Question

The cost \(C\), in dollars, to tow a car is modeled by the function \(C(x)=2.5 x+85,\) where \(x\) is the number of miles towed. (a) What is the cost of towing a car 40 miles? (b) If the cost of towing a car is \(\$ 245,\) how many miles was it towed? (c) Suppose that you have only \(\$ 150 .\) What is the maximum number of miles that you can be towed? (d) What is the domain of \(C ?\)

Short answer

(a) \$185 (b) 64 miles (c) 26 miles (d) \(x \geq 0\)}

Step-by-step solution

Identify the given function

The towing cost function is given by \(C(x) = 2.5x + 85\), where \(x\) is the number of miles towed.

Calculate the cost for towing 40 miles

Substitute \(x = 40\) into the function \(C(x)\): \[ C(40) = 2.5(40) + 85 = 100 + 85 = 185. \] Thus, the cost of towing a car 40 miles is \$185.

Determine the miles for a cost of \$245

Set \(C(x) = 245\). This gives the equation \[ 2.5x + 85 = 245. \] Subtract 85 from both sides: \[ 2.5x = 160. \] Divide both sides by 2.5: \[ x = 64. \] So, the car was towed 64 miles.

Calculate the maximum miles within a \$150 budget

Set \(C(x) \leq 150\). This gives the inequality: \[ 2.5x + 85 \leq 150. \] Subtract 85 from both sides: \[ 2.5x \leq 65. \] Divide both sides by 2.5: \[ x \leq 26. \] Therefore, the maximum number of miles that can be towed for \$150 is 26 miles.

Determine the domain of the function

The domain represents all possible values of \(x\). Since \(x\) is the number of miles towed, it must be a non-negative value. Therefore, the domain is \(x \geq 0\).

Key concepts

linear function

A linear function is a type of mathematical function where the relationship between two variables forms a straight line when graphed. In this exercise, the function is represented as \(C(x) = 2.5x + 85\). Here, \(C(x)\) represents the cost of towing a car, and \(x\) is the number of miles towed.

The structure of a linear function is always similar to \(y = mx + b\), where:

• \(y\) (or \(C(x)\) in our example) is the dependent variable
• \(m\) (2.5 in our case) is the slope, indicating the cost per mile
• \(x\) is the independent variable
• \(b\) (85 here) is the y-intercept, showing the initial charge

Understanding the slope and y-intercept helps in recognizing how the cost increases with the distance towed.

solving inequalities

Inequalities are mathematical statements that compare two expressions. They use signs like \(\leq\), \(<\), \(>\), and \(\geq\). In our exercise, we deal with the inequality to find how far a car can be towed within a certain budget.

We solved an inequality \(2.5x + 85 \leq 150\). Here are the steps we followed:

• Subtract 85 from both sides to isolate the term with \(x\): \(2.5x \leq 65\)
• Divide each side by the slope (2.5) to solve for \(x\): \(x \leq 26\)

This result shows that, with a budget of \$150, the maximum distance a car can be towed is 26 miles.

domain and range

In functions, the domain refers to all possible input values, and the range refers to all possible output values. For our towing cost function \(C(x) = 2.5x + 85\):

• The domain is the set of all possible values of \(x\), which represents the number of miles towed. Since you can't tow a negative distance, \(x\) must be non-negative, making the domain \(x \geq 0\).
• The range is the set of all possible values of \(C(x)\). Given the domain, the lowest cost is when \(x=0\), resulting in \(C(0) = 85\). Thus, the range is \(C(x) \geq 85\).

Knowing the domain and range helps you understand the limits and possibilities of the function.

algebraic manipulation

Algebraic manipulation involves using algebraic operations to rearrange and solve equations or inequalities. In our exercise, we performed several manipulations:

• For calculating specific values, we substituted numbers into the function.
• For solving \(2.5x + 85 = 245\), we rearranged the equation by subtracting 85 and dividing by 2.5 to find \(x\).
• For solving the inequality \(2.5x + 85 \leq 150\), we used similar subtraction and division steps to isolate \(x\).

These manipulations help transform complex expressions into simpler ones, making it easier to find solutions.