To determine which of the given functions is an odd function, we need to recall the definition of an odd function. A function [tex]\( f(x) \)[/tex] is odd if and only if [tex]\( f(-x) = -f(x) \)[/tex] for all [tex]\( x \)[/tex] in the domain of [tex]\( f \)[/tex].
We will check each function one by one to see if this property holds.
### Function 1: [tex]\( f(x) = x^3 + 5x^2 + x \)[/tex]
1. Calculate [tex]\( f(-x) \)[/tex]:
[tex]\[
f(-x) = (-x)^3 + 5(-x)^2 + -x = -x^3 + 5x^2 - x
\][/tex]
2. Check if [tex]\( f(-x) = -f(x) \)[/tex]:
[tex]\[
-f(x) = -(x^3 + 5x^2 + x) = -x^3 - 5x^2 - x
\][/tex]
3. Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[
-x^3 + 5x^2 - x \neq -x^3 - 5x^2 - x
\][/tex]
This function is not odd.
### Function 2: [tex]\( f(x) = \sqrt{x} \)[/tex]
1. Calculate [tex]\( f(-x) \)[/tex]:
[tex]\[
f(-x) = \sqrt{-x}
\][/tex]
Note that [tex]\( \sqrt{-x} \)[/tex] is not defined for real numbers when [tex]\( x > 0 \)[/tex]. Therefore, we cannot proceed with checking [tex]\( f(-x) = -f(x) \)[/tex] in the realm of real numbers.
This function is not odd.
### Function 3: [tex]\( f(x) = x^2 + x \)[/tex]
1. Calculate [tex]\( f(-x) \)[/tex]:
[tex]\[
f(-x) = (-x)^2 + (-x) = x^2 - x
\][/tex]
2. Check if [tex]\( f(-x) = -f(x) \)[/tex]:
[tex]\[
-f(x) = -(x^2 + x) = -x^2 - x
\][/tex]
3. Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[
x^2 - x \neq -x^2 - x
\][/tex]
This function is not odd.
### Function 4: [tex]\( f(x) = -x \)[/tex]
1. Calculate [tex]\( f(-x) \)[/tex]:
[tex]\[
f(-x) = -(-x) = x
\][/tex]
2. Check if [tex]\( f(-x) = -f(x) \)[/tex]:
[tex]\[
-f(x) = -(-x) = x
\][/tex]
3. Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[
x = x
\][/tex]
This function is odd.
Based on this analysis, the only odd function among the given options is:
[tex]\[ f(x) = -x \][/tex]
