## srw1496 Group Title Why isn't the inverse of a function always a function? one year ago one year ago

1. ghazi Group Title

okay just suppose that you have a function lets say y= sin x now inverse of it will be $x= \sin^{-1}y$ so does the definition of function applies in inverse, try it ?

2. ghazi Group Title

did you get it?

3. srw1496 Group Title

uhm no,

4. ghazi Group Title

for defining a function it should have one output to the one input but when you will reverse the condition that is inverse , you will have same x for two Ys which contradicts the definition of function

5. ghazi Group Title

and in some cases it is possible the inverse of a function could be a function like if i say y=f(x)=x its inverse could be a function but if i say y= x^3 its a function but inverse is not a function

6. ghazi Group Title

still confused? @srw1496

7. srw1496 Group Title

uh yeah...

8. satellite73 Group Title

i am confused if $$f(x)=x^3$$ then $$f^{-1}(x)=\sqrt[3]{x}$$ which is a function

9. ghazi Group Title

@satellite73 i was missing you and i knew this

10. ghazi Group Title

@srw1496 look fundamentally just check out the definition of function and if the inverse of function gives two values for same input then it is not a function

11. ghazi Group Title

and yes my apology for choosing wrong example

12. satellite73 Group Title

on the other hand, lets look at $$f(x)=x^2$$ here we have $$f(2)=2^2=4$$ and $$f(-2)=(-2)^2=4$$ as well so both $$(-2,4)$$ and $$2,4)$$ are on the graph but if you switch the ordered pairs, then you get both $$(4,-2)$$ and $$(4,2)$$ so the inverse is not a function

13. satellite73 Group Title

the difference is $$x^3$$ is a "one to one" function, which means two different inputs give two different outputs but $$x^2$$ is not a one to one function, different inputs can give the same output

14. ghazi Group Title

y=x^2 is not a function

15. Kainui Group Title

q is not a letter

16. satellite73 Group Title

????!!!!

17. Kainui Group Title

What are you talking about, seriously?

18. Kainui Group Title

First you tell me arcsine isn't a function and now this? lol

19. satellite73 Group Title

this is not a pipe

20. ghazi Group Title

yes plot y=x^2 and use straight line test , straight line cuts the parabola twice so its not a function

21. ghazi Group Title

lets play

22. Kainui Group Title

|dw:1356208624058:dw|

23. malevolence19 Group Title

y=x^2 is a function. x=sqrt(y) is not for all y. Only positive y.

24. ghazi Group Title

25. ghazi Group Title

LOL

26. srw1496 Group Title

uhhh i have no idea what is going on...

27. malevolence19 Group Title

It's not injective. That doesn't mean it isn't a function...

28. ghazi Group Title

we are having useless debate but @srw1496 i think @satellite73 has explained everything that is all you need to know

29. Kainui Group Title

You cut yours the wrong way, kid. Study my picture carefully and you'll see.

30. ghazi Group Title

31. satellite73 Group Title

@srw1496 of course you are confused, this was a silly discussion post again, ignore all this

32. ghazi Group Title

here you go , the first one , or else you can google it @Kainui

33. malevolence19 Group Title

Its as simple as this. For a function you put in an x, you get out a y. The say X will not give you more than 1 Y. To test for it being "one to one" we do the "horizontal line test". So if we look at y=x^2 we get:|dw:1356208787580:dw| Since we know an inverse is just rotating it around the line y=x we have the inverse **would** look like this: |dw:1356208824688:dw| But the vertical line test (how we test if its a function) FAILS! To get around this we usually just define it over a certain domain. So,y=sqrt(x) for positive x.

34. ghazi Group Title

@Kainui this is the another one http://www.sparknotes.com/math/algebra1/quadratics/section1.rhtml , so before having any debate , do know what we are talking about

35. malevolence19 Group Title

for positive y***.

36. dpaInc Group Title

lol...

37. Kainui Group Title

You said y=x^2 is not a function lol

38. ghazi Group Title

yes i said that and i proved it

39. Kainui Group Title

No you proved x=y^2 is not a function

40. ghazi Group Title

no no no dont change the point please, go through the comments and the links i have mentioned , its not a function that is y=x^2 and i am done , thank you :)

41. Kainui Group Title

Lol alright but you're backwards man

42. malevolence19 Group Title

ghazi you're a fluttering idiot. Get out of here. y=x^2 is a function, stop trying to be a fluttering troll.

43. Kainui Group Title

I love how it censors the fluttering words with owl pellet

44. ghazi Group Title

@malevolence19 look dude i am not violent yet and dont provoke me to become violent be like a good kid and again for your satisfaction here is your answer http://answers.yahoo.com/question/index?qid=20071023192855AAs2E9z mind your tongue guys

45. malevolence19 Group Title

AGAIN. THAT IS X=Y^2 NOT Y=X^2 dumbflutter

46. ghazi Group Title

thanks, i am not accepting, do accept what you want

47. malevolence19 Group Title

I get a warning? Really, get this guy out of here who is flat out spreading BS.

48. malevolence19 Group Title

At least my response is right.

49. ghazi Group Title

just plot y=x^2 and see the result , secondly you are not supposed to use bad words

50. malevolence19 Group Title

I have plotted it, hundreds of times, YOU plot it. http://www.wolframalpha.com/input/?i=y%3Dx^2

51. AccessDenied Group Title

My very first note of the thread is seeing that language I have not even read the whole question yet.

52. ghazi Group Title

:P

53. AccessDenied Group Title

It would be correct, however, to say y = x^2 is a function because of the vertical line test. Its inverse, y = + or - sqrt(x), is not a function because it fails the vertical line test. The lines must be vertical because we are only concerned that the function does not output two y-values for any single x-value. Horizontal lines only decide if the inverse is a function.

54. Kainui Group Title

Yeah I would have to agree with @malevolence19 that ghazi is intentionally misleading.

55. ghazi Group Title

alright guys its over, why are you dragging it , it doesnt sound fun to me at all

56. ParthKohli Group Title

As I said in the other one, let$\{(1,2),(2,2),(3,2)\cdots\}$be a function. Then the inverse is$\{(2,1),(2,2),(2,3)\cdots\}$which overrules the definition of a function