Answer
a) ${{f}^{-1}}\left( x \right)=\frac{-x-3}{x-2}$ ,$x\ne 2$.
b) they are inverses
Work Step by Step
(a)
Consider the function:
$f\left( x \right)=\frac{2x-3}{x+1}$
Follow the steps to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)=\frac{2x-3}{x+1}$.
Step 1: Replace the function $f\left( x \right)$ with y in $f\left( x \right)$.
$y=\frac{2x-3}{x+1}$
Step 2: Interchange the variables x and y.
$x=\frac{2y-3}{y+1}$
Step 3: Solve the equation for y.
The variable y has to be isolated. Multiply by $\left( y+1 \right)$ on both sides of the equation to clear fractions. So, the equation becomes,
$x\left( y+1 \right)=\left( \frac{2y-3}{y+1} \right)\left( y+1 \right)$
Simplify the equation. So,
$x\left( y+1 \right)=2y-3$
Use the distributive property $a\left( b+c \right)=ab+ac$ to remove parentheses.
$xy+x=2y-3$
Subtract $2y$ and subtract $x$ to both the sides of the equation.
$\begin{align}
& xy+x-2y-x=2y-3-2y-x \\
& xy-2y=-x-3
\end{align}$
Factor out $y$ from the left-hand side of the equation to isolate $y$.
$y\left( x-2 \right)=-x-3$
Divide by $\left( x-2 \right)$ on both the sides of the equation for $x\ne 2$.
$\begin{align}
& \frac{y\left( x-2 \right)}{\left( x-2 \right)}=\frac{-x-3}{\left( x-2 \right)} \\
& y=\frac{-x-3}{x-2}
\end{align}$
Because the obtained equation defines y as a function of x for $x\ne 2$ , 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,
${{f}^{-1}}\left( x \right)=\frac{-x-3}{x-2}$
Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\frac{-x-3}{x-2}$ ,$x\ne 2$.
(b)
Consider the function: $f\left( x \right)=\frac{2x-3}{x+1}$.
The inverse of the function is obtained as ${{f}^{-1}}\left( x \right)=\frac{-x-3}{x-2}$ 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 ${{f}^{-1}}\left( x \right)=\frac{-x-3}{x-2}$ and evaluate the function f.
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \frac{-x-3}{x-2} \right) \\
& =\frac{2\left( \frac{-x-3}{x-2} \right)-3}{\left( \frac{-x-3}{x-2} \right)+1}
\end{align}$
Take the least common denominator as $x-2$ in both numerator and denominator. So,
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=\frac{\frac{2\left( -x-3 \right)-3\left( x-2 \right)}{x-2}}{\frac{-x-3+1\left( x-2 \right)}{x-2}} \\
& =\frac{2\left( -x-3 \right)-3\left( x-2 \right)}{-x-3+1\left( x-2 \right)}
\end{align}$
Use the distributive property $a\left( b+c \right)=ab+ac$ and combine like terms.
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=\frac{-2x-6-3x+6}{-x-3+x-2} \\
& =\frac{-5x}{-5} \\
& =x
\end{align}$
Now, calculate ${{f}^{-1}}\left( f\left( x \right) \right)$. Replace $f\left( x \right)=\frac{2x-3}{x+1}$ and evaluate the inverse function ${{f}^{-1}}$.
$\begin{align}
& {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( \frac{2x-3}{x+1} \right) \\
& =\frac{-\left( \frac{2x-3}{x+1} \right)-3}{\left( \frac{2x-3}{x+1} \right)-2}
\end{align}$
Take the least common denominator as $x+1$ in both numerator and denominator. So,
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=\frac{\frac{-\left( 2x-3 \right)-3\left( x+1 \right)}{x+1}}{\frac{2x-3-2\left( x+1 \right)}{x+1}} \\
& =\frac{-\left( 2x-3 \right)-3\left( x+1 \right)}{2x-3-2\left( x+1 \right)}
\end{align}$
Use the distributive property $a\left( b+c \right)=ab+ac$ and combine like terms.
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=\frac{-2x+3-3x-3}{2x-3-2x-2} \\
& =\frac{-5x}{-5} \\
& =x
\end{align}$
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 ${{f}^{-1}}\left( x \right)=\frac{-x-3}{x-2}$ ,$x\ne 2$ is correct.
Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.
