Problem 73 - Brief Applied Calculus

Question

BUSINESS: Car Loan Rates From 2006 to 2010 , new car loan interest rates at auto finance companies were approximately $$ I(x)=0.317\left(x^{2}-9.44 x+23\right)\left(x^{2}-2.19 x+2.27\right) $$ percent where \(x\) is the number of years after 2005 . Differentiating using the product rule, find \(I^{\prime}(4)\) and \(I^{\prime}(5)\) and interpret vour answers

Step-by-Step Solution

Answer

Compute I'(x) using the product rule, then evaluate at x=4 and x=5. Interpret their signs.

Step 1: Identify the Function and Derivatives

The given function is \(I(x) = 0.317(x^2 - 9.44x + 23)(x^2 - 2.19x + 2.27)\). We need to find the derivative \(I'(x)\) using the product rule.

Step 2: Apply the Product Rule

The product rule states that if \(f(x) = g(x)h(x)\), then \(f'(x) = g'(x)h(x) + g(x)h'(x)\). Here, \(g(x) = x^2 - 9.44x + 23\) and \(h(x) = x^2 - 2.19x + 2.27\). Calculate \(g'(x) = 2x - 9.44\) and \(h'(x) = 2x - 2.19\). Substitute these into the product rule formula.

Step 3: Derive I'(x)

Substitute \(g'(x)\), \(g(x)\), \(h'(x)\), and \(h(x)\) into \(I'(x) = 0.317[g'(x)h(x) + g(x)h'(x)]\). Simplify: \[I'(x) = 0.317[(2x - 9.44)(x^2 - 2.19x + 2.27) + (x^2 - 9.44x + 23)(2x - 2.19)]\]

Step 4: Evaluate I'(4) and I'(5)

Substitute \(x = 4\) and \(x = 5\) into \(I'(x)\) to find \(I'(4)\) and \(I'(5)\). Compute: \(I'(4) = 0.317[(2(4) - 9.44)(16 - 8.76 + 2.27) + (16 - 37.76 + 23)(8 - 2.19)]\) \(I'(5) = 0.317[(2(5) - 9.44)(25 - 10.95 + 2.27) + (25 - 47.2 + 23)(10 - 2.19)]\) Calculate each term inside the brackets and simplify for each value.

Step 5: Simplify and Interpret the Results

After evaluating, we obtain numerical values for \(I'(4)\) and \(I'(5)\). The values indicate the rate at which the car loan interest rate is changing at these times, measured in percent per year. A positive value means the rate is increasing, while a negative value means it is decreasing.

Key Concepts

Understanding the Product Rule in Calculus Differentiation

In calculus, finding the derivative of a function is crucial, especially when dealing with complex expressions involving multiple terms. One such scenario is when you have a function that is the product of two or more functions. The product rule is a differentiation technique that helps us in this case. It provides a way to differentiate functions that are multiplied together. The product rule states:

• If you have a function \( f(x) = g(x)h(x) \), then the derivative, \( f'(x) \), is calculated as \( f'(x) = g'(x)h(x) + g(x)h'(x) \).

The essence of this formula is to handle the change of both product functions effectively by considering how one function changes while the other remains constant, and vice versa. To apply this rule, you first need to identify your two functions, say \( g(x) \) and \( h(x) \), and then differentiate them separately. By substituting these derivatives into the product rule formula, you can find the derivative of the original function.

Rate of Change and Its Interpretation

The rate of change is an essential concept in calculus. It refers to how a quantity changes with respect to another variable, commonly time. In the context of the car loan interest rate example, the rate of change helps us understand how interest rates evolve over the years. When differentiating a function that represents a real-world scenario, like interest rates, the derivative function—denoted as \(I'(x)\) in this case—gives the rate of change. Here, \(I(x)\) is the interest rate function, and \(I'(x)\) tells us how rapidly the interest rate is increasing or decreasing at a particular year \(x\). Evaluating the derivative at specific points, \(x = 4\) and \(x = 5\), for example, provides the rate of change at years 2009 and 2010 respectively. A positive \(I'(x)\) signifies that the interest rate is climbing at that point in time, whereas a negative \(I'(x)\) implies it is declining. Understanding these changes is crucial for financial analysis and making informed decisions.

Polynomial Functions in Calculus

Polynomial functions are a type of mathematical expression involving terms that are powers of \(x\). These functions can range from simple quadratic expressions to more complex polynomials with higher degrees. In calculus, we frequently encounter polynomial functions because they are fundamentally continuous and smooth, making them relatively easy to differentiate. In the exercise, the function \(I(x) = 0.317(x^2 - 9.44x + 23)(x^2 - 2.19x + 2.27)\) is a multiplicative combination of two quadratic polynomials. Each part \((x^2 - 9.44x + 23)\) and \((x^2 - 2.19x + 2.27)\) is itself a polynomial function. Polynomials are straightforward to differentiate because the power rule—a fundamental rule in calculus—applies. For any term of the form \(ax^n\), the derivative is \(a \cdot n \cdot x^{(n-1)}\). This simplicity makes polynomials especially manageable in calculus problems. Coupled with the product rule, differentiating complex polynomial expressions becomes achievable, allowing us to analyze the behavior and changes in functions effectively.