mathmath333
  • mathmath333
Graphical functions
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} &a.) \ f(x)=-f(x) \hspace{.33em}\\~\\ &b.) \ f(x)=f(-x) \hspace{.33em}\\~\\ &c.) \ \normalsize \text{neither even nor odd function} \hspace{.33em}\\~\\ &d.) \ f(x)\ \normalsize \text{doesn't exist at atleast one point of the domain.} \hspace{.33em}\\~\\ \end{align}}\) \(\large \color{black}{\begin{align} \normalsize \text{}\hspace{.33em}\\~\\ \end{align}}\)
anonymous
  • anonymous
i guess we are missing \(f(x)\)
mathmath333
  • mathmath333
|dw:1435239903652:dw|

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anonymous
  • anonymous
looks symmetric wrt the y axis to me, how about you?
mathmath333
  • mathmath333
u mean it even function?
anonymous
  • anonymous
yes, if it is symmetric wrt the y axis, it is even
mathmath333
  • mathmath333
but it fails the vertical line test i think.
mathmath333
  • mathmath333
@satellite73
ikram002p
  • ikram002p
why u think it fails ? like its not 1-1 function also not onto but u can see the symmetric
mathmath333
  • mathmath333
|dw:1435240841975:dw|
mathmath333
  • mathmath333
vertical line test verdict : failed.
ganeshie8
  • ganeshie8
yeah the graph fails to be a function x=-2, 2
mathmath333
  • mathmath333
book gave it as even gunction
ganeshie8
  • ganeshie8
maybe x=-2, 2 are vertical asymptotes ?
ganeshie8
  • ganeshie8
thats the only way the textbook answer makes sense i guess
ikram002p
  • ikram002p
huh dont use vertical line test its hmmm not our point the function itself is not a function but indeed its an even relation
ganeshie8
  • ganeshie8
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)
ikram002p
  • ikram002p
so that would be half circle
ganeshie8
  • ganeshie8
something like this |dw:1435242199499:dw|
mathmath333
  • mathmath333
this is even funxtion
ganeshie8
  • ganeshie8
Yes the graph that you showed must be like this to be called a "function"

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