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srw1496

  • 3 years ago

Why isn't the inverse of a function always a function?

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  1. ghazi
    • 3 years ago
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    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
    • 3 years ago
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    did you get it?

  3. srw1496
    • 3 years ago
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    uhm no,

  4. ghazi
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    still confused? @srw1496

  7. srw1496
    • 3 years ago
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    uh yeah...

  8. anonymous
    • 3 years ago
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    i am confused if \(f(x)=x^3\) then \(f^{-1}(x)=\sqrt[3]{x}\) which is a function

  9. ghazi
    • 3 years ago
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    @satellite73 i was missing you and i knew this

  10. ghazi
    • 3 years ago
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    @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
    • 3 years ago
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    and yes my apology for choosing wrong example

  12. anonymous
    • 3 years ago
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    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. anonymous
    • 3 years ago
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    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
    • 3 years ago
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    y=x^2 is not a function

  15. Kainui
    • 3 years ago
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    q is not a letter

  16. anonymous
    • 3 years ago
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    ????!!!!

  17. Kainui
    • 3 years ago
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    What are you talking about, seriously?

  18. Kainui
    • 3 years ago
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    First you tell me arcsine isn't a function and now this? lol

  19. anonymous
    • 3 years ago
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    this is not a pipe

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  20. ghazi
    • 3 years ago
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    yes plot y=x^2 and use straight line test , straight line cuts the parabola twice so its not a function

  21. ghazi
    • 3 years ago
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    lets play

  22. Kainui
    • 3 years ago
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    |dw:1356208624058:dw|

  23. malevolence19
    • 3 years ago
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    y=x^2 is a function. x=sqrt(y) is not for all y. Only positive y.

  24. ghazi
    • 3 years ago
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    |dw:1356208670981:dw| function @Kainui dad

  25. ghazi
    • 3 years ago
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    LOL

  26. srw1496
    • 3 years ago
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    uhhh i have no idea what is going on...

  27. malevolence19
    • 3 years ago
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    It's not injective. That doesn't mean it isn't a function...

  28. ghazi
    • 3 years ago
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    we are having useless debate but @srw1496 i think @satellite73 has explained everything that is all you need to know

  29. Kainui
    • 3 years ago
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    You cut yours the wrong way, kid. Study my picture carefully and you'll see.

  30. ghazi
    • 3 years ago
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    okay sir, i am upload plenty of stuff for your convenience

  31. anonymous
    • 3 years ago
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    @srw1496 of course you are confused, this was a silly discussion post again, ignore all this

  32. ghazi
    • 3 years ago
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    here you go , the first one , or else you can google it @Kainui

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  33. malevolence19
    • 3 years ago
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    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
    • 3 years ago
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    @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
    • 3 years ago
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    for positive y***.

  36. dpaInc
    • 3 years ago
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    lol...

  37. Kainui
    • 3 years ago
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    You said y=x^2 is not a function lol

  38. ghazi
    • 3 years ago
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    yes i said that and i proved it

  39. Kainui
    • 3 years ago
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    No you proved x=y^2 is not a function

  40. ghazi
    • 3 years ago
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    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
    • 3 years ago
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    Lol alright but you're backwards man

  42. malevolence19
    • 3 years ago
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    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
    • 3 years ago
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    I love how it censors the fluttering words with owl pellet

  44. ghazi
    • 3 years ago
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    @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
    • 3 years ago
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    AGAIN. THAT IS X=Y^2 NOT Y=X^2 dumbflutter

  46. ghazi
    • 3 years ago
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    thanks, i am not accepting, do accept what you want

  47. malevolence19
    • 3 years ago
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    I get a warning? Really, get this guy out of here who is flat out spreading BS.

  48. malevolence19
    • 3 years ago
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    At least my response is right.

  49. ghazi
    • 3 years ago
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    just plot y=x^2 and see the result , secondly you are not supposed to use bad words

  50. malevolence19
    • 3 years ago
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    I have plotted it, hundreds of times, YOU plot it. http://www.wolframalpha.com/input/?i=y%3Dx^2

  51. AccessDenied
    • 3 years ago
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    My very first note of the thread is seeing that language I have not even read the whole question yet.

  52. ghazi
    • 3 years ago
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    :P

  53. AccessDenied
    • 3 years ago
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    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
    • 3 years ago
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    Yeah I would have to agree with @malevolence19 that ghazi is intentionally misleading.

  55. ghazi
    • 3 years ago
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    alright guys its over, why are you dragging it , it doesnt sound fun to me at all

  56. ParthKohli
    • 3 years ago
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    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

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