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[tex]$f(x)=x^5$[/tex] has which type of symmetry?

A. no symmetry
B. even symmetry
C. odd symmetry
D. both even and odd symmetry



Answer :

GinnyAnswer
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]

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