Question

Find the inverse of x/sqrtx^2+7

Answer 1

\[\frac{ x }{ \sqrt{x ^{2}+7}}\]

Answer 2

seems unlikely that this is a one to one function, so it will not have an inverse

Answer 3

I know, it looks weird but apparently there is an inverse

Answer 4

apparently the inverse is \[\frac{ \sqrt{7x} }{ \sqrt{1-x^{2}} }\]

Answer 5

well i assure you it does not
on the other hand you can write
\[x=\frac{y}{\sqrt{y^2+7}}\] and solve for \(y\)

Answer 6

we can do it if you like

Answer 7

that's what I did

Answer 8

still couldn't find it

Answer 9

nevermind, I got it lol

Answer 10

i skipped a step in my head so i couldn't find out what i did wrong on the paper

Answer 11

\[x=\frac{y}{\sqrt{y^2+7}}\]
\[x\sqrt{y^2+7}=y\] square both sides
\[x^2(y^2+7)=y^2\] distribute
\[x^2y^2+7x^2=y^2\] put all terms with \(y\) on one side, all else on the other
\[x^2y^2-y^2=-7x^2\] factor
\[y^2(x^2-1)=-7x^2\] divide
\[y^2=\frac{-7x^2}{x^2-1}\]

Answer 12

or if you prefer
\[y^2=\frac{7x^2}{1-x^2}\] but now how are you going to solve that for \(y\)??

Answer 13

yeah i see it now, thanks for the help

Answer 14

square root both sides

Answer 15

you get
\[y=\pm\sqrt{\frac{7x^2}{1-x^2}}\] which is certainly not a function

Answer 16

don't forget the \(\pm\)

Answer 17

well you could actually get \[\frac{ \sqrt{7x} }{ \sqrt{1-x ^{2}} }\] which is a function