How $g(f(x))=f(g(x))=x$ makes $f(x)$ and $g(f(x))$ inverse of each other?

Question

If f(x) and g(x) are two functions, they would be inverse of each other when the following holds,$$ g(f(x))=f(g(x))=x $$ If we pick $f(x) = x+5$ and $g(x)= x-5$ we can see the truth of this fact. But I am thinking for some simple linguistic explanation of why actually this equation holds for all inverse functions.

Answer 1

Just imagine there is an arrow (taken by $f$) from $x$ to $f(x)$, as $g(f(x))=x$, this means that there is an arrow (taken by $g$) from $f(x)$ to $x$, the arrow by $g$ is just a reverse arrow taken by $f$, so can you see that at the end $x$ stays at the original place?

Answer 2

By definition of $f^{-1}$ we have:$$y= f(x)\Longleftrightarrow x =f^{-1}(y)$$

Since we have $g(f(x))=x$ we have $g(y)=x$ which makes $g=f^{-1}$

Answer 3

You're simply showing that the function g takes the output y of f(x) and gives back the input x.

Answer 4

Worth a bit of caution about, for example, trigonometric functions, when the nice relation needs some adjustment when $x$ is outside some standard interval.

When $- \pi/2 < x < \pi/2,$ we do get $\arctan \tan x = x.$ However, for $ \pi/2 < x < 3 \pi/2,$ we do get $\arctan \tan x = x - \pi.$ So, for instance, $\tan 5 \pi / 4 = 1.$ But $\arctan 1 = \pi/4 = 5\pi/4 - \pi$

Most important, fairly standard, functions, have some such difficulty. The pair $e^x, \log x$ are unusually well behaved

Answer 5

We have $f(f^{-1}(x))=x$, substitute $ x \rightarrow f(x)$, so $f(f^{-1}(f(x)))=f(x)$ now apply $f^{-1}$ to this \begin{eqnarray*} f^{-1}(f(\color{red}{f^{-1}(f(x))}))= \color{red}{f^{-1}(f(x))} \end{eqnarray*} Now replace $ f^{-1}(f(y)) \rightarrow y$ and we have $ f^{-1}(f(y))=y$.