To determine which of the following functions is an even function, we must recall that a function [tex]\( f(x) \)[/tex] is considered even if it satisfies the condition:
[tex]\[ f(-x) = f(x) \][/tex]
We will analyze each given function step-by-step.
1. [tex]\( f(x) = (x - 1)^2 \)[/tex]:
[tex]\[
f(-x) = (-x - 1)^2 = (-(x + 1))^2 = (x + 1)^2
\][/tex]
Now compare [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[
f(x) = (x - 1)^2
\][/tex]
Clearly, [tex]\( (x + 1)^2 \neq (x - 1)^2 \)[/tex]. Therefore, [tex]\( f(x) = (x - 1)^2 \)[/tex] is not an even function.
2. [tex]\( f(x) = 8x \)[/tex]:
[tex]\[
f(-x) = 8(-x) = -8x
\][/tex]
Now compare [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[
f(x) = 8x
\][/tex]
Clearly, [tex]\(-8x \neq 8x \)[/tex]. Therefore, [tex]\( f(x) = 8x \)[/tex] is not an even function.
3. [tex]\( f(x) = x^2 - x \)[/tex]:
[tex]\[
f(-x) = (-x)^2 - (-x) = x^2 + x
\][/tex]
Now compare [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[
f(x) = x^2 - x
\][/tex]
Clearly, [tex]\( x^2 + x \neq x^2 - x \)[/tex]. Therefore, [tex]\( f(x) = x^2 - x \)[/tex] is not an even function.
4. [tex]\( f(x) = 7 \)[/tex]:
[tex]\[
f(-x) = 7
\][/tex]
Compare [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[
f(x) = 7
\][/tex]
Clearly, [tex]\( 7 = 7 \)[/tex]. Therefore, [tex]\( f(x) = 7 \)[/tex] is an even function.
So, out of the given functions, the only even function is:
[tex]\[ f(x) = 7 \][/tex]
