If $g$ is the inverse function of $f$, then prove that $f(g(x)) = x$ 
Problem : If $g$ is the inverse function of $f$, then prove that $f(g(x)) = x$.
Solution :

I know it is well known result.
But I have no idea how to prove it.
 A: I think we all need some enlightenment on what it means to have an inverse for functions. 
Let $f : X \to Y$ be a function defined by some expression like $x \mapsto f(x)$ (for instance, when $X=Y= \mathbb R$, we could let $x \mapsto f(x) = 2x+1$). 
This function $f$ is said to have a left inverse if there exists a function $g : Y \to X$ such that for every $x \in X$, $g(f(x)) = x$. Similarly, $f$ has a right inverse if there exists $g : Y \to X$ such that for all $y \in Y$, $f(g(y)) = y$. 
If we take this definition as the inverse, then if $f$ has a left inverse $g$, if $f(x) = y$, we have $x = g(f(x)) = g(y)$. If $f$ has a right inverse, then if $g(y) = x$, we have $f(g(y)) = f(x) = y$. 
Therefore the fact that $f(x)=y \quad \Longleftrightarrow \quad g(y) = x$ is true if and only if $g$ is a left and right inverse for $f$. 
If $g$ is simultaneously a left AND right inverse, we can say that $g$ is the inverse of $f$ and denote it by $f^{-1}$ because $f^{-1}(f(x)) = x$ for all $x \in X$ and $f(f^{-1}(y)) = y$ for every $y \in Y$. Let me give some examples to be clearer. 
Consider $f : \mathbb R \to [0,\infty[$ defined by $f(x) = x^2$. If $g : [0,\infty[ \to \mathbb R$ is defined by $g(y) = \sqrt y$, then $g$ is a right inverse for $f$, because $f(g(y)) = f(\sqrt{y}) = \sqrt{y}^2 = y$ (because we assume $y \ge 0$ since this is the domain of $g$). Notice that $g$ is not a left inverse for $f$ because $g(f(x)) = \sqrt{x^2} = |x| \neq x$ if $x$ is negative in the domain of $f$.
However, if we look at the function $f : [0,\infty[ \to [0,\infty[$ defined by $f(x) = x^2$ (note that the definition of the map is the same, but the domain is different!), then the function $g : [0,\infty[ \to [0, \infty[$ defined by $g(y) = \sqrt y$ is a left and right inverse for $f$, because then for every positive $x$, $\sqrt{x^2} = x$ and for every positive $y$, $(\sqrt y)^2 = y$.
Feel free to ask any more questions.
Hope that helps,
A: Functions $f(x)$ and $g(x)$ are inverses of one another if:
$f(g(x)) = x$   and   $g(f(x)) = x,$
for all values of $x$ in their respective domains.
The above property is what inverse functions are defined to be. I don't think a proof is needed for the definition.  
A: If $g$ is inverse function of $f$
Then $f(x)=y$ and $g(y)=x$ 
So $g(f(x))=g(y)=x$
Another Way
If $g$ is inverse function of $f$
Then $f(x)=y$ and $g(y)=x$ 
So $f(g(y))=f(x)=y$
or $f(g(x))=x$
A: Take an easy function: $f(x) = 2x + 1$. 
the inverse of y = 2x + 1 is equal to $$g(x)= (x-1)/2$$
so $f(g(x))$ = 
$    2(g(x)) + 1 = f(g(x)) therefore, $2((x-1)/2) + 1 = $f(g(x))$
if you multiply the first part you get $x-1 + 1 = f(g(x))$, and then by subtracting the $1
's$ you get $x = f(g(x))$.
Hope that helps. 
(this isn't a proof but it shows the concept).   
Also, the definition of an inverse is that if g(x) is the inverse, then f(g(x)) = x.
