To determine which of the given functions is an odd function, we need to verify whether each function satisfies the property of odd functions. A function [tex]\( f(x) \)[/tex] is considered odd if [tex]\( f(-x) = -f(x) \)[/tex] for all [tex]\( x \)[/tex] in the domain of the function.
Let's analyze each function:
1. [tex]\( f(x) = x^3 + 5x^2 + x \)[/tex]
To check if this function is odd:
[tex]\[ f(-x) = (-x)^3 + 5(-x)^2 + (-x) = -x^3 + 5x^2 - x \][/tex]
Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(x^3 + 5x^2 + x) = -x^3 - 5x^2 - x \][/tex]
Since [tex]\( f(-x) \ne -f(x) \)[/tex], the function [tex]\( f(x) = x^3 + 5x^2 + x \)[/tex] is not an odd function.
2. [tex]\( f(x) = \sqrt{x} \)[/tex]
To check if this function is odd:
[tex]\[ f(-x) = \sqrt{-x} \][/tex]
Since the square root function is only defined for non-negative values of [tex]\( x \)[/tex], [tex]\( f(-x) \)[/tex] is not defined for [tex]\( x > 0 \)[/tex] in the real numbers. Thus, [tex]\( f(x) = \sqrt{x} \)[/tex] cannot be an odd function.
3. [tex]\( f(x) = x^2 + x \)[/tex]
To check if this function is odd:
[tex]\[ f(-x) = (-x)^2 + (-x) = x^2 - x \][/tex]
Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(x^2 + x) = -x^2 - x \][/tex]
Since [tex]\( f(-x) \ne -f(x) \)[/tex], the function [tex]\( f(x) = x^2 + x \)[/tex] is not an odd function.
4. [tex]\( f(x) = -x \)[/tex]
To check if this function is odd:
[tex]\[ f(-x) = -(-x) = x \][/tex]
Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(-x) = x \][/tex]
Since [tex]\( f(-x) = -f(x) \)[/tex], the function [tex]\( f(x) = -x \)[/tex] is an odd function.
Therefore, the function [tex]\( f(x) = -x \)[/tex] is the odd function among the given options.
