To determine the type of symmetry for the function [tex]\( f(x) = x^5 \)[/tex], we need to check for either even or odd symmetry.
1. Even Symmetry: A function [tex]\( f(x) \)[/tex] is even if [tex]\( f(-x) = f(x) \)[/tex] for all [tex]\( x \)[/tex] in the domain.
2. Odd Symmetry: A function [tex]\( f(x) \)[/tex] is odd if [tex]\( f(-x) = -f(x) \)[/tex] for all [tex]\( x \)[/tex] in the domain.
Let's check for odd symmetry:
- Calculate [tex]\( f(x) \)[/tex] for [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = 1^5 = 1
\][/tex]
- Calculate [tex]\( f(-x) \)[/tex] for [tex]\( x = 1 \)[/tex]:
[tex]\[
f(-1) = (-1)^5 = -1
\][/tex]
Now, compare [tex]\( f(-x) \)[/tex] and [tex]\(-f(x) \)[/tex]:
- [tex]\(-f(x)\)[/tex] for [tex]\( x = 1 \)[/tex]:
[tex]\[
-f(1) = -1
\][/tex]
Since [tex]\( f(-1) = -1 \)[/tex] is equal to [tex]\(-f(1) = -1 \)[/tex]:
[tex]\[
f(-x) = -f(x)
\][/tex]
This confirms that the function [tex]\( f(x) = x^5 \)[/tex] is odd.
Therefore, [tex]\( f(x) \)[/tex] has odd symmetry, which is also known as origin or rotational symmetry.
The correct answer is:
[tex]\[ \text{odd} \][/tex]
