, 17, 18} → {1, 6, 7, 12, 13, 17} which is not injective. 3. f: {5, 15, 17, 18} → {12, 13, 17} which is surjective. 4. f: {5, 15, 17, 18} - → {12, 13, 17} which is not surjective. f(5) f(15) = f(17) = f(18) = = f(5) = f(15) = f(17) = f(18) = = f(5) = f(15) = f(17) = f(18) = f(5) = f(15) = f(17) = f(18) : =

Define a function satisfying the given condition by defining its value on each element of the domain. Note that there are many correct answers for these questions. 1. f: {5, 15, 17, 18} → {1, 6, 7, 12, 13, 17} which is injective. 2. f: {5, 15, 17, 18} → {1, 6, 7, 12, 13, 17} which is not injective. 3. f: {5, 15, 17, 18} → {12, 13, 17} which is surjective. 4. f: {5, 15, 17, 18} - → {12, 13, 17} which is not surjective. f(5) f(15) = f(17) = f(18) = = f(5) = f(15) = f(17) = f(18) = = f(5) = f(15) = f(17) = f(18) = f(5) = f(15) = f(17) = f(18) : =

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Define a function satisfying the given condition by defining its value on each element of the domain. Note that there are many correct answers for these questions. 1. f : {5, 15, 17, 18} → {1, 6, 7, 12, 13, 17} which is injective. 2. f: {5, 15, 17, 18} → {1, 6, 7, 12, 13, 17} which is not injective. 3. f: {5, 15, 17, 18} - › {12, 13, 17} which is surjective. 4. f : {5, 15, 17, 18} → {12, 13, 17} which is not surjective. ƒ(5) = f(15) = f(17) ƒ(18) = = ƒ(5) = f(15) ƒ(17) = f(18) = = T = ƒ(5) = f(15) = f(17) = f(18) = = f(5) f(15) = f(17) = f(18) = =
Transcribed Image Text:Define a function satisfying the given condition by defining its value on each element of the domain. Note that there are many correct answers for these questions. 1. f : {5, 15, 17, 18} → {1, 6, 7, 12, 13, 17} which is injective. 2. f: {5, 15, 17, 18} → {1, 6, 7, 12, 13, 17} which is not injective. 3. f: {5, 15, 17, 18} - › {12, 13, 17} which is surjective. 4. f : {5, 15, 17, 18} → {12, 13, 17} which is not surjective. ƒ(5) = f(15) = f(17) ƒ(18) = = ƒ(5) = f(15) ƒ(17) = f(18) = = T = ƒ(5) = f(15) = f(17) = f(18) = = f(5) f(15) = f(17) = f(18) = =
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