Proving $f(f^{-1}(x))=f^{-1}(f(x))$ I have to prove $f^{-1}(f(x))=f(f^{-1}(x))$ for the following function :
$$f(x)=-\sqrt{x-1}$$
But I got $-x$ on the one side and $x$ on the other.
What am I missing here?
 A: Actually,
$$f(f^{-1}(x))=f^{-1}(f(x))$$
doesn't make sense. The domain and the range of $f(x)$ are $[1,\infty)$ and $(-\infty,0]$, respectively. So its inverse is
$$ f^{-1}(x)=x^2+1 $$
whose domain and range are $(-\infty,0]$ and $[1,\infty)$. For $x\in(-\infty,0]$, $\sqrt{x^2}=-x$ and hence
$$ f(f^{-1}(x))=f(x^2+1)=-\sqrt{x^2}=-(-x)=x. $$
For $x\in[1,\infty)$,
$$ f^{-1}(f(x))=f^{-1}(-\sqrt{x-1})=(-\sqrt{x-1})^2=x. $$
A: $f(x)=-\sqrt{x-1}\;\;,\qquad f:\left[1,+\infty\right[\to\left]-\infty,0\right]\;.$
In order to obtain the inverse function $\;f^{-1}(x)\;,\;$ we have to solve the following equation for $\;x\in\left[1,+\infty\right[\;$ and then substitute $\;x\;$ for $\;f^{-1}(x)\;$ and $\;y\;$ for $\;x\;.$
$y=-\sqrt{x-1}$
$y^2=x-1$
$x=y^2+1$
Hence
$f^{-1}(x)=x^2+1\;\;,\qquad f^{-1}:\left]-\infty,0\right]\to\left[1,+\infty\right[\;.$
Consequently we get that
\begin{align}
f^{-1}\left(f(x)\right)&=f^2(x)+1=\left(-\sqrt{x-1}\right)^2+1=\left|x-1\right|+1=\\
&=x-1+1=x
\end{align}
for all $\;x\in\left[1,+\infty\right[\;.$
Moreover
\begin{align}
f\left(f^{-1}(x)\right)&=-\sqrt{f^{-1}(x)-1}=-\sqrt{x^2+1-1}=-\sqrt{x^2}=\\
&=-|x|=x
\end{align}
for all $\;x\in\left]-\infty,0\right]\;.$
Therefore
$f^{-1}\left(f(x)\right)=x\;\;$ for all $\;x\in\left[1,+\infty\right[\;,$
$f\left(f^{-1}(x)\right)=x\;\;$ for all $\;x\in\left]-\infty,0\right]\;.$
Nevertheless there does not exist any $\;x\in\left]-\infty,+\infty\right[\;$ such that $\;f^{-1}\left(f(x)\right)=f\left(f^{-1}(x)\right)\;$ indeed $\;f^{-1}\left(f(x)\right)\;$ is defined for $\;x\ge1\;,\;$ where as $\;f\left(f^{-1}(x)\right)\;$ is defined for $\;x\le0\;.$
