## mathmath333 one year ago Graphical functions

1. 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}}

2. anonymous

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

3. mathmath333

|dw:1435239903652:dw|

4. anonymous

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

5. mathmath333

u mean it even function?

6. anonymous

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

7. mathmath333

but it fails the vertical line test i think.

8. mathmath333

@satellite73

9. ikram002p

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

10. mathmath333

|dw:1435240841975:dw|

11. mathmath333

vertical line test verdict : failed.

12. ganeshie8

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

13. mathmath333

book gave it as even gunction

14. ganeshie8

maybe x=-2, 2 are vertical asymptotes ?

15. ganeshie8

thats the only way the textbook answer makes sense i guess

16. 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

17. 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)

18. ikram002p

so that would be half circle

19. ganeshie8

something like this |dw:1435242199499:dw|

20. mathmath333

this is even funxtion

21. ganeshie8

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