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If [tex]$g(x)$[/tex] is an odd function, which function must be an even function?

[tex]\[
\begin{array}{l}
A. \ f(x)=g(x)+g(x) \\
B. \ f(x)=-g(x) \\
C. \ f(x)=g(x)+2 \\
D. \ f(x)=g(x)^2 \\
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
To determine which of the given functions is an even function when [tex]\( g(x) \)[/tex] is an odd function, let's first recall the definitions:

- A function [tex]\( g(x) \)[/tex] is odd if [tex]\( g(-x) = -g(x) \)[/tex].
- A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex].

Now, let's analyze each option step-by-step.

### Option 1: [tex]\( f(x) = g(x) + g(x) \)[/tex]

Here, [tex]\( f(x) = 2g(x) \)[/tex].
- Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( f(-x) = 2g(-x) = 2(-g(x)) = -2g(x) \)[/tex].
- This means [tex]\( f(-x) \neq f(x) \)[/tex], so [tex]\( f(x) \)[/tex] is not even.

### Option 2: [tex]\( f(x) = -g(x) \)[/tex]

Here, [tex]\( f(x) = -g(x) \)[/tex].
- Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( f(-x) = -g(-x) = -(-g(x)) = g(x) \)[/tex].
- This means [tex]\( f(-x) \neq f(x) \)[/tex], so [tex]\( f(x) \)[/tex] is not even.

### Option 3: [tex]\( f(x) = g(x) + 2 \)[/tex]

Here, [tex]\( f(x) = g(x) + 2 \)[/tex].
- Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( f(-x) = g(-x) + 2 = -g(x) + 2 \)[/tex].
- This means [tex]\( f(-x) \neq f(x) \)[/tex], so [tex]\( f(x) \)[/tex] is not even.

### Option 4: [tex]\( f(x) = g(x)^2 \)[/tex]

Here, [tex]\( f(x) = [g(x)]^2 \)[/tex].
- Since [tex]\( g(x) \)[/tex] is odd, [tex]\( g(-x) = -g(x) \)[/tex].
- Therefore, [tex]\( f(-x) = [g(-x)]^2 = [-g(x)]^2 = [g(x)]^2 \)[/tex].
- This means [tex]\( f(-x) = f(x) \)[/tex], so [tex]\( f(x) \)[/tex] is indeed an even function.

### Conclusion

The function [tex]\( f(x) = [g(x)]^2 \)[/tex], which is Option 4, must be an even function when [tex]\( g(x) \)[/tex] is an odd function.
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