Answer :
To determine which of the given functions is an odd function, we need to verify whether each function satisfies the condition for oddness. A function [tex]\( f(x) \)[/tex] is odd if and only if [tex]\( f(-x) = -f(x) \)[/tex].
Let's go through each function one by one.
1. Function: [tex]\( f(x) = x^3 + 5x^2 + x \)[/tex]
- Finding [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (-x)^3 + 5(-x)^2 + (-x) = -x^3 + 5x^2 - x \][/tex]
- Comparing [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]\( f(-x) \neq -f(x) \)[/tex], the function [tex]\( f(x) \)[/tex] is not odd.
2. Function: [tex]\( f(x) = \sqrt{x} \)[/tex]
- Finding [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \sqrt{-x} \][/tex]
- Comparing [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -\sqrt{x} \][/tex]
- Since [tex]\( f(-x) \neq -f(x) \)[/tex] (also note that [tex]\( \sqrt{-x} \)[/tex] is not defined for all real numbers [tex]\( x \)[/tex]), the function [tex]\( f(x) \)[/tex] is not odd.
3. Function: [tex]\( f(x) = x^2 + x \)[/tex]
- Finding [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (-x)^2 + (-x) = x^2 - x \][/tex]
- Comparing [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(x^2 + x) = -x^2 - x \][/tex]
- Since [tex]\( f(-x) \neq -f(x) \)[/tex], the function [tex]\( f(x) \)[/tex] is not odd.
4. Function: [tex]\( f(x) = -x \)[/tex]
- Finding [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = -(-x) = x \][/tex]
- Comparing [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]\( f(x) \)[/tex] is odd.
Based on the analysis, the only odd function among the given functions is [tex]\( f(x) = -x \)[/tex].
Let's go through each function one by one.
1. Function: [tex]\( f(x) = x^3 + 5x^2 + x \)[/tex]
- Finding [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (-x)^3 + 5(-x)^2 + (-x) = -x^3 + 5x^2 - x \][/tex]
- Comparing [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]\( f(-x) \neq -f(x) \)[/tex], the function [tex]\( f(x) \)[/tex] is not odd.
2. Function: [tex]\( f(x) = \sqrt{x} \)[/tex]
- Finding [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \sqrt{-x} \][/tex]
- Comparing [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -\sqrt{x} \][/tex]
- Since [tex]\( f(-x) \neq -f(x) \)[/tex] (also note that [tex]\( \sqrt{-x} \)[/tex] is not defined for all real numbers [tex]\( x \)[/tex]), the function [tex]\( f(x) \)[/tex] is not odd.
3. Function: [tex]\( f(x) = x^2 + x \)[/tex]
- Finding [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (-x)^2 + (-x) = x^2 - x \][/tex]
- Comparing [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(x^2 + x) = -x^2 - x \][/tex]
- Since [tex]\( f(-x) \neq -f(x) \)[/tex], the function [tex]\( f(x) \)[/tex] is not odd.
4. Function: [tex]\( f(x) = -x \)[/tex]
- Finding [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = -(-x) = x \][/tex]
- Comparing [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]\( f(x) \)[/tex] is odd.
Based on the analysis, the only odd function among the given functions is [tex]\( f(x) = -x \)[/tex].
