Let's determine which of the given functions are even. An even function satisfies the property [tex]\( f(x) = f(-x) \)[/tex] for all [tex]\( x \)[/tex].
1. Consider the function [tex]\( f(x) = (x-1)^2 \)[/tex]:
[tex]\[
f(x) = (x-1)^2
\][/tex]
To check if this is even, we evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[
f(-x) = (-x-1)^2 = (-(x+1))^2 = (x+1)^2
\][/tex]
Comparing [tex]\( f(x) \)[/tex] and [tex]\( f(-x) \)[/tex]:
[tex]\[
(x-1)^2 \neq (x+1)^2
\][/tex]
Therefore, [tex]\( f(x) = (x-1)^2 \)[/tex] is not an even function.
2. Consider the function [tex]\( f(x) = 8x \)[/tex]:
[tex]\[
f(x) = 8x
\][/tex]
To check if this is even, we evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[
f(-x) = 8(-x) = -8x
\][/tex]
Comparing [tex]\( f(x) \)[/tex] and [tex]\( f(-x) \)[/tex]:
[tex]\[
8x \neq -8x
\][/tex]
Therefore, [tex]\( f(x) = 8x \)[/tex] is not an even function.
3. Consider the function [tex]\( f(x) = x^2 - x \)[/tex]:
[tex]\[
f(x) = x^2 - x
\][/tex]
To check if this is even, we evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[
f(-x) = (-x)^2 - (-x) = x^2 + x
\][/tex]
Comparing [tex]\( f(x) \)[/tex] and [tex]\( f(-x) \)[/tex]:
[tex]\[
x^2 - x \neq x^2 + x
\][/tex]
Therefore, [tex]\( f(x) = x^2 - x \)[/tex] is not an even function.
4. Consider the function [tex]\( f(x) = 7 \)[/tex]:
[tex]\[
f(x) = 7
\][/tex]
To check if this is even, we evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[
f(-x) = 7
\][/tex]
Comparing [tex]\( f(x) \)[/tex] and [tex]\( f(-x) \)[/tex]:
[tex]\[
7 = 7
\][/tex]
Therefore, [tex]\( f(x) = 7 \)[/tex] is an even function.
So, the function that is even is:
[tex]\[ f(x) = 7 \][/tex]
