To determine which of the given functions are even, let's recall the definition of an even function. A function [tex]\( f(x) \)[/tex] is even if for all [tex]\( x \)[/tex] in its domain, [tex]\( f(-x) = f(x) \)[/tex].
Let's check each function one by one:
### 1. [tex]\( f(x) = (x-1)^2 \)[/tex]
For [tex]\( f(x) \)[/tex] to be even, [tex]\( f(-x) \)[/tex] must equal [tex]\( f(x) \)[/tex].
[tex]\[ f(x) = (x-1)^2 \][/tex]
[tex]\[ f(-x) = (-x-1)^2 \][/tex]
Now, let's simplify [tex]\( f(-x) \)[/tex]:
[tex]\[ (-x-1)^2 = ((-x-1)(-x-1)) = (-x-1)(-x-1) =(x^2+2x+1)+2x = x^2 + 2x + 1 \][/tex]
However,
[tex]\[f(x) = (x-1)^2 = x^2 - 2x + 1\][/tex]
Since [tex]\( f(-x) \neq f(x) \)[/tex], [tex]\( (x-1)^2 \)[/tex] is not an even function.
### 2. [tex]\( f(x) = 8x \)[/tex]
For [tex]\( f(x) \)[/tex] to be even, [tex]\( f(-x) \)[/tex] must equal [tex]\( f(x) \)[/tex].
[tex]\[ f(x) = 8x \][/tex]
[tex]\[ f(-x) = 8(-x) = -8x \][/tex]
Since [tex]\( f(-x) \neq f(x) \)[/tex], [tex]\( 8x \)[/tex] is not an even function.
### 3. [tex]\( f(x) = x^2 - x \)[/tex]
For [tex]\( f(x) \)[/tex] to be even, [tex]\( f(-x) \)[/tex] must equal [tex]\( f(x) \)[/tex].
[tex]\[ f(x) = x^2 - x \][/tex]
[tex]\[ f(-x) = (-x)^2 - (-x) = x^2 + x \][/tex]
Since [tex]\( f(-x) \neq f(x) \)[/tex], [tex]\( x^2 - x \)[/tex] is not an even function.
### 4. [tex]\( f(x) = 7 \)[/tex]
For [tex]\( f(x) \)[/tex] to be even, [tex]\( f(-x) \)[/tex] must equal [tex]\( f(x) \)[/tex].
[tex]\[ f(x) = 7 \][/tex]
[tex]\[ f(-x) = 7 \][/tex]
Since [tex]\( f(-x) = f(x) \)[/tex], [tex]\( 7 \)[/tex] is an even function.
Thus, among the given functions, the only even function is [tex]\( f(x) = 7 \)[/tex].
