Chapter 6.3, Problem 53HP

(a)

To determine

To Write:

Write an equation of a relation that contains a radical and its inverse such that the original relation is a function, and its inverse is not a function

Explanation of Solution

Concept Used:

• A function is :

A relation from a set of inputs to a set of possible outputs where each input is related to exactly one output.

So, for 1 value of x , there should be only 1 value of y .

• A relation is a function if it passes the vertical line test. That is , any vertical line must intersect the graph of the relation at exactly one point for the relation to be function.
• The inverse of a relation is a function , if the graph of the relation passes the horizontal line test.

That is , any horizontal line must intersect the graph of the relation at exactly one point for the inverse of the relation to be a function.

Calculation:

  $\begin{array}{l}Consider\\ y=x^2+7\\ \text{Since for each value of x, there is exactly one y-value}\text{.}\\ \text{So this relation is a function}\text{.}\\ \text{Also , it passes the vertical line test since any vertical }\\ \text{line intersects the graph of the equation at exactly one point}\text{.}\\ \text{Find its inverse:}\\ y=x^2+7\\ \text{Replace x by y and y by x:}\\ x=y^2+7\\ \text{Solve for y:}\\ x-7=y^2\mathrm{.........}[\text{subtract 7 from each side]}\\ y=\pm \sqrt{x-7}\mathrm{......}[\text{take square root each side]}\\ \text{So, the inverse is :}\\ y=\pm \sqrt{x-7}\\ but\text{ it is not a funcion , }\\ \text{since for each value of x , there are two values of y}\\ \pm \sqrt{x-7}.\\ So,\text{ the inverse is not a function}\text{.}\end{array}$

(b)

To determine

To Write:

Write an equation of a relation that contains a radical and its inverse such that the original relation is not a function, and its inverse is a function

Explanation of Solution

Concept Used:

• A function is :

A relation from a set of inputs to a set of possible outputs where each input is related to exactly one output.

So, for 1 value of x , there should be only 1 value of y .

• A relation is a function if it passes the vertical line test. That is , any vertical line must intersect the graph of the relation at exactly one point for the relation to be function.
• The inverse of a relation is a function , if the graph of the relation passes the horizontal line test.

That is , any horizontal line must intersect the graph of the relation at exactly one point for the inverse of the relation to be a function.

Calculation:

  $\begin{array}{l}Consider\\ y=\pm \sqrt{x}\\ \text{For each value of x , there are two values of y }\pm \sqrt{x}.\\ So,the\text{ relation is not a function }\text{.}\\ \text{Find the inverse:}\\ y=\pm \sqrt{x}\\ \text{Replace x by y and y by x:}\\ x=\pm \sqrt{y}\\ Solve\text{ for y:}\\ y=x^2\mathrm{..........}[\text{square each side]}\\ \text{Since for each value of x , there is exactly one y-value}\text{.}\\ Also,the\text{ graph of the relation passes the vertical }\\ \text{line test }\text{.}\\ \text{So, it is a function}\text{.}\\ \text{Hence, the inverse of given relation is a function}\text{.}\end{array}$