Answer :
To determine whether a given function is an even function, we need to check if it satisfies the condition \( f(-x) = f(x) \) for all \( x \).
Let's check each function one by one.
### 1. \( f(x) = (x-1)^2 \)
First, we calculate \( f(-x) \):
[tex]\[ f(-x) = (-x - 1)^2 = (-(x + 1))^2 = (x + 1)^2 \][/tex]
Then, we compare \( f(x) \) and \( f(-x) \):
[tex]\[ (x - 1)^2 \neq (x + 1)^2 \][/tex]
Thus, \( f(x) = (x - 1)^2 \) is not an even function.
### 2. \( f(x) = 8x \)
First, we calculate \( f(-x) \):
[tex]\[ f(-x) = 8(-x) = -8x \][/tex]
Then, we compare \( f(x) \) and \( f(-x) \):
[tex]\[ 8x \neq -8x \][/tex]
Thus, \( f(x) = 8x \) is not an even function.
### 3. \( f(x) = x^2 - x \)
First, we compute \( f(-x) \):
[tex]\[ f(-x) = (-x)^2 - (-x) = x^2 + x \][/tex]
Then, we compare \( f(x) \) and \( f(-x) \):
[tex]\[ x^2 - x \neq x^2 + x \][/tex]
Thus, \( f(x) = x^2 - x \) is not an even function.
### 4. \( f(x) = 7 \)
First, we calculate \( f(-x) \):
[tex]\[ f(-x) = 7 \][/tex]
Then, we compare \( f(x) \) and \( f(-x) \):
[tex]\[ f(x) = f(-x) = 7 \][/tex]
Since \( f(x) = 7 \), \( f(-x) = 7 \). We see that these are equal, which means:
[tex]\[ f(-x) = f(x) \][/tex]
Thus, \( f(x) = 7 \) is an even function.
### Summary
Among the given functions, the only even function is \( f(x) = 7 \).
So, the even function is:
[tex]\[ f(x) = 7 \][/tex]
Let's check each function one by one.
### 1. \( f(x) = (x-1)^2 \)
First, we calculate \( f(-x) \):
[tex]\[ f(-x) = (-x - 1)^2 = (-(x + 1))^2 = (x + 1)^2 \][/tex]
Then, we compare \( f(x) \) and \( f(-x) \):
[tex]\[ (x - 1)^2 \neq (x + 1)^2 \][/tex]
Thus, \( f(x) = (x - 1)^2 \) is not an even function.
### 2. \( f(x) = 8x \)
First, we calculate \( f(-x) \):
[tex]\[ f(-x) = 8(-x) = -8x \][/tex]
Then, we compare \( f(x) \) and \( f(-x) \):
[tex]\[ 8x \neq -8x \][/tex]
Thus, \( f(x) = 8x \) is not an even function.
### 3. \( f(x) = x^2 - x \)
First, we compute \( f(-x) \):
[tex]\[ f(-x) = (-x)^2 - (-x) = x^2 + x \][/tex]
Then, we compare \( f(x) \) and \( f(-x) \):
[tex]\[ x^2 - x \neq x^2 + x \][/tex]
Thus, \( f(x) = x^2 - x \) is not an even function.
### 4. \( f(x) = 7 \)
First, we calculate \( f(-x) \):
[tex]\[ f(-x) = 7 \][/tex]
Then, we compare \( f(x) \) and \( f(-x) \):
[tex]\[ f(x) = f(-x) = 7 \][/tex]
Since \( f(x) = 7 \), \( f(-x) = 7 \). We see that these are equal, which means:
[tex]\[ f(-x) = f(x) \][/tex]
Thus, \( f(x) = 7 \) is an even function.
### Summary
Among the given functions, the only even function is \( f(x) = 7 \).
So, the even function is:
[tex]\[ f(x) = 7 \][/tex]