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) $\begin{aligned}f(g(x) ) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right) } \\&=\frac{2}{2+\left(\frac{1}{x}\right) } \\&=\frac{1}{\left(\frac{1}{x}\right) } \\&=1 \cdot \frac{x}{1} \\&=x\end{aligned}$

$\begin{aligned}g(f(x) ) &=\frac{2-2\left(\frac{2}{2+x}\right) }{\left(\frac{2}{2+x}\right) } \\\\&=\frac{0-\left(\frac{2}{2+x}\right) }{\left(\frac{2}{2+x}\right) } \\\\&=\frac{\frac{-2}{2+x}}{\left(\frac{2}{2+x}\right) } \\\\&=\left(\frac{-2}{2+x}\right) \left(\frac{2+x}{2}\right) \\\\&=\frac{2 x+2}{2+x}\\&=x\end{aligned}$
B) $\begin{aligned}f(g(x) ) &=\frac{2}{2+\left(\frac{2-2 x}{x}\right) } \\&=\frac{1}{1+\frac{2-2 x}{x}} \\&=\frac{1}{\left(\frac{0}{x}\right) } \\&=x\end{aligned}$

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

$\begin{array}{l}g(f(x) ) =\frac{2-2\left(\frac{2}{2+x}\right) }{\left(\frac{2}{2+x}\right) }\\\\=\frac{2-\left(\frac{4}{2+x}\right) }{\left(\frac{2}{2+x}\right) }\\\\=\frac{\frac{4+2 x-4}{2+x}}{\left(\frac{2}{2+x}\right) }\\\\=\frac{\frac{2 x}{2+x}}{\left(\frac{2}{2+x}\right) }\\\\=\frac{2 x}{2}\\\\=x\end{array}$