## Precalculus (6th Edition) Blitzer

a) ${{f}^{-1}}\left( x \right)=\frac{6x+5}{x-1}$ , $x\ne 1$. b) they are inverses
(a) Consider the function: $f\left( x \right)=\frac{x+5}{x-6}$ Follow the procedure to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)=\frac{x+5}{x-6}$. Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$. $y=\frac{x+5}{x-6}$ Step 2: Interchange the variables x and y. $x=\frac{y+5}{y-6}$ Step 3: Solve the equation for y. The variable y has to be isolated. Multiply by $\left( y-6 \right)$ on both sides of the equation to clear fractions. So, the equation becomes, $x\left( y-6 \right)=\left( \frac{y+5}{y-6} \right)\left( y-6 \right)$ Simplify the equation. So, $x\left( y-6 \right)=y+5$ Use the distributive property $a\left( b+c \right)=ab+ac$ to remove parentheses. $xy-6x=y+5$ Subtract $y$ and add $6x$ to both the sides of the equation. \begin{align} & xy-6x-y+6x=y+5-y+6x \\ & xy-y=6x+5 \end{align} Factor out $y$ from the left-hand side of the equation to isolate $y$. $y\left( x-1 \right)=6x+5$ 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{6x+5}{\left( x-1 \right)} \\ & y=\frac{6x+5}{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, ${{f}^{-1}}\left( x \right)=\frac{6x+5}{x-1}$ Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\frac{6x+5}{x-1}$ ,$x\ne 1$. (b) Consider the function: $f\left( x \right)=\frac{x+5}{x-6}$. The inverse of the function is obtained as ${{f}^{-1}}\left( x \right)=\frac{6x+5}{x-1}$ 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 ${{f}^{-1}}\left( x \right)=\frac{6x+5}{x-1}$ and evaluate the function f. \begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \frac{6x+5}{x-1} \right) \\ & =\frac{\frac{6x+5}{x-1}+5}{\frac{6x+5}{x-1}-6} \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{6x+5+5\left( x-1 \right)}{x-1}}{\frac{6x+5-6\left( x-1 \right)}{x-1}} \\ & =\frac{6x+5+5\left( x-1 \right)}{6x+5-6\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{6x+5+5x-5}{6x+5-6x+6} \\ & =\frac{11x}{11} \\ & =x \end{align} Now, calculate ${{f}^{-1}}\left( f\left( x \right) \right)$. Replace $f\left( x \right)=\frac{x+5}{x-6}$ and evaluate the inverse function ${{f}^{-1}}$. \begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( \frac{x+5}{x-6} \right) \\ & =\frac{6\left( \frac{x+5}{x-6} \right)+5}{\left( \frac{x+5}{x-6} \right)-1} \end{align} Take the least common denominator as $x-6$ in both numerator and denominator. So, \begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)=\frac{\frac{6\left( x+5 \right)+5\left( x-6 \right)}{x-6}}{\frac{x+5-1\left( x-6 \right)}{x-6}} \\ & =\frac{6\left( x+5 \right)+5\left( x-6 \right)}{x+5-1\left( x-6 \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+30+5x-30}{x+5-x+6} \\ & =\frac{11x}{11} \\ & =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 ${{f}^{-1}}\left( x \right)=\frac{6x+5}{x-1}$ ,$x\ne 1$ is correct. Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.