Question

Graphical functions

Answer 1

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

\(\large \color{black}{\begin{align} \normalsize \text{}\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

i guess we are missing \(f(x)\)

Answer 3

looks symmetric wrt the y axis to me, how about you?

Answer 4

u mean it even function?

Answer 5

yes, if it is symmetric wrt the y axis, it is even

Answer 6

but it fails the vertical line test i think.

Answer 7

@satellite73

Answer 8

why u think it fails ? like its not 1-1 function also not onto but u can see the symmetric

Answer 9

vertical line test
verdict : failed.

Answer 10

yeah the graph fails to be a function x=-2, 2

Answer 11

book gave it as even gunction

Answer 12

maybe x=-2, 2 are vertical asymptotes ?

Answer 13

thats the only way the textbook answer makes sense i guess

Answer 14

huh dont use vertical line test its hmmm not our point the function itself is not a function but indeed its an even relation

Answer 15

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)

Answer 16

so that would be half circle

Answer 17

something like this

Answer 18

this is even funxtion

Answer 19

Yes the graph that you showed must be like this to be called a "function"