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Sagot :
To determine if the given functions are odd, we need to verify if they satisfy the property of an odd function, which is:
[tex]\[ f(-x) = -f(x) \][/tex]
We will check each function one by one to see if this condition holds true.
### Function 1: [tex]\( f(x) = x^3 - x^2 \)[/tex]
1. Substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = (-x)^3 - (-x)^2 = -x^3 - x^2 \][/tex]
2. Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(x^3 - x^2) = -x^3 + x^2 \][/tex]
3. Since [tex]\( f(-x) \)[/tex] is not equal to [tex]\( -f(x) \)[/tex]:
[tex]\[ -x^3 - x^2 \neq -x^3 + x^2 \][/tex]
Therefore, [tex]\( f(x) = x^3 - x^2 \)[/tex] is not an odd function.
### Function 2: [tex]\( f(x) = x^5 - 3x^3 + 2x \)[/tex]
1. Substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = (-x)^5 - 3(-x)^3 + 2(-x) = -x^5 + 3x^3 - 2x \][/tex]
2. Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(x^5 - 3x^3 + 2x) = -x^5 + 3x^3 - 2x \][/tex]
3. Since [tex]\( f(-x) \)[/tex] is equal to [tex]\( -f(x) \)[/tex]:
[tex]\[ -x^5 + 3x^3 - 2x = -x^5 + 3x^3 - 2x \][/tex]
Therefore, [tex]\( f(x) = x^5 - 3x^3 + 2x \)[/tex] is an odd function.
### Function 3: [tex]\( f(x) = 4x + 9 \)[/tex]
1. Substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = 4(-x) + 9 = -4x + 9 \][/tex]
2. Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(4x + 9) = -4x - 9 \][/tex]
3. Since [tex]\( f(-x) \)[/tex] is not equal to [tex]\( -f(x) \)[/tex]:
[tex]\[ -4x + 9 \neq -4x - 9 \][/tex]
Therefore, [tex]\( f(x) = 4x + 9 \)[/tex] is not an odd function.
### Function 4: [tex]\( f(x) = \frac{1}{x} \)[/tex]
1. Substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = \frac{1}{-x} = -\frac{1}{x} \][/tex]
2. Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -\left( \frac{1}{x} \right) = -\frac{1}{x} \][/tex]
3. Since [tex]\( f(-x) \)[/tex] is equal to [tex]\( -f(x) \)[/tex]:
[tex]\[ -\frac{1}{x} = -\frac{1}{x} \][/tex]
Therefore, [tex]\( f(x) = \frac{1}{x} \)[/tex] is an odd function.
### Summary
- [tex]\( f(x) = x^3 - x^2 \)[/tex] is not an odd function.
- [tex]\( f(x) = x^5 - 3x^3 + 2x \)[/tex] is an odd function.
- [tex]\( f(x) = 4x + 9 \)[/tex] is not an odd function.
- [tex]\( f(x) = \frac{1}{x} \)[/tex] is an odd function.
Thus, the functions that are odd are:
[tex]\[ f(x) = x^5 - 3x^3 + 2x \][/tex]
[tex]\[ f(x) = \frac{1}{x} \][/tex]
[tex]\[ f(-x) = -f(x) \][/tex]
We will check each function one by one to see if this condition holds true.
### Function 1: [tex]\( f(x) = x^3 - x^2 \)[/tex]
1. Substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = (-x)^3 - (-x)^2 = -x^3 - x^2 \][/tex]
2. Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(x^3 - x^2) = -x^3 + x^2 \][/tex]
3. Since [tex]\( f(-x) \)[/tex] is not equal to [tex]\( -f(x) \)[/tex]:
[tex]\[ -x^3 - x^2 \neq -x^3 + x^2 \][/tex]
Therefore, [tex]\( f(x) = x^3 - x^2 \)[/tex] is not an odd function.
### Function 2: [tex]\( f(x) = x^5 - 3x^3 + 2x \)[/tex]
1. Substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = (-x)^5 - 3(-x)^3 + 2(-x) = -x^5 + 3x^3 - 2x \][/tex]
2. Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(x^5 - 3x^3 + 2x) = -x^5 + 3x^3 - 2x \][/tex]
3. Since [tex]\( f(-x) \)[/tex] is equal to [tex]\( -f(x) \)[/tex]:
[tex]\[ -x^5 + 3x^3 - 2x = -x^5 + 3x^3 - 2x \][/tex]
Therefore, [tex]\( f(x) = x^5 - 3x^3 + 2x \)[/tex] is an odd function.
### Function 3: [tex]\( f(x) = 4x + 9 \)[/tex]
1. Substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = 4(-x) + 9 = -4x + 9 \][/tex]
2. Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(4x + 9) = -4x - 9 \][/tex]
3. Since [tex]\( f(-x) \)[/tex] is not equal to [tex]\( -f(x) \)[/tex]:
[tex]\[ -4x + 9 \neq -4x - 9 \][/tex]
Therefore, [tex]\( f(x) = 4x + 9 \)[/tex] is not an odd function.
### Function 4: [tex]\( f(x) = \frac{1}{x} \)[/tex]
1. Substitute [tex]\( -x \)[/tex] for [tex]\( x \)[/tex]:
[tex]\[ f(-x) = \frac{1}{-x} = -\frac{1}{x} \][/tex]
2. Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -\left( \frac{1}{x} \right) = -\frac{1}{x} \][/tex]
3. Since [tex]\( f(-x) \)[/tex] is equal to [tex]\( -f(x) \)[/tex]:
[tex]\[ -\frac{1}{x} = -\frac{1}{x} \][/tex]
Therefore, [tex]\( f(x) = \frac{1}{x} \)[/tex] is an odd function.
### Summary
- [tex]\( f(x) = x^3 - x^2 \)[/tex] is not an odd function.
- [tex]\( f(x) = x^5 - 3x^3 + 2x \)[/tex] is an odd function.
- [tex]\( f(x) = 4x + 9 \)[/tex] is not an odd function.
- [tex]\( f(x) = \frac{1}{x} \)[/tex] is an odd function.
Thus, the functions that are odd are:
[tex]\[ f(x) = x^5 - 3x^3 + 2x \][/tex]
[tex]\[ f(x) = \frac{1}{x} \][/tex]
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