Which functions are odd? Choose two correct answers.

[tex]\[
\begin{array}{c}
A. \ f(x)=4x+9 \\
B. \ f(x)=x^5-3x^2+2x \\
C. \ f(x)=\frac{1}{x} \\
D. \ f(x)=x^3-x^2
\end{array}
\][/tex]



Answer :

To determine whether the given functions are odd, we need to verify if they satisfy the property of odd functions: [tex]\( f(-x) = -f(x) \)[/tex]. Let's analyze each function step-by-step.

1. For [tex]\( f(x) = 4x + 9 \)[/tex]:
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = 4(-x) + 9 = -4x + 9 \][/tex]
- Compare it with [tex]\(-f(x)\)[/tex]:
[tex]\[ -f(x) = - (4x + 9) = -4x - 9 \][/tex]
- Since [tex]\( f(-x) \neq -f(x) \)[/tex], the function [tex]\( f(x) = 4x + 9 \)[/tex] is not odd.

2. For [tex]\( f(x) = x^5 - 3x^2 + 2x \)[/tex]:
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (-x)^5 - 3(-x)^2 + 2(-x) = -x^5 - 3x^2 - 2x \][/tex]
- Compare it with [tex]\(-f(x)\)[/tex]:
[tex]\[ -f(x) = - (x^5 - 3x^2 + 2x) = -x^5 + 3x^2 - 2x \][/tex]
- Since [tex]\( f(-x) \neq -f(x) \)[/tex], the function [tex]\( f(x) = x^5 - 3x^2 + 2x \)[/tex] is not odd.

3. For [tex]\( f(x) = \frac{1}{x} \)[/tex]:
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = \frac{1}{-x} = -\frac{1}{x} \][/tex]
- Compare it with [tex]\(-f(x)\)[/tex]:
[tex]\[ -f(x) = -\left(\frac{1}{x}\right) = -\frac{1}{x} \][/tex]
- Since [tex]\( f(-x) = -f(x) \)[/tex], the function [tex]\( f(x) = \frac{1}{x} \)[/tex] is odd.

4. For [tex]\( f(x) = x^3 - x^2 \)[/tex]:
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = (-x)^3 - (-x)^2 = -x^3 - x^2 \][/tex]
- Compare it with [tex]\(-f(x)\)[/tex]:
[tex]\[ -f(x) = - (x^3 - x^2) = -x^3 + x^2 \][/tex]
- Since [tex]\( f(-x) \neq -f(x) \)[/tex], the function [tex]\( f(x) = x^3 - x^2 \)[/tex] is not odd.

Given the results of our analysis, only the function [tex]\( f(x) = \frac{1}{x} \)[/tex] is odd.

So the correct answer should be:
```
[3]
```