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Which functions are even? Choose two correct answers.

[tex]
\begin{array}{l}
f(x)=|x| \\
f(x)=x^2-3x+2
\end{array}
[/tex]


Sagot :

To determine whether a function is even, we need to check if the function satisfies the condition [tex]\( f(-x) = f(x) \)[/tex] for all [tex]\( x \)[/tex].

### Step-by-Step Analysis

1. Function [tex]\( f(x) = |x| \)[/tex]:
- We need to evaluate [tex]\( f(-x) \)[/tex].
- Since the absolute value function is defined as [tex]\( |x| \)[/tex] being the distance from [tex]\( x \)[/tex] to zero, it does not change if [tex]\( x \)[/tex] is negated:
[tex]\[ f(-x) = |-x| = |x| \][/tex]
- Therefore, [tex]\( f(-x) = f(x) \)[/tex] for all [tex]\( x \)[/tex].
- Hence, [tex]\( f(x) = |x| \)[/tex] is an even function.

2. Function [tex]\( f(x) = x^2 - 3x + 2 \)[/tex]:
- We need to evaluate [tex]\( f(-x) \)[/tex].
- Substitute [tex]\( -x \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(-x) = (-x)^2 - 3(-x) + 2 = x^2 + 3x + 2 \][/tex]
- Compare [tex]\( f(-x) \)[/tex] with [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = x^2 - 3x + 2 \][/tex]
[tex]\[ f(-x) = x^2 + 3x + 2 \][/tex]
- Clearly, [tex]\( f(-x) \neq f(x) \)[/tex] for all [tex]\( x \)[/tex] (since [tex]\( -3x \neq 3x \)[/tex]).

### Conclusion

Given these steps, we can conclude:
- [tex]\( f(x) = |x| \)[/tex] is an even function.
- [tex]\( f(x) = x^2 - 3x + 2 \)[/tex] is not an even function.

Thus, the only correct answer is:
- [tex]\( f(x) = |x| \)[/tex]

The function [tex]\( f(x) = x^2 - 3x + 2 \)[/tex] is not one of the even functions.