## Precalculus (6th Edition) Blitzer

a) ${{f}^{-1}}\left( x \right)=\frac{3x+1}{x-2}$ ,$x\ne 2$. b) they are inverses
(a) Consider the function: $f\left( x \right)=\frac{2x+1}{x-3}$ Follow the steps to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)=\frac{2x+1}{x-3}$. Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$. $y=\frac{2x+1}{x-3}$ Step 2: Interchange the variables x and y. $x=\frac{2y+1}{y-3}$ Step 3: Solve the equation for y. The variable y has to be isolated. Multiply by $\left( y-3 \right)$ on both sides of the equation to clear fractions. So, the equation becomes, $x\left( y-3 \right)=\left( \frac{2y+1}{y-3} \right)\left( y-3 \right)$ Simplify the equation. So, $x\left( y-3 \right)=2y+1$ Use the distributive property $a\left( b+c \right)=ab+ac$ to remove parentheses. $xy-3x=2y+1$ Subtract $2y$ and add $3x$ to both the sides of the equation. \begin{align} & xy-3x-2y+3x=2y+1-2y+3x \\ & xy-2y=3x+1 \end{align} Factor out $y$ from the left-hand side of the equation to isolate $y$. $y\left( x-2 \right)=3x+1$ 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{3x+1}{\left( x-2 \right)} \\ & y=\frac{3x+1}{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, ${{f}^{-1}}\left( x \right)=\frac{3x+1}{x-2}$ Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\frac{3x+1}{x-2}$ ,$x\ne 2$. (b) Consider the function: $f\left( x \right)=\frac{2x+1}{x-3}$. The inverse of the function is obtained as ${{f}^{-1}}\left( x \right)=\frac{3x+1}{x-2}$ 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 ${{f}^{-1}}\left( x \right)=\frac{3x+1}{x-2}$ and evaluate the function f. \begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \frac{3x+1}{x-2} \right) \\ & =\frac{2\left( \frac{3x+1}{x-2} \right)+1}{\left( \frac{3x+1}{x-2} \right)-3} \end{align} Take the least common denominator as $x-2$ in both the numerator and denominator. So, \begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)=\frac{\frac{6x+2+1\left( x-2 \right)}{x-2}}{\frac{3x+1-3\left( x-2 \right)}{x-2}} \\ & =\frac{6x+2+1\left( x-2 \right)}{3x+1-3\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{6x+2+x-2}{3x+1-3x+6} \\ & =\frac{7x}{7} \\ & =x \end{align} Now, calculate ${{f}^{-1}}\left( f\left( x \right) \right)$. Replace $f\left( x \right)=\frac{2x+1}{x-3}$ and evaluate the inverse function ${{f}^{-1}}$. \begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( \frac{2x+1}{x-3} \right) \\ & =\frac{3\left( \frac{2x+1}{x-3} \right)+1}{\left( \frac{2x+1}{x-3} \right)-2} \end{align} Take the least common denominator as $x-3$ in both numerator and denominator. So, \begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)=\frac{\frac{3\left( 2x+1 \right)+1\left( x-3 \right)}{x-3}}{\frac{2x+1-2\left( x-3 \right)}{x-3}} \\ & =\frac{3\left( 2x+1 \right)+1\left( x-3 \right)}{2x+1-2\left( x-3 \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{6x+3+x-3}{2x+1-2x+6} \\ & =\frac{7x}{7} \\ & =x \end{align} Thus, it verifies that $f\left( {{f}^{-1}}\left( x \right) \right)=x$ and ${{f}^{-1}}\left( f\left( x \right) \right)=x\text{ }$. Therefore, the obtained inverse function ${{f}^{-1}}\left( x \right)=\frac{3x+1}{x-2}$ ,$x\ne 2$ is correct. Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.