Question

A business is studying the cost to remove a pollutant from the ground at its site. The function \(y=\frac{15 x}{1.1-x}\) models the estimated cost \(y\) (in thousands of dollars) to remove \(x\) percent (expressed as a decimal) of the pollutant. a. Graph the function. Describe a reasonable domain and range. b. How much does it cost to remove \(20 \%\) of the pollutant? \(40 \%\) of the pollutant? \(80 \%\) of the pollutant? Does doubling the percentage of the pollutant removed double the cost? Explain.

Short answer

The graph of the function is a hyperbola in quadrant I, with x approaching 1.1 as an asymptote. The domain is all real numbers except 1.1, and the range is all positive real numbers. The costs associated with removing \(20\%, 40\%, 80\%\), of the pollutant can be calculated by substituting these values (in decimal form) into the given equation. Doubling the percentage of pollutant removal does not necessarily double the cost, which can be understood by comparing the costs found.

Step-by-step solution

Graphing the Function

To graph the function \(y=\frac{15 x}{1.1-x}\), you'll have to plot a variety of x values and find their corresponding y values. For the domain, consider the values for which this function is defined. The denominator of the function is zero when \(x = 1.1\), so all real numbers except \(x = 1.1\) constitute the domain. The range, given \(x < 1.1\), will be all positive real numbers.

Use the Function to Find Specific Costs

To calculate the cost of removing \(20\%, 40\%, 80\%\) of the pollutant, you have to plug these percentages into x, which is expressed as a decimal. Hence, find the values of y for \(x = 0.20\), \(x = 0.40\), and \(x = 0.80\).

Analyze the Relationship

By comparing the respective costs of removing \(20\%, 40\%, 80\%\) of the pollutant, you can analyze if doubling the percentage of removal, doubles the cost or not.

Key concepts

Domain and Range of Rational Functions

When dealing with rational functions like \(y = \frac{15x}{1.1-x}\), understanding domain and range is crucial. The **domain** refers to the set of all possible input values (\(x\) values) that the function can accept. Since any value making the denominator zero is undefined for the function, the domain for this function excludes \(x = 1.1\). Thus, the domain consists of all real numbers except \(x = 1.1\).
On the other hand, the **range** is all possible output values (\(y\) values) that the function can generate for the given range of \(x\). For \(x < 1.1\), the range will be the set of all positive real numbers because \(y\) becomes exceedingly large as \(x\) approaches 1.1 from the left. This understanding helps to know the limitations and extent of the function.

Graphing Rational Functions

**Graphing** a rational function requires plotting based on specific points and understanding the function's behavior. Begin by choosing several \(x\) values within the domain, ignoring \(x = 1.1\) where the function is undefined. Calculate and plot the corresponding \(y\) values using the given equation.

• Plot the points graphically, for instance, using values like \(x = 0.2, 0.4, 0.8\), etc., to get \(y\).
• As \(x\) approaches \(1.1\), \(y\) tends to infinity, which means the graph will show a vertical asymptote at \(x = 1.1\).

Understand that the graph of \(y = \frac{15x}{1.1-x}\) will have a gradual curve increasing rapidly as it nears the undefined \(x\)-value of 1.1. Drawing this graph offers a visual representation of the function's behavior and helps solidify your understanding of the domain and range.

Cost Analysis of Removing Pollutants

Cost analysis in this context refers to evaluating how the cost (\(y\)) of removing a certain percentage of pollutant (Using the given function, you determine the cost for specific scenarios:

• For \(20\%\), substitute \(x = 0.20\) to find \(y\).
• For \(40\%\), substitute \(x = 0.40\).
• And for \(80\%\), use \(x = 0.80\) to calculate \(y\).

This analysis allows you to see how costs escalate as more pollutants are removed. You'll notice that doubling \(x\) doesn't result in a doubled cost. The costs grow more than proportionately due to the nature of the rational function, indicating it is a non-linear relationship. This exponential increase in cost as \(x\) grows is particularly important for financial planning in pollution cleanup efforts.