Answer
See below:
Work Step by Step
(a)
Consider the function:
$f\left( x \right)={{x}^{2}}-4$
Follow the steps to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)={{x}^{2}}-4$.
Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$.
$y={{x}^{2}}-4$
Step 2: Interchange the variables x and y.
$x={{y}^{2}}-4$
Step 3: Solve the equation for y.
The variable y has to be isolated. Add $4$ to both sides of the equation. So, the equation becomes,
$x+4={{y}^{2}}-4+4$
Combine like terms.
$x+4={{y}^{2}}$
Take square root on both sides of the equation.
$\begin{align}
& \sqrt{x+4}=\sqrt{{{y}^{2}}} \\
& \sqrt{x+4}=y \\
\end{align}$
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,
${{f}^{-1}}\left( x \right)=\sqrt{x+4}$
Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\sqrt{x+4}$.
(b) Given above.
(c)
The domain and range of the functions in interval notation are:
\[\begin{align}
& \text{domain of }f=\text{range of }{{f}^{-1}}=\left[ 0,\infty \right) \\
& \text{range of }f=\text{domain of }{{f}^{-1}}=\left[ -4,\infty \right)
\end{align}\]
