Proving $f(f^{-1}(x))=f^{-1}(f(x))$

Question

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?

Answer 1

$f(x)=-\sqrt{x-1}\;\;,\qquad f:\left[1,+\infty\right[\to\left]-\infty,0\right]\;.$

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 $\;x\le0\;.$

Answer 2

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 $x\in[1,\infty)$, $$ f^{-1}(f(x))=f^{-1}(-\sqrt{x-1})=(-\sqrt{x-1})^2=x. $$