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steve816

  • one year ago

Determine whether this function is even or odd

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  1. steve816
    • one year ago
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    \[h(x)=\frac{ -x^3 }{ 3x^2-9 }\]

  2. FireKat97
    • one year ago
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    Use these two rules- if h(x) = h(-x) the function is considered even. if -h(x) = h(-x) the function is considered odd.

  3. steve816
    • one year ago
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    I plugged in -x and got neither. Is that right?

  4. FireKat97
    • one year ago
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    so what exactly did you get when you plugged in -x?

  5. steve816
    • one year ago
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    \[h(x)=\frac{ x^3 }{ 3x^2-9 }\]

  6. FireKat97
    • one year ago
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    okay and now try and find the function -h(x)

  7. steve816
    • one year ago
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    \[h(x)=\frac{ -x^3 }{ -3x^2+9 }\]

  8. FireKat97
    • one year ago
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    No, when you multiply by a negative, you don't have to multiply both the numerator and denominator by the negative. So in this case, you should get \[-\frac{ -x^3 }{ 3x^2 -9}\] and the two negatives should cancel so you get left with \[\frac{ x^3 }{ 3x^2-9 }\]

  9. FireKat97
    • one year ago
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    now that you have found h(x) and -h(x) what do you notice

  10. steve816
    • one year ago
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    oooooh silly mistake, so it is ODD!

  11. FireKat97
    • one year ago
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    yup :D

  12. steve816
    • one year ago
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    THanks for the help

  13. FireKat97
    • one year ago
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    no problem

  14. ytrewqmiswi
    • one year ago
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