Let's determine which of the given functions are even. Recall that a function [tex]\( f(x) \)[/tex] is even if it satisfies the condition [tex]\( f(x) = f(-x) \)[/tex] for all [tex]\( x \)[/tex].
1. Function: [tex]\( g(x) = (x-1)^2 + 1 \)[/tex]
To check if this function is even, we compute [tex]\( g(-x) \)[/tex]:
[tex]\[
g(-x) = ((-x) - 1)^2 + 1 = (-x - 1)^2 + 1 = (-(x + 1))^2 + 1 = (x + 1)^2 + 1
\][/tex]
Since [tex]\( (x + 1)^2 \neq (x - 1)^2 \)[/tex] for all [tex]\( x \)[/tex], we have [tex]\( g(-x) \neq g(x) \)[/tex]. Therefore, [tex]\( g(x) = (x-1)^2 + 1 \)[/tex] is not an even function.
2. Function: [tex]\( g(x) = 2x^2 + 1 \)[/tex]
To check if this function is even, we compute [tex]\( g(-x) \)[/tex]:
[tex]\[
g(-x) = 2(-x)^2 + 1 = 2x^2 + 1
\][/tex]
Here, [tex]\( g(-x) = g(x) \)[/tex]. Therefore, [tex]\( g(x) = 2x^2 + 1 \)[/tex] is an even function.
3. Function: [tex]\( g(x) = 4x + 2 \)[/tex]
To check if this function is even, we compute [tex]\( g(-x) \)[/tex]:
[tex]\[
g(-x) = 4(-x) + 2 = -4x + 2
\][/tex]
Since [tex]\( -4x + 2 \neq 4x + 2 \)[/tex] for all [tex]\( x \)[/tex], we have [tex]\( g(-x) \neq g(x) \)[/tex]. Therefore, [tex]\( g(x) = 4x + 2 \)[/tex] is not an even function.
4. Function: [tex]\( g(x) = 2x \)[/tex]
To check if this function is even, we compute [tex]\( g(-x) \)[/tex]:
[tex]\[
g(-x) = 2(-x) = -2x
\][/tex]
Since [tex]\( -2x \neq 2x \)[/tex] for all [tex]\( x \neq 0 \)[/tex], we have [tex]\( g(-x) \neq g(x) \)[/tex]. Therefore, [tex]\( g(x) = 2x \)[/tex] is not an even function.
Based on this analysis:
- [tex]\( g(x) = 2x^2 + 1 \)[/tex] is the only even function among the options given.
