Show Algebraically That the Functions $f$ And $g$ Shown Below Are Inverse Functions

Multiple Choice

Show algebraically that the functions $f$ and $g$ shown below are inverse functions.
$$f ( x ) = \frac { 2 } { 2 + x } , x \geq 0 , \quad g ( x ) = \frac { 2 - 2 x } { x } , 0 < x \leq 1$$

A)
$\begin{array}{l}\begin{array}{lll}f(g(x) ) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right) } \\&=\frac{2}{2+\left(\frac{1}{x}\right) } \\&=\frac{1}{\left(\frac{1}{x}\right) } \\&=1 \cdot \frac{x}{1} \\&=x\end{array}\begin{array}{lll}g(f(x) ) &=\frac{2-2\left(\frac{2}{2+x}\right) }{\left(\frac{2}{2+x}\right) } \\&=\frac{0-\left(\frac{2}{2+x}\right) }{\left(\frac{2}{2+x}\right) } \\&=\frac{\frac{-2}{2+x}}{\left(\frac{2}{2+x}\right) } \\&=\left(\frac{-2}{2+x}\right) \left(\frac{2+x}{2}\right) \\&=\frac{2 x+2}{2+x} \\&=x\end{array}\end{array}$
B)
$\begin{array}{l}\begin{array}{lll}f(g(x) ) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right) } \\&=\frac{1}{1+\frac{2-2 x}{x}} \\&=\frac{1}{\left(\frac{0}{x}\right) } \\&=x\end{array}\begin{array}{lll}g(f(x) ) &=\frac{2-2\left(\frac{2}{2+x}\right) }{\left(\frac{2}{2+x}\right) } \\&=\frac{2-\left(\frac{4}{2+x}\right) }{\left(\frac{2}{2+x}\right) } \\&=\frac{\frac{4+2 x-4}{2+x}}{\left(\frac{2}{2+x}\right) } \\&=\frac{2 x}{\left(\frac{x}{2+x}\right) .} \\&=\frac{2 x}{x} \\&= x\end{array}\end{array}$
C)
$\begin{array}{l}\begin{array}{lll}f(g(x) ) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right) } \\&=\frac{4}{\frac{2-2 x}{x}} \\&=\frac{2}{\left(\frac{2 x}{x}\right) } \\&=2 \cdot \frac{x}{2} \\&=x\end{array}\begin{array}{lll}g(f(x) ) &=\frac{2-2\left(\frac{2}{2+x}\right) }{\left(\frac{2}{2+x}\right) } \\&=\frac{\left(\frac{2}{2+2 x}\right) }{\left(\frac{2}{2+x}\right) } \\&=\left(\frac{2}{2+2 x}\right) \left(\frac{2+x}{2}\right) \\&=\frac{2+x}{2+2 x} \\&=\frac{x}{2 x} \\&=x\end{array}\end{array}$
D)
$\begin{array}{l}\begin{array}{lll}f(g(x) ) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right) } \\&=\frac{2}{\frac{2 x+2-2 x}{x}} \\&=\frac{2-2 x}{\left(\frac{2}{x}\right) } \\&=\frac{1-x}{1} \\&=x\end{array}\begin{array}{lll}g(f(x) ) &=\frac{2-2\left(\frac{2}{2+x}\right) }{\left(\frac{2}{2+x}\right) } \\&=\frac{\left(\frac{4}{2+x}\right) }{\left(\frac{2}{2+x}\right) } \\&=\left(\frac{4}{2+x}\right) \left(\frac{2+x}{2}\right) \\&=\frac{2(2+x) }{2+x} \\&=\frac{2 x}{2}\\&=x\end{array}\end{array}$
E)
$\begin{array}{l}\begin{array}{lll}f(g(x) ) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right) } \\&=\frac{2}{\frac{2 x+2-2 x}{x}} \\&=\frac{2}{\left(\frac{2}{x}\right) } \\&=2 \cdot \frac{x}{2} \\&=x\end{array}\begin{array}{lll}g(f(x) ) &=\frac{2-2\left(\frac{2}{2+x}\right) }{\left(\frac{2}{2+x}\right) } \\&=\frac{2-\left(\frac{4}{2+x}\right) }{\left(\frac{2}{2+x}\right) } \\&=\frac{\frac{4+2 x-4}{2+x}}{\left(\frac{2}{2+x}\right) } \\&=\frac{\frac{2 x}{2+x}}{\left(\frac{2}{2+x}\right) } \\&=\frac{2 x}{2}\end{array}\end{array}$