To determine which functions are even, we need to check if each function [tex]\( f(x) \)[/tex] satisfies the condition for even functions: [tex]\( f(x) = f(-x) \)[/tex].
Let's analyze each function step-by-step:
1. Function [tex]\( f(x) = x^4 - x^2 \)[/tex]
We will substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[
f(-x) = (-x)^4 - (-x)^2
\][/tex]
Simplify the expression:
[tex]\[
(-x)^4 = x^4, \quad (-x)^2 = x^2
\][/tex]
So,
[tex]\[
f(-x) = x^4 - x^2
\][/tex]
Since [tex]\( f(-x) = f(x) \)[/tex], the function [tex]\( f(x) = x^4 - x^2 \)[/tex] is even.
2. Function [tex]\( f(x) = x^2 - 3x + 2 \)[/tex]
Next, we will substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[
f(-x) = (-x)^2 - 3(-x) + 2
\][/tex]
Simplify the expression:
[tex]\[
(-x)^2 = x^2, \quad -3(-x) = 3x
\][/tex]
So,
[tex]\[
f(-x) = x^2 + 3x + 2
\][/tex]
Since [tex]\( f(-x) \neq f(x) \)[/tex], the function [tex]\( f(x) = x^2 - 3x + 2 \)[/tex] is not even.
3. Function [tex]\( f(x) = \sqrt{x} - 2 \)[/tex]
We will substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[
f(-x) = \sqrt{-x} - 2
\][/tex]
The square root of a negative number is not real (unless we consider complex numbers), which changes the nature of the function. Thus,
[tex]\[
f(-x) \neq f(x)
\][/tex]
Therefore, [tex]\( f(x) = \sqrt{x} - 2 \)[/tex] is not even.
4. Function [tex]\( f(x) = |x| \)[/tex]
Finally, we will substitute [tex]\(-x\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[
f(-x) = |-x|
\][/tex]
Simplify the expression:
[tex]\[
|-x| = |x|
\][/tex]
Since [tex]\( f(-x) = f(x) \)[/tex], the function [tex]\( f(x) = |x| \)[/tex] is even.
By evaluating each function, we conclude that the even functions are:
- [tex]\( f(x) = x^4 - x^2 \)[/tex]
- [tex]\( f(x) = |x| \)[/tex]
Thus, the functions that are even are:
[tex]\[ \boxed{f(x) = x^4 - x^2 \quad \text{and} \quad f(x) = |x|} \][/tex]
