Chapter 1.6, Problem 65E

To determine

To state: that whether the function is one-to-one or not and verify graphically.

Answer to Problem 65E

Yes, the function is one-to-one and the inverse function is $f^{-1}\left(x\right)=\sqrt{x}-3$.

Explanation of Solution

Given information : thefunction is,

  $f\left(x\right)={\left(x+3\right)}^2,x\ge -3$

let $f\left(x\right)=f\left(y\right)$ ,

  $\begin{array}{l}f\left(x\right)=f\left(y\right)\\ {\left(x+3\right)}^2={\left(y+3\right)}^2\text{ [substitute }f\left(x\right)={\left(x+3\right)}^2]\\ x=y\text{ [simplify]}\end{array}$

Since $x\ge -3$ ,so There isno possibilities that $x\text{ and }y$ , are not equal. So the value of y has occur onlyone time for x -value. So the function is one-to-one.

to graph the given function first create a table for plotting the points and then make a curve smoothly by jointing the points.

x	y
-3	0
-2	1
-1	4
0	9

The graph can be obtained as:

  

From the above graph, it can be observed that the graph of the function hasno horizontal lines that intercept the graph more than one times, so the function isa one-to-one and its inverse functionexists.

Now for inverse function, substitute $f\left(x\right)=y$ in the function,

  $\begin{array}{l}f\left(x\right)={\left(x+3\right)}^2\\ y={\left(x+3\right)}^2\text{ [substitute }f\left(x\right)=y]\\ x+3=\sqrt{y}\text{ [square root]}\\ x=\sqrt{y}-3\text{ [subtract 3 from both side]}\\ f^{-1}\left(x\right)=\sqrt{x}-3\text{ [simplify]}\end{array}$

Thus, the inverse function is $f^{-1}\left(x\right)=\sqrt{x}-3$.