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To determine which function is even, we need to understand the definition of an even function. A function [tex]\( f(x) \)[/tex] is classified as even if it satisfies the condition [tex]\( f(x) = f(-x) \)[/tex] for all [tex]\( x \)[/tex] in its domain.
Let's analyze each given function step by step:
1. Function [tex]\( f(x) = x^3 + 2 \)[/tex]:
- Substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex] in the function:
[tex]\[ f(-x) = (-x)^3 + 2 = -x^3 + 2 \][/tex]
- Check if [tex]\( f(x) = f(-x) \)[/tex]:
[tex]\[ x^3 + 2 \neq -x^3 + 2 \][/tex]
- Since [tex]\( f(x) \neq f(-x) \)[/tex], this function is not even.
2. Function [tex]\( f(x) = x^2 + 1 \)[/tex]:
- Substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex] in the function:
[tex]\[ f(-x) = (-x)^2 + 1 = x^2 + 1 \][/tex]
- Check if [tex]\( f(x) = f(-x) \)[/tex]:
[tex]\[ x^2 + 1 = x^2 + 1 \][/tex]
- Since [tex]\( f(x) = f(-x) \)[/tex], this function is even.
3. Function [tex]\( f(x) = |x + 2| \)[/tex]:
- Substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex] in the function:
[tex]\[ f(-x) = |-x + 2| = |-(x - 2)| = |x - 2| \][/tex]
- Check if [tex]\( f(x) = f(-x) \)[/tex]:
[tex]\[ |x + 2| \neq |x - 2| \][/tex]
- Since [tex]\( f(x) \neq f(-x) \)[/tex], this function is not even.
4. Function [tex]\( f(x) = \sin(2x) \)[/tex]:
- Substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex] in the function:
[tex]\[ f(-x) = \sin(2(-x)) = \sin(-2x) \][/tex]
- Recall that [tex]\( \sin(-\theta) = -\sin(\theta) \)[/tex]:
[tex]\[ \sin(-2x) = -\sin(2x) \][/tex]
- Check if [tex]\( f(x) = f(-x) \)[/tex]:
[tex]\[ \sin(2x) \neq -\sin(2x) \][/tex]
- Since [tex]\( f(x) \neq f(-x) \)[/tex], this function is not even.
Based on the analysis, the function that is even is:
[tex]\[ \boxed{2} \][/tex]
Let's analyze each given function step by step:
1. Function [tex]\( f(x) = x^3 + 2 \)[/tex]:
- Substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex] in the function:
[tex]\[ f(-x) = (-x)^3 + 2 = -x^3 + 2 \][/tex]
- Check if [tex]\( f(x) = f(-x) \)[/tex]:
[tex]\[ x^3 + 2 \neq -x^3 + 2 \][/tex]
- Since [tex]\( f(x) \neq f(-x) \)[/tex], this function is not even.
2. Function [tex]\( f(x) = x^2 + 1 \)[/tex]:
- Substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex] in the function:
[tex]\[ f(-x) = (-x)^2 + 1 = x^2 + 1 \][/tex]
- Check if [tex]\( f(x) = f(-x) \)[/tex]:
[tex]\[ x^2 + 1 = x^2 + 1 \][/tex]
- Since [tex]\( f(x) = f(-x) \)[/tex], this function is even.
3. Function [tex]\( f(x) = |x + 2| \)[/tex]:
- Substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex] in the function:
[tex]\[ f(-x) = |-x + 2| = |-(x - 2)| = |x - 2| \][/tex]
- Check if [tex]\( f(x) = f(-x) \)[/tex]:
[tex]\[ |x + 2| \neq |x - 2| \][/tex]
- Since [tex]\( f(x) \neq f(-x) \)[/tex], this function is not even.
4. Function [tex]\( f(x) = \sin(2x) \)[/tex]:
- Substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex] in the function:
[tex]\[ f(-x) = \sin(2(-x)) = \sin(-2x) \][/tex]
- Recall that [tex]\( \sin(-\theta) = -\sin(\theta) \)[/tex]:
[tex]\[ \sin(-2x) = -\sin(2x) \][/tex]
- Check if [tex]\( f(x) = f(-x) \)[/tex]:
[tex]\[ \sin(2x) \neq -\sin(2x) \][/tex]
- Since [tex]\( f(x) \neq f(-x) \)[/tex], this function is not even.
Based on the analysis, the function that is even is:
[tex]\[ \boxed{2} \][/tex]
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