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Easyaspi314
 one year ago
Best ResponseYou've already chosen the best response.2As you are correct, x^2  2x has no inverse. BUT, if you restrict the domain, it would have an inverse.

Easyaspi314
 one year ago
Best ResponseYou've already chosen the best response.2What we mean "it has no inverse"...that when you find the inverse, it will not be a function. So we say it has no inverse.

johnny101
 one year ago
Best ResponseYou've already chosen the best response.0Ok thankyou that's what I thought, its a multiplce choice homework but I was confused because it can and cant have one

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0hmmm if you complete the square on the inverse, I do get an inverse function

Easyaspi314
 one year ago
Best ResponseYou've already chosen the best response.2Can you demonstrate that?

johnny101
 one year ago
Best ResponseYou've already chosen the best response.0x^2x2 is a parabola graph therefore passes vertical line test and has inverse?

Easyaspi314
 one year ago
Best ResponseYou've already chosen the best response.2Johnny.......you can find the inverse of x^2  2...no one is doubting that...but the inverse will not be a function..we call that the function not being "invertible".

johnny101
 one year ago
Best ResponseYou've already chosen the best response.0so the question is determine if the function have an inverse f^1, so in that regards it does? Now im confused

Easyaspi314
 one year ago
Best ResponseYou've already chosen the best response.2Johnny...that language is a very touchy area...many textbooks say that it does not have an inverse.....other textbooks say it has an inverse, whose inverse is not a function. It really is a question of symmantics. So a question has to worded very carefully to accomodate the many texts out there.

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0ahemm, I see.... the inverse expression I should call it, will not meet the criterion of a "function" since it'd be a horizontal parabola and thus not a function that will pass the vertical line test

Easyaspi314
 one year ago
Best ResponseYou've already chosen the best response.2But, everyone will agree that if we have a function f(x) = x^2 , and we restrict the domain to the positive real numbers, then the function has an inverse, no if's and's or but's.

Easyaspi314
 one year ago
Best ResponseYou've already chosen the best response.2Its inverse will be + sqrt(x)

johnny101
 one year ago
Best ResponseYou've already chosen the best response.0so for the sake of a multiple choice homework, worded does it contain the function f^1, x^22x does not because it then becomes a horizontal parabola i.e failing the vertical line test?

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0notice the picture, the "red inverse" would not pass the vertical line test

Easyaspi314
 one year ago
Best ResponseYou've already chosen the best response.2I prefer not to answer that question as I do not know which textbook you are using and how the author feels about the issue. But, I can tell you, that on a national exam, etc.. such a question would have to be worded so carefully as to accomodate all textbboks.

Easyaspi314
 one year ago
Best ResponseYou've already chosen the best response.2Why not ask your teacher?

johnny101
 one year ago
Best ResponseYou've already chosen the best response.01 quick question, does it matter if its f(x) = x^22x or f(y)= x^22x?

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf y = x^22x\qquad inverse\implies x = y^22y\\ \quad \\ x = y^22y+11\implies x = (y1)^21\implies \sqrt{x+1}+1=y\)

Easyaspi314
 one year ago
Best ResponseYou've already chosen the best response.2Again, thats a touchy matter...so I prefer to stay away from there; I dont want to mislead you, as I dont know your teachers' preferences, books style, etc.

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.0@johnny101 so though you can simplify it and solve for "y", the resulting expression doesn't not meet the criterion of a "function", thus is not a function per se, thus no inverse \(\bf function\)
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