Answer
a) ${{f}^{-1}}\left( x \right)=\frac{2x+4}{x-1}$, $x\ne 1$.
b) they are inverses
Work Step by Step
(a)
Consider the function:
$f\left( x \right)=\frac{x+4}{x-2}$
Follow the procedure to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)=\frac{x+4}{x-2}$.
Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$.
$y=\frac{x+4}{x-2}$
Step 2: Interchange the variables x and y.
$x=\frac{y+4}{y-2}$
Step 3: Solve the equation for y.
The variable y has to be isolated. Multiply by $\left( y-2 \right)$ on both the sides of the equation to clear fractions. So, the equation becomes,
$x\left( y-2 \right)=\left( \frac{y+4}{y-2} \right)\left( y-2 \right)$
Simplify the equation. So,
$x\left( y-2 \right)=y+4$
Use the distributive property $a\left( b+c \right)=ab+ac$ to remove parentheses.
$xy-2x=y+4$
Subtract $y$ and add $2x$ to both the sides of the equation.
$\begin{align}
& xy-2x-y+2x=y+4-y+2x \\
& xy-y=2x+4
\end{align}$
Factor out $y$ from the left-hand side of the equation to isolate $y$.
$y\left( x-1 \right)=2x+4$
Divide by $\left( x-1 \right)$ on both the sides of the equation for $x\ne 1$.
$\begin{align}
& \frac{y\left( x-1 \right)}{\left( x-1 \right)}=\frac{2x+4}{x-1} \\
& y=\frac{2x+4}{x-1}
\end{align}$
Because the obtained equation defines y as a function of x for $x\ne 1$ , 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,
${{f}^{-1}}\left( x \right)=\frac{2x+4}{x-1}$
Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\frac{2x+4}{x-1}$ ,$x\ne 1$.
(b)
Consider the function: $f\left( x \right)=\frac{x+4}{x-2}$.
The inverse of the function is obtained as ${{f}^{-1}}\left( x \right)=\frac{2x+4}{x-1}$ 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 ${{f}^{-1}}\left( x \right)=\frac{2x+4}{x-1}$ and evaluate the function f.
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \frac{2x+4}{x-1} \right) \\
& =\frac{\frac{2x+4}{x-1}+4}{\frac{2x+4}{x-1}-2}
\end{align}$
Take the least common denominator as $x-1$ in both numerator and denominator. So,
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=\frac{\frac{2x+4+4\left( x-1 \right)}{x-1}}{\frac{2x+4-2\left( x-1 \right)}{x-1}} \\
& =\frac{2x+4+4\left( x-1 \right)}{2x+4-2\left( x-1 \right)}
\end{align}$
Use the distributive property $a\left( b+c \right)=ab+ac$ and combine like terms.
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=\frac{2x+4+4x-4}{2x+4-2x+2} \\
& =\frac{6x}{6} \\
& =x
\end{align}$
Now, calculate ${{f}^{-1}}\left( f\left( x \right) \right)$. Replace $f\left( x \right)=\frac{x+4}{x-2}$ and evaluate the inverse function ${{f}^{-1}}$.
$\begin{align}
& {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( \frac{x+4}{x-2} \right) \\
& =\frac{2\left( \frac{x+4}{x-2} \right)+4}{\left( \frac{x+4}{x-2} \right)-1}
\end{align}$
Take the least common denominator as $x-2$ in both numerator and denominator. So,
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=\frac{\frac{2\left( x+4 \right)+4\left( x-2 \right)}{x-2}}{\frac{x+4-1\left( x-2 \right)}{x-2}} \\
& =\frac{2\left( x+4 \right)+4\left( x-2 \right)}{x+4-1\left( x-2 \right)}
\end{align}$
Use the distributive property $a\left( b+c \right)=ab+ac$ and combine like terms.
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=\frac{2x+8+4x-8}{x+4-x+2} \\
& =\frac{6x}{6} \\
& =x
\end{align}$
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 ${{f}^{-1}}\left( x \right)=\frac{2x+4}{x-1}$ ,$x\ne 1$ is correct.
Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.
