To determine which of the given functions are even, we need to recall that a function [tex]\( f(x) \)[/tex] is considered even if [tex]\( f(-x) = f(x) \)[/tex] for all [tex]\( x \)[/tex] in its domain. Let's analyze each function step-by-step to see if they satisfy this condition.
1. [tex]\( g(x) = (x - 1)^2 + 1 \)[/tex]:
We need to evaluate [tex]\( g(-x) \)[/tex]:
[tex]\[ g(-x) = ((-x) - 1)^2 + 1 \][/tex]
[tex]\[ = (-x - 1)^2 + 1 \][/tex]
[tex]\[ = (x + 1)^2 + 1 \][/tex]
[tex]\[ = (x^2 + 2x + 1) + 1 \][/tex]
[tex]\[ = x^2 + 2x + 2 \][/tex]
Now compare this to [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = (x - 1)^2 + 1 \][/tex]
[tex]\[ = (x^2 - 2x + 1) + 1 \][/tex]
[tex]\[ = x^2 - 2x + 2 \][/tex]
Clearly,
[tex]\[ x^2 - 2x + 2 \neq x^2 + 2x + 2 \][/tex]
Hence, [tex]\( g(x) = (x - 1)^2 + 1 \)[/tex] is not an even function.
2. [tex]\( g(x) = 2x^2 + 1 \)[/tex]:
Evaluate [tex]\( g(-x) \)[/tex]:
[tex]\[ g(-x) = 2(-x)^2 + 1 \][/tex]
[tex]\[ = 2x^2 + 1 \][/tex]
This matches [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = 2x^2 + 1 \][/tex]
So,
[tex]\[ g(-x) = g(x) \][/tex]
Hence, [tex]\( g(x) = 2x^2 + 1 \)[/tex] is an even function.
3. [tex]\( g(x) = 4x + 2 \)[/tex]:
Evaluate [tex]\( g(-x) \)[/tex]:
[tex]\[ g(-x) = 4(-x) + 2 \][/tex]
[tex]\[ = -4x + 2 \][/tex]
Now compare this to [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = 4x + 2 \][/tex]
Clearly,
[tex]\[ 4x + 2 \neq -4x + 2 \][/tex]
Hence, [tex]\( g(x) = 4x + 2 \)[/tex] is not an even function.
4. [tex]\( g(x) = 2x \)[/tex]:
Evaluate [tex]\( g(-x) \)[/tex]:
[tex]\[ g(-x) = 2(-x) \][/tex]
[tex]\[ = -2x \][/tex]
Now compare this to [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = 2x \][/tex]
Clearly,
[tex]\[ 2x \neq -2x \][/tex]
Hence, [tex]\( g(x) = 2x \)[/tex] is not an even function.
To summarize, the only even function among the given options is:
[tex]\[ g(x) = 2x^2 + 1 \][/tex]
