IDNLearn.com: Where curiosity meets clarity and questions find their answers. Join our platform to receive prompt and accurate responses from experienced professionals in various fields.
Sagot :
To determine which of the given functions is an odd function, we need to verify whether [tex]\( f(-x) = -f(x) \)[/tex] for each function.
An odd function must satisfy the condition [tex]\( f(-x) = -f(x) \)[/tex].
Let's analyze each function step-by-step:
1. Function: [tex]\( f(x) = 3x^2 + x \)[/tex]
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = 3(-x)^2 + (-x) = 3x^2 - x \][/tex]
- Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(3x^2 + x) = -3x^2 - x \][/tex]
- Conclusion:
[tex]\[ f(-x) = 3x^2 - x \quad \text{is not equal to} \quad -f(x) = -3x^2 - x \][/tex]
So, [tex]\( f(x) = 3x^2 + x \)[/tex] is not an odd function.
2. Function: [tex]\( f(x) = 4x^3 + 7 \)[/tex]
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = 4(-x)^3 + 7 = -4x^3 + 7 \][/tex]
- Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(4x^3 + 7) = -4x^3 - 7 \][/tex]
- Conclusion:
[tex]\[ f(-x) = -4x^3 + 7 \quad \text{is not equal to} \quad -f(x) = -4x^3 - 7 \][/tex]
So, [tex]\( f(x) = 4x^3 + 7 \)[/tex] is not an odd function.
3. Function: [tex]\( f(x) = 5x^2 + 9 \)[/tex]
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = 5(-x)^2 + 9 = 5x^2 + 9 \][/tex]
- Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(5x^2 + 9) = -5x^2 - 9 \][/tex]
- Conclusion:
[tex]\[ f(-x) = 5x^2 + 9 \quad \text{is not equal to} \quad -f(x) = -5x^2 - 9 \][/tex]
So, [tex]\( f(x) = 5x^2 + 9 \)[/tex] is not an odd function.
4. Function: [tex]\( f(x) = 6x^3 + 2x \)[/tex]
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = 6(-x)^3 + 2(-x) = -6x^3 - 2x \][/tex]
- Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(6x^3 + 2x) = -6x^3 - 2x \][/tex]
- Conclusion:
[tex]\[ f(-x) = -6x^3 - 2x \quad \text{is equal to} \quad -f(x) = -6x^3 - 2x \][/tex]
So, [tex]\( f(x) = 6x^3 + 2x \)[/tex] is an odd function.
Based on the analysis, the odd function among the given options is:
[tex]\[ f(x) = 6x^3 + 2x \][/tex]
An odd function must satisfy the condition [tex]\( f(-x) = -f(x) \)[/tex].
Let's analyze each function step-by-step:
1. Function: [tex]\( f(x) = 3x^2 + x \)[/tex]
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = 3(-x)^2 + (-x) = 3x^2 - x \][/tex]
- Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(3x^2 + x) = -3x^2 - x \][/tex]
- Conclusion:
[tex]\[ f(-x) = 3x^2 - x \quad \text{is not equal to} \quad -f(x) = -3x^2 - x \][/tex]
So, [tex]\( f(x) = 3x^2 + x \)[/tex] is not an odd function.
2. Function: [tex]\( f(x) = 4x^3 + 7 \)[/tex]
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = 4(-x)^3 + 7 = -4x^3 + 7 \][/tex]
- Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(4x^3 + 7) = -4x^3 - 7 \][/tex]
- Conclusion:
[tex]\[ f(-x) = -4x^3 + 7 \quad \text{is not equal to} \quad -f(x) = -4x^3 - 7 \][/tex]
So, [tex]\( f(x) = 4x^3 + 7 \)[/tex] is not an odd function.
3. Function: [tex]\( f(x) = 5x^2 + 9 \)[/tex]
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = 5(-x)^2 + 9 = 5x^2 + 9 \][/tex]
- Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(5x^2 + 9) = -5x^2 - 9 \][/tex]
- Conclusion:
[tex]\[ f(-x) = 5x^2 + 9 \quad \text{is not equal to} \quad -f(x) = -5x^2 - 9 \][/tex]
So, [tex]\( f(x) = 5x^2 + 9 \)[/tex] is not an odd function.
4. Function: [tex]\( f(x) = 6x^3 + 2x \)[/tex]
- Evaluate [tex]\( f(-x) \)[/tex]:
[tex]\[ f(-x) = 6(-x)^3 + 2(-x) = -6x^3 - 2x \][/tex]
- Compare [tex]\( f(-x) \)[/tex] with [tex]\( -f(x) \)[/tex]:
[tex]\[ -f(x) = -(6x^3 + 2x) = -6x^3 - 2x \][/tex]
- Conclusion:
[tex]\[ f(-x) = -6x^3 - 2x \quad \text{is equal to} \quad -f(x) = -6x^3 - 2x \][/tex]
So, [tex]\( f(x) = 6x^3 + 2x \)[/tex] is an odd function.
Based on the analysis, the odd function among the given options is:
[tex]\[ f(x) = 6x^3 + 2x \][/tex]
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find precise solutions at IDNLearn.com. Thank you for trusting us with your queries, and we hope to see you again.
