## Precalculus (6th Edition) Blitzer

a) ${{f}^{-1}}\left( x \right)=\frac{1}{x}$ b) they are inverses
(a) Consider the function: $f\left( x \right)=\frac{1}{x}$ Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$. $y=\frac{1}{x}$ Step 2: Interchange the variables x and y. $x=\frac{1}{y}$ Step 3: Solve the equation for y. The variable y has to be isolated. Take the reciprocal of both sides of the equation. So, the equation becomes, $\frac{1}{x}=y$ 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, ${{f}^{-1}}\left( x \right)=\frac{1}{x}$ Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\frac{1}{x}$. (b) Consider the function: $f\left( x \right)=\frac{1}{x}$ and ${{f}^{-1}}\left( x \right)=\frac{1}{x}$ The inverse of the function is obtained as ${{f}^{-1}}\left( x \right)=\frac{1}{x}$ 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 ${{f}^{-1}}\left( x \right)=\frac{1}{x}$ and evaluate the function f. \begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \frac{1}{x} \right) \\ & =\frac{1}{\frac{1}{x}} \\ & =x \end{align} Now, calculate ${{f}^{-1}}\left( f\left( x \right) \right)$. Replace $f\left( x \right)=\frac{1}{x}$ and evaluate the inverse function ${{f}^{-1}}$. \begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( \frac{1}{x} \right) \\ & =\frac{1}{\frac{1}{x}} \\ & =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 ${{f}^{-1}}\left( x \right)=\frac{1}{x}$ is correct. Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.