Which of the following is an even function?

A. [tex]f(x)=(x-1)^2[/tex]

B. [tex]f(x)=8x[/tex]

C. [tex]f(x)=x^2-x[/tex]

D. [tex]f(x)=7[/tex]



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]