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ghaziBest ResponseYou've already chosen the best response.1
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 ?
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
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
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
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
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
still confused? @srw1496
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
i am confused if \(f(x)=x^3\) then \(f^{1}(x)=\sqrt[3]{x}\) which is a function
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
@satellite73 i was missing you and i knew this
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
@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
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
and yes my apology for choosing wrong example
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
What are you talking about, seriously?
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
First you tell me arcsine isn't a function and now this? lol
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
this is not a pipe
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
yes plot y=x^2 and use straight line test , straight line cuts the parabola twice so its not a function
 one year ago

malevolence19Best ResponseYou've already chosen the best response.3
y=x^2 is a function. x=sqrt(y) is not for all y. Only positive y.
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
dw:1356208670981:dw function @Kainui dad
 one year ago

srw1496Best ResponseYou've already chosen the best response.0
uhhh i have no idea what is going on...
 one year ago

malevolence19Best ResponseYou've already chosen the best response.3
It's not injective. That doesn't mean it isn't a function...
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
we are having useless debate but @srw1496 i think @satellite73 has explained everything that is all you need to know
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
You cut yours the wrong way, kid. Study my picture carefully and you'll see.
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
okay sir, i am upload plenty of stuff for your convenience
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
@srw1496 of course you are confused, this was a silly discussion post again, ignore all this
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
here you go , the first one , or else you can google it @Kainui
 one year ago

malevolence19Best ResponseYou've already chosen the best response.3
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.
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
@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
 one year ago

malevolence19Best ResponseYou've already chosen the best response.3
for positive y***.
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
You said y=x^2 is not a function lol
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
yes i said that and i proved it
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
No you proved x=y^2 is not a function
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
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 :)
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
Lol alright but you're backwards man
 one year ago

malevolence19Best ResponseYou've already chosen the best response.3
ghazi you're a fluttering idiot. Get out of here. y=x^2 is a function, stop trying to be a fluttering troll.
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
I love how it censors the fluttering words with owl pellet
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
@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
 one year ago

malevolence19Best ResponseYou've already chosen the best response.3
AGAIN. THAT IS X=Y^2 NOT Y=X^2 dumbflutter
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
thanks, i am not accepting, do accept what you want
 one year ago

malevolence19Best ResponseYou've already chosen the best response.3
I get a warning? Really, get this guy out of here who is flat out spreading BS.
 one year ago

malevolence19Best ResponseYou've already chosen the best response.3
At least my response is right.
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
just plot y=x^2 and see the result , secondly you are not supposed to use bad words
 one year ago

malevolence19Best ResponseYou've already chosen the best response.3
I have plotted it, hundreds of times, YOU plot it. http://www.wolframalpha.com/input/?i=y%3Dx^2
 one year ago

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

AccessDeniedBest ResponseYou've already chosen the best response.0
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 yvalues for any single xvalue. Horizontal lines only decide if the inverse is a function.
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
Yeah I would have to agree with @malevolence19 that ghazi is intentionally misleading.
 one year ago

ghaziBest ResponseYou've already chosen the best response.1
alright guys its over, why are you dragging it , it doesnt sound fun to me at all
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
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
 one year ago
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