Chapter 1.9, Problem 3E

To determine

To complete the given statement.

Answer to Problem 3E

The domain of $f$ is the range of $f^{-1}$ , and the domain of $f^{-1}$ is the range of $f$ .

Explanation of Solution

Given information:

The domain of $f$ is the $\underset{\_}{}$ of $f^{-1}$ , and the $\underset{\_}{}$ of $f^{-1}$ is the range of $f$ .

Calculation:

This statement is correct if,

The domain of $f$ is the range of $f^{-1}$ , and the domain of $f^{-1}$ is the range of $f$ .

Let $p\left(t\right)=t+1$

The function $p\left(t\right)$ is defined for all t.

So, the domain of $p\left(t\right)$ is $\left(-\infty ,\infty \right)$ .

The inverse of $p\left(t\right)$ is,

  $p\left(t\right)=t+1$

Subtract 1 from both sides,

  $\begin{array}{l}p\left(t\right)-1=t+1-1\\ p\left(t\right)-1=t\end{array}$

So, the inverse of $p\left(t\right)$ is,

  $p^{-1}=t-1$

The function is defined for all values of dependent variable $p^{-1}$ .

So, the range of $p^{-1}$ is $\left(-\infty ,\infty \right)$ .

So, from the above explanation, it conclude that the domain of $f$ is the range of $f^{-1}$ , and vice-versa.