Question

Help!!!!!!!!!!
Find the inverse of f(x)= x^2 over x^2 +1 and show that f(f^-1(x))=x

Answer 1

unlikely that this function has an inverse function, because it is not one to one
on the other hand, you can get a choice by solving
\[x=\frac{y^2}{y^2+1}\] for \(y\)

Answer 2

damn typo
start with
\[x(y^2+1)=y^2\] then
\[xy^2+x=y^2\] so
\[xy^2-y^2=-x\] factor to get
\[(x-1)y^2=-x\] so
\[y^2=\frac{-x}{x-1}=\frac{x}{1-x}\]

Answer 3

but that is \(y^2\) not \(y\) so you need
\[y=\pm\sqrt{\frac{x}{1-x}}\]
the plus or minus part means it is not a function