Answer
a) ${{f}^{-1}}\left( x \right)=\frac{1}{x}$
b) they are inverses
Work Step by Step
(a)
Consider the function:
$f\left( x \right)=\frac{1}{x}$
Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$.
$y=\frac{1}{x}$
Step 2: Interchange the variables x and y.
$x=\frac{1}{y}$
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,
$\frac{1}{x}=y$
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,
${{f}^{-1}}\left( x \right)=\frac{1}{x}$
Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\frac{1}{x}$.
(b)
Consider the function:
$f\left( x \right)=\frac{1}{x}$ and ${{f}^{-1}}\left( x \right)=\frac{1}{x}$
The inverse of the function is obtained as ${{f}^{-1}}\left( x \right)=\frac{1}{x}$ 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 ${{f}^{-1}}\left( x \right)=\frac{1}{x}$ and evaluate the function f.
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \frac{1}{x} \right) \\
& =\frac{1}{\frac{1}{x}} \\
& =x
\end{align}$
Now, calculate ${{f}^{-1}}\left( f\left( x \right) \right)$. Replace $f\left( x \right)=\frac{1}{x}$ and evaluate the inverse function ${{f}^{-1}}$.
$\begin{align}
& {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( \frac{1}{x} \right) \\
& =\frac{1}{\frac{1}{x}} \\
& =x
\end{align}$
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 ${{f}^{-1}}\left( x \right)=\frac{1}{x}$ is correct.
Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.
