Graphical functions

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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

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Other answers:

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

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