Chapter 1.6, Problem 88E

To determine

To find: the inverse of the given function by restricting the domain of the function f and

then to state their domains.

Answer to Problem 88E

The inverses function $f^{-1}(x)=\pm \sqrt{2(x-1)}$ .

Domains of f is $[0,\infty )$ and $f^{-1}(x)$ is $[1,\infty )$

range of f is $[1,\infty )$ and $f^{-1}(x)$ is $[0,\infty )$

Explanation of Solution

Given information:

The given equation is $f(x)={(x+3)}^2$

Concept used:

The one to one functions only invertible.

Horizontal line test:

The horizontal line drawn to the graph of the function intersect at only once, then the

function is one to one. Otherwise not.

Calculation:

Apply horizontal line test:

  

The given function is not one to one function. Since, it fails horizontal line test.

To make the given function one to one, restrict the domain of f to $[-3,\infty )$

Find the inverse:

  $\begin{array}{l}$ y=f(x)={(x+3)}^2$$ y={(x+3)}^2$\sqrt[]{y}-3=x\\ x=\sqrt{y}-3\\ f^{-1}(y)=\sqrt{y}-3\\ f^{-1}(x)=\sqrt{x}-3\end{array}$

Hence, the inverses function $f^{-1}(x)=\sqrt{x}-3$ .

Domains of f is $[-3,\infty )$ and $f^{-1}(x)$ is $[0,\infty )$

range of f is $[0,\infty )$ and $f^{-1}(x)$ is $[-3,\infty )$