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mathmath333

  • one year ago

Graphical functions

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} &a.) \ f(x)=-f(x) \hspace{.33em}\\~\\ &b.) \ f(x)=f(-x) \hspace{.33em}\\~\\ &c.) \ \normalsize \text{neither even nor odd function} \hspace{.33em}\\~\\ &d.) \ f(x)\ \normalsize \text{doesn't exist at atleast one point of the domain.} \hspace{.33em}\\~\\ \end{align}}\) \(\large \color{black}{\begin{align} \normalsize \text{}\hspace{.33em}\\~\\ \end{align}}\)

  2. anonymous
    • one year ago
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    i guess we are missing \(f(x)\)

  3. mathmath333
    • one year ago
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    |dw:1435239903652:dw|

  4. anonymous
    • one year ago
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    looks symmetric wrt the y axis to me, how about you?

  5. mathmath333
    • one year ago
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    u mean it even function?

  6. anonymous
    • one year ago
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    yes, if it is symmetric wrt the y axis, it is even

  7. mathmath333
    • one year ago
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    but it fails the vertical line test i think.

  8. mathmath333
    • one year ago
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    @satellite73

  9. ikram002p
    • one year ago
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    why u think it fails ? like its not 1-1 function also not onto but u can see the symmetric

  10. mathmath333
    • one year ago
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    |dw:1435240841975:dw|

  11. mathmath333
    • one year ago
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    vertical line test verdict : failed.

  12. ganeshie8
    • one year ago
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    yeah the graph fails to be a function x=-2, 2

  13. mathmath333
    • one year ago
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    book gave it as even gunction

  14. ganeshie8
    • one year ago
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    maybe x=-2, 2 are vertical asymptotes ?

  15. ganeshie8
    • one year ago
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    thats the only way the textbook answer makes sense i guess

  16. ikram002p
    • one year ago
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    huh dont use vertical line test its hmmm not our point the function itself is not a function but indeed its an even relation

  17. ganeshie8
    • one year ago
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    then the vertical line test passes because that x=2, -2 lines don't really exist, think that the graph is defined only in the interval (-2, 2)

  18. ikram002p
    • one year ago
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    so that would be half circle

  19. ganeshie8
    • one year ago
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    something like this |dw:1435242199499:dw|

  20. mathmath333
    • one year ago
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    this is even funxtion

  21. ganeshie8
    • one year ago
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    Yes the graph that you showed must be like this to be called a "function"

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