Answer
a) ${{f}^{-1}}\left( x \right)=\frac{2}{x}$.
b) they are inverses
Work Step by Step
(a)
Consider the function:
$f\left( x \right)=\frac{2}{x}$
Follow the steps to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)=\frac{2}{x}$.
Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$.
$y=\frac{2}{x}$
Step 2: Interchange the variables x and y.
$x=\frac{2}{y}$
Step 3: Solve the equation for y.
The variable y has to be isolated. Take the reciprocal of both sides of the equation. So, the equation becomes,
$\frac{1}{x}=\frac{y}{2}$
Multiply by $2$ on both the sides of the equation.
$\begin{align}
& \frac{1}{x}\times 2=\frac{y}{2}\times 2 \\
& \frac{2}{x}=y
\end{align}$
Because the obtained equation defines y as a function of x, then the inverse function exists for the function f.
Step 4: Replace y by ${{f}^{-1}}\left( x \right)$ in the equation obtained in step 3.
Thus, the inverse function ${{f}^{-1}}\left( x \right)$ is obtained as,
${{f}^{-1}}\left( x \right)=\frac{2}{x}$
Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\frac{2}{x}$.
(b)
Consider the function: $f\left( x \right)=\frac{2}{x}$. The inverse of the function is obtained as ${{f}^{-1}}\left( x \right)=\frac{2}{x}$ in part (a).
The inverse of an equation can be verified by showing that $f\left( {{f}^{-1}}\left( x \right) \right)=x\text{ and }{{f}^{-1}}\left( f\left( x \right) \right)=x\text{ }$. First calculate $f\left( {{f}^{-1}}\left( x \right) \right)$.
Replace ${{f}^{-1}}\left( x \right)=\frac{2}{x}$ and evaluate the function f.
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \frac{2}{x} \right) \\
& =\frac{2}{\frac{2}{x}} \\
& =2\times \frac{x}{2} \\
& =x
\end{align}$
Now, calculate ${{f}^{-1}}\left( f\left( x \right) \right)$. Replace $f\left( x \right)=\frac{2}{x}$ and evaluate the inverse function ${{f}^{-1}}$.
$\begin{align}
& {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( \frac{2}{x} \right) \\
& =\frac{2}{\frac{2}{x}} \\
& =2\times \frac{x}{2} \\
& =x
\end{align}$
Thus, it verifies that $f\left( {{f}^{-1}}\left( x \right) \right)=x\text{ and }{{f}^{-1}}\left( f\left( x \right) \right)=x\text{ }$. Therefore, the obtained inverse function ${{f}^{-1}}\left( x \right)=\frac{2}{x}$ is correct.
Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.
