Question

*Urgent*

Answer 1

To the risk of sounding redundant.

$\bf \begin{cases} f(x)=\cfrac{2}{x}\\\\ g(x)=\cfrac{2}{x} \end{cases} \\\\\\ f(~~g(x)~~)=\cfrac{2}{g(x)}\implies \cfrac{2}{\frac{2}{x}}\implies \cfrac{\quad \frac{2}{1}\quad }{\frac{2}{x}}\implies \cfrac{2}{1}\cdot \cfrac{x}{2}\implies x \\\\\\ g(~~f(x)~~)=\cfrac{2}{f(x)}\implies \cfrac{2}{\frac{2}{x}}\implies \cfrac{\quad \frac{2}{1}\quad }{\frac{2}{x}}\implies \cfrac{2}{1}\cdot \cfrac{x}{2}\implies x$

Answer 2

Let's find f(g(x)):

f(x) = 2/x.  Let's replace the two instances of x with g(x) = 2/x:


                        2
f( g(x) ) = --------------- = x.  Thus, we have shown that f and g are inverses.
                     (2/x)