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ryoblck

  • 2 years ago

7. Which if the following is always true of odd functions? I. f (−x) = −f(x) II. f(|x|) is even III. |f(x)| is even A. All of these are true. B. None of these are true. C. I only. D. II only. E. I and III only.

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  1. Mr.Math
    • 2 years ago
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    What do you think?

  2. ryoblck
    • 2 years ago
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    I think all of them are true but not too sure. I get confused a lot with these odd and even function questions.

  3. Mr.Math
    • 2 years ago
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    I will help you out. The definition of an odd function only tells us that a function \(f(x)\) is odd if it satisfies the property: \(f(-x)=-f(x)\), for all x in the domain of \(f\). Now we know that "I" is true, and we should use to make a conclusion about the other two statements. Would you like to try?

  4. ryoblck
    • 2 years ago
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    Well wouldn't it still be positive because of the absolute values? And if it is not negative then it is not an odd function.

  5. Mr.Math
    • 2 years ago
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    In II, we will only have \(f(|-x|)=f(|x|)=-f(|x|)\), and we can't say it's even. However in III, we have \(|f(-x)|=|-f(x)|=|f(x)|\), and therefore it's even.

  6. ryoblck
    • 2 years ago
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    Wait why is that? Because once we put the negative within the absolute values in "II", it should turn positive right?

  7. Mr.Math
    • 2 years ago
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    Yes; \(|-x|=|x|\).

  8. ryoblck
    • 2 years ago
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    Then how is the outcome still negative?

  9. Mr.Math
    • 2 years ago
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    It can be. Suppose you have \(f(x)=-x^3\). Then \(f(|-x|)=-(|-x|)^3=-x^3\). Right?

  10. ryoblck
    • 2 years ago
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    Oh so it doesn't matter if the absolute value only covers x? It can still be negative. But if the absolute value covers the whole function, then it will be positive? Correct?

  11. Mr.Math
    • 2 years ago
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    Yes.

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