Find f−1(x)f^{-1}(x)f−1(x) if f(x)=\acx+1x−1f(x) = \ac{x+1}{x-1}f(x)=\acx+1x−1 (assume xeq1x eq 1xeq1).
f−1(x)=f(x)f^{-1}(x) = f(x)f−1(x)=f(x)
f−1(x)=\acx−1x+1f^{-1}(x) = \ac{x-1}{x+1}f−1(x)=\acx−1x+1
f−1(x)=1/xf^{-1}(x) = 1/xf−1(x)=1/x
No inverse