Consider the following function: f: Z → N given by f(x) = (x+4)² + 8; where, as usual, Z and N, denote the set of all integers and the set of all natural numbers, respectively. You are given that the function f is not an injection. In order to show that f is not an injection, state two distinct elements of the domain of f, denoted by x1, x2, such that f(x₁) = f(x₂). • State 1: State ₂: You are given that the function f is not a surjection. In order to show that f is not a surjection, state an element of the codomain of f, denoted by y that is not in the image of f. • State y: I

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.5: Graphs Of Functions
Problem 32E
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Consider the following function:
f : Z → N given by ƒ(x) = (x + 4)² + 8;
where, as usual, Z and N, denote the set of all integers and the set of all natural numbers,
respectively.
You are given that the function f is not an injection.
In order to show that f is not an injection, state two distinct elements of the domain of f,
denoted by x₁, x2, such that ƒ(x1) = f(x2).
• State 1:
• State x₂:
You are given that the function f is not a surjection.
In order to show that f is not a surjection, state an element of the codomain of ƒ, denoted by y
that is not in the image of f.
I
• State y:
Transcribed Image Text:Consider the following function: f : Z → N given by ƒ(x) = (x + 4)² + 8; where, as usual, Z and N, denote the set of all integers and the set of all natural numbers, respectively. You are given that the function f is not an injection. In order to show that f is not an injection, state two distinct elements of the domain of f, denoted by x₁, x2, such that ƒ(x1) = f(x2). • State 1: • State x₂: You are given that the function f is not a surjection. In order to show that f is not a surjection, state an element of the codomain of ƒ, denoted by y that is not in the image of f. I • State y:
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