To determine which of the given functions is even, we need to verify if [tex]\( g(x) = g(-x) \)[/tex] for each function.
1. [tex]\( g(x) = (x-1)^2 + 1 \)[/tex]
Evaluate [tex]\( g(-x) \)[/tex]:
[tex]\[
g(-x) = ((-x) - 1)^2 + 1 = (-x - 1)^2 + 1
\][/tex]
[tex]\[
= (x + 1)^2 + 1
\][/tex]
[tex]\[
= x^2 + 2x + 1 + 1
\][/tex]
[tex]\[
= x^2 + 2x + 2
\][/tex]
Compare this with [tex]\( g(x) \)[/tex]:
[tex]\[
g(x) = x^2 - 2x + 1 + 1
\][/tex]
[tex]\[
= x^2 - 2x + 2
\][/tex]
Since [tex]\( g(x) \neq g(-x) \)[/tex], this function is not even.
2. [tex]\( g(x) = 2x^2 + 1 \)[/tex]
Evaluate [tex]\( g(-x) \)[/tex]:
[tex]\[
g(-x) = 2(-x)^2 + 1 = 2x^2 + 1
\][/tex]
Since [tex]\( g(x) = g(-x) \)[/tex], this function is even.
3. [tex]\( g(x) = 4x + 2 \)[/tex]
Evaluate [tex]\( g(-x) \)[/tex]:
[tex]\[
g(-x) = 4(-x) + 2 = -4x + 2
\][/tex]
Compare this with [tex]\( g(x) \)[/tex]:
[tex]\[
g(x) = 4x + 2
\][/tex]
Since [tex]\( g(x) \neq g(-x) \)[/tex], this function is not even.
4. [tex]\( g(x) = 2x \)[/tex]
Evaluate [tex]\( g(-x) \)[/tex]:
[tex]\[
g(-x) = 2(-x) = -2x
\][/tex]
Compare this with [tex]\( g(x) \)[/tex]:
[tex]\[
g(x) = 2x
\][/tex]
Since [tex]\( g(x) \neq g(-x) \)[/tex], this function is not even.
Upon evaluation, only the function [tex]\( g(x) = 2x^2 + 1 \)[/tex] is even. Therefore, the correct answer is [tex]\( \boxed{2} \)[/tex].
