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anonymous
 3 years ago
Why isn't the inverse of a function always a function?
anonymous
 3 years ago
Why isn't the inverse of a function always a function?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay 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 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0still confused? @srw1496

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i am confused if \(f(x)=x^3\) then \(f^{1}(x)=\sqrt[3]{x}\) which is a function

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@satellite73 i was missing you and i knew this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and yes my apology for choosing wrong example

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0on 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y=x^2 is not a function

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0What are you talking about, seriously?

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0First you tell me arcsine isn't a function and now this? lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes plot y=x^2 and use straight line test , straight line cuts the parabola twice so its not a function

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y=x^2 is a function. x=sqrt(y) is not for all y. Only positive y.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356208670981:dw function @Kainui dad

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0uhhh i have no idea what is going on...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's not injective. That doesn't mean it isn't a function...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we are having useless debate but @srw1496 i think @satellite73 has explained everything that is all you need to know

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0You cut yours the wrong way, kid. Study my picture carefully and you'll see.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay sir, i am upload plenty of stuff for your convenience

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@srw1496 of course you are confused, this was a silly discussion post again, ignore all this

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0here you go , the first one , or else you can google it @Kainui

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Its 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0You said y=x^2 is not a function lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes i said that and i proved it

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0No you proved x=y^2 is not a function

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no 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 :)

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0Lol alright but you're backwards man

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ghazi you're a fluttering idiot. Get out of here. y=x^2 is a function, stop trying to be a fluttering troll.

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0I love how it censors the fluttering words with owl pellet

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0AGAIN. THAT IS X=Y^2 NOT Y=X^2 dumbflutter

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks, i am not accepting, do accept what you want

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I get a warning? Really, get this guy out of here who is flat out spreading BS.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0At least my response is right.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just plot y=x^2 and see the result , secondly you are not supposed to use bad words

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have plotted it, hundreds of times, YOU plot it. http://www.wolframalpha.com/input/?i=y%3Dx^2

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.0My very first note of the thread is seeing that language I have not even read the whole question yet.

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.0It 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 yvalues for any single xvalue. Horizontal lines only decide if the inverse is a function.

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah I would have to agree with @malevolence19 that ghazi is intentionally misleading.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright guys its over, why are you dragging it , it doesnt sound fun to me at all

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.0As 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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