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

question 96

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

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