Chapter 1.6, Problem 68E

To determine

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

Answer to Problem 68E

Yes, the function is one-to-one and the inverse function is $f^{-1}\left(x\right)=x^2\text{+2}$.

Explanation of Solution

Given information : thefunction is,

  $f\left(x\right)=\sqrt{x-2}$

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

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

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
2	0
3	1
4	$\sqrt{2}$
6	2

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,

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

Thus, the inverse function is $f^{-1}\left(x\right)=x^2\text{+2}$.