To determine which of the given functions are even, we need to recall the definition of an even function. A function [tex]\( f(x) \)[/tex] is even if and only if:
[tex]\[ f(-x) = f(x) \text{ for all } x \][/tex]
Let's analyze each function one by one.
### 1. [tex]\( f(x) = (x-1)^2 \)[/tex]
First, substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = (-x - 1)^2 \][/tex]
Now, simplify [tex]\( f(-x) \)[/tex]:
[tex]\[ (-x - 1)^2 = (-x - 1)(-x - 1) = x^2 + 2x + 1 \][/tex]
Compare [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = (x - 1)^2 = x^2 - 2x + 1 \][/tex]
Here, [tex]\( f(-x) \neq f(x) \)[/tex], so [tex]\((x-1)^2\)[/tex] is not an even function.
### 2. [tex]\( f(x) = 8x \)[/tex]
First, substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = 8(-x) \][/tex]
Now, simplify [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = -8x \][/tex]
Compare [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 8x \][/tex]
Here, [tex]\( f(-x) \neq f(x) \)[/tex], so [tex]\( 8x \)[/tex] is not an even function.
### 3. [tex]\( f(x) = x^2 - x \)[/tex]
First, substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = (-x)^2 - (-x) \][/tex]
Now, simplify [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = x^2 + x \][/tex]
Compare [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = x^2 - x \][/tex]
Here, [tex]\( f(-x) \neq f(x) \)[/tex], so [tex]\( x^2 - x \)[/tex] is not an even function.
### 4. [tex]\( f(x) = 7 \)[/tex]
First, substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = 7 \][/tex]
Since this function is a constant, it does not change when [tex]\( x \)[/tex] is replaced with [tex]\( -x \)[/tex]. Therefore:
[tex]\[ f(-x) = f(x) = 7 \][/tex]
Thus, [tex]\( 7 \)[/tex] is an even function.
### Conclusion
Among the given functions, only [tex]\( f(x) = 7 \)[/tex] is an even function.
