Chapter 4.7, Problem 37PPS

a.

To determine

To find the function for the area $A(x)$ of the rectangle.

Answer to Problem 37PPS

Function for the area $A(x)$ of the rectangle $A(x)=x^2-3x$ .

Explanation of Solution

Given information: Given rectangle has length x -3 and width x .

Formula:

Area of the rectangle=length $\times$ width.

Calculation:

  $\begin{array}{l}\text{length}=(x-3)\text{and}\text{width}=x.\\ A(x)=(x-3)\times x.\\ A(x)=x^2-3x.\end{array}$

Function for the area $A(x)$ of the rectangle $A(x)=x^2-3x$ .

b.

To determine

To describe the domain and range of A( x ).

Answer to Problem 37PPS

The domain of $A(x)$ is all real numbers greater than 3, and the range of $A(x)$ is all positive real numbers.

Explanation of Solution

Given information: Given rectangle has length x -3 and width x .

Formula:

Area of the rectangle=length $\times$ width.

Calculation:

  $\begin{array}{l}\text{length}=(x-3)\text{and}\text{width}=x.\\ A(x)=(x-3)\times x.\\ A(x)=x^2-3x.\end{array}$

Function for the area $A(x)$ of the rectangle $A(x)=x^2-3x$ .

The domain represents possible values of x . The range represents the area of the rectangle and must be positive. This means that the domain of $A(x)$ is all real numbers greater than 3, and the range of $A(x)$ is all positive real numbers.

c.

To determine

To find the inverse of $A(x)$ and what do x and $A(x)$ represent .

Answer to Problem 37PPS

  $A^{-1}(x)=(1/8)x+3$.

x is the area of the rectangle and $A^{-1}(x)$ is the value of x in the expression for the

length of the side of the rectangle x - 3.

Explanation of Solution

Given information: Given rectangle has length x -3 and width x .

Formula:

Area of the rectangle=length $\times$ width.

Calculation:

  $\begin{array}{l}\text{length}=(x-3)\text{and}\text{width}=x.\\ A(x)=(x-3)\times x.\\ A(x)=x^2-3x.\end{array}$

Function for the area $A(x)$ of the rectangle $A(x)=x^2-3x$ .

  $A^{-1}(x)=(1/8)x+3$.

x is the area of the rectangle and $A^{-1}(x)$ is the value of x in the expression for the

length of the side of the rectangle x - 3.

d.

To determine

To describe the domain and range of $A^{-1}(x)$ ).

Answer to Problem 37PPS

The domain of $A^{-1}(x)$ is all positive real numbers, and the range of $A^{-1}(x)$ is all real numbers greater than 3.

.

Explanation of Solution

Given information: Given rectangle has length x -3 and width x .

Formula:

Area of the rectangle=length $\times$ width.

Calculation:

  $\begin{array}{l}\text{length}=(x-3)\text{and}\text{width}=x.\\ A(x)=(x-3)\times x.\\ A(x)=x^2-3x.\end{array}$

Function for the area $A(x)$ of the rectangle $A(x)=x^2-3x$ .

  $A^{-1}(x)=(1/8)x+3$.

x is the area of the rectangle and $A^{-1}(x)$ is the value of x in the expression for the

length of the side of the rectangle x - 3.

The domain represents the area of the rectangle and must be positive. The range represents possible values for x in the expression x −3. This means that the domain of $A^{-1}(x)$ is all positive real numbers, and the range of $A^{-1}(x)$ is all real numbers greater than 3.

e.

To determine

To find the relationship between the domains and ranges of A( x ) and $A^{-1}(x)$ .

Answer to Problem 37PPS

The domain of $A(x)$ is the range of $A^{-1}(x)$ and the range of $A(x)$ is the domain of

  $A^{-1}(x)$ .

Explanation of Solution

Given information: Given rectangle has length x -3 and width x .

Formula:

Area of the rectangle=length $\times$ width.

Calculation:

  $\begin{array}{l}\text{length}=(x-3)\text{and}\text{width}=x.\\ A(x)=(x-3)\times x.\\ A(x)=x^2-3x.\end{array}$

Function for the area $A(x)$ of the rectangle $A(x)=x^2-3x$ .

The domain represents possible values of x . The range represents the area of the rectangle and must be positive. This means that the domain of $A(x)$ is all real numbers greater than 3, and the range of $A(x)$ is all positive real numbers.

  $A^{-1}(x)=(1/8)x+3$.

x is the area of the rectangle and $A^{-1}(x)$ is the value of x in the expression for the

length of the side of the rectangle x - 3.

The domain of $A(x)$ is the range of $A^{-1}(x)$ and the range of $A(x)$ is the domain of

  $A^{-1}(x)$ .