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To determine which of the provided functions must be an even function given that [tex]\( g(x) \)[/tex] is an odd function, let's analyze each of the options step by step:
1. Option 1: [tex]\( f(x) = g(x) + 2 \)[/tex]:
- An odd function [tex]\( g(x) \)[/tex] satisfies the property [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( f(x) = g(x) + 2 \)[/tex] would become:
[tex]\[ f(-x) = g(-x) + 2 = -g(x) + 2 \][/tex]
- Since [tex]\( -g(x) + 2 \neq g(x) + 2 \)[/tex] for all [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] is not an even function.
2. Option 2: [tex]\( f(x) = g(x) + g(x) \)[/tex]:
- Simplifying, we get [tex]\( f(x) = 2g(x) \)[/tex].
- For an odd function [tex]\( g(x) \)[/tex], [tex]\( g(-x) = -g(x) \)[/tex].
- Thus,
[tex]\[ f(-x) = 2g(-x) = 2(-g(x)) = -2g(x) \][/tex]
- Since [tex]\( -2g(x) \neq 2g(x) \)[/tex] for all [tex]\( x \)[/tex], [tex]\( f(x) = 2g(x) \)[/tex] is still an odd function, and therefore, it is not an even function.
3. Option 3: [tex]\( f(x) = g(x)^2 \)[/tex]:
- An odd function [tex]\( g(x) \)[/tex] satisfies [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, for [tex]\( f(x) = g(x)^2 \)[/tex], we have:
[tex]\[ f(-x) = (g(-x))^2 = (-g(x))^2 = (g(x))^2 = f(x) \][/tex]
- Since [tex]\( f(-x) = f(x) \)[/tex], [tex]\( f(x) = g(x)^2 \)[/tex] is an even function.
4. Option 4: [tex]\( f(x) = -g(x) \)[/tex]:
- For an odd function [tex]\( g(x) \)[/tex], [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( f(x) = -g(x) \)[/tex] would become:
[tex]\[ f(-x) = -g(-x) = -(-g(x)) = g(x) \][/tex]
- So, [tex]\( f(-x) = g(x) \)[/tex] does not equal [tex]\( -g(x) \)[/tex] unless [tex]\( g(x) = 0 \)[/tex], thus [tex]\( f(x) \)[/tex] is still an odd function and not an even function.
Among the given options, the third function [tex]\( f(x) = g(x)^2 \)[/tex] must be an even function.
1. Option 1: [tex]\( f(x) = g(x) + 2 \)[/tex]:
- An odd function [tex]\( g(x) \)[/tex] satisfies the property [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( f(x) = g(x) + 2 \)[/tex] would become:
[tex]\[ f(-x) = g(-x) + 2 = -g(x) + 2 \][/tex]
- Since [tex]\( -g(x) + 2 \neq g(x) + 2 \)[/tex] for all [tex]\( x \)[/tex], [tex]\( f(x) \)[/tex] is not an even function.
2. Option 2: [tex]\( f(x) = g(x) + g(x) \)[/tex]:
- Simplifying, we get [tex]\( f(x) = 2g(x) \)[/tex].
- For an odd function [tex]\( g(x) \)[/tex], [tex]\( g(-x) = -g(x) \)[/tex].
- Thus,
[tex]\[ f(-x) = 2g(-x) = 2(-g(x)) = -2g(x) \][/tex]
- Since [tex]\( -2g(x) \neq 2g(x) \)[/tex] for all [tex]\( x \)[/tex], [tex]\( f(x) = 2g(x) \)[/tex] is still an odd function, and therefore, it is not an even function.
3. Option 3: [tex]\( f(x) = g(x)^2 \)[/tex]:
- An odd function [tex]\( g(x) \)[/tex] satisfies [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, for [tex]\( f(x) = g(x)^2 \)[/tex], we have:
[tex]\[ f(-x) = (g(-x))^2 = (-g(x))^2 = (g(x))^2 = f(x) \][/tex]
- Since [tex]\( f(-x) = f(x) \)[/tex], [tex]\( f(x) = g(x)^2 \)[/tex] is an even function.
4. Option 4: [tex]\( f(x) = -g(x) \)[/tex]:
- For an odd function [tex]\( g(x) \)[/tex], [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( f(x) = -g(x) \)[/tex] would become:
[tex]\[ f(-x) = -g(-x) = -(-g(x)) = g(x) \][/tex]
- So, [tex]\( f(-x) = g(x) \)[/tex] does not equal [tex]\( -g(x) \)[/tex] unless [tex]\( g(x) = 0 \)[/tex], thus [tex]\( f(x) \)[/tex] is still an odd function and not an even function.
Among the given options, the third function [tex]\( f(x) = g(x)^2 \)[/tex] must be an even function.
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