91Ó°ÊÓ

In calculus, the rate of change helps us understand how a function changes with respect to one of its variables. It is particularly useful to analyze how modifications in one variable affect the overall function. In our exercise, the function \( R(x) = 5x^2 - 35x + 104 \) represents the death rate per 10,000 person-years of Harvard alumni based on their weekly exercise measured in kilocalories. When we talk about the rate of change in this context, we're asking how sensitive the death rate is to changes in the amount of exercise performed. By examining this rate, policymakers and health practitioners can make informed decisions about recommended exercise levels.

The rate of change is represented by the derivative of a function, which provides a powerful way to approximate changes over small intervals. This is done through two primary calculations: the derivative, which gives us an instantaneous rate of change, and the actual change in function value over a given interval. Understanding both helps in comparing theoretical predictions to actual outcomes, adding depth to our analysis.

To find how the death rate function \( R(x) \) changes with the amount of exercise, we calculate its derivative. The derivative, denoted as \( R'(x) \), tells us the slope of the tangent line to the curve at any point \( x \). It represents the rate at which the death rate changes per unit change in exercise kilocalories.

• Start by differentiating \( R(x) = 5x^2 - 35x + 104 \).
• Using the power rule, which states the derivative of \( ax^n \) is \( anx^{n-1} \), we find: \( R'(x) = 10x - 35 \).

This result indicates that at any value of \( x \), the rate at which the death rate changes is influenced by \( 10x - 35 \). In the exercise, with \( x = 2 \), the derivative calculates as \( R'(2) = 10 \times 2 - 35 = -15 \). Hence, the immediate rate of change at this exercise level is \(-15\), meaning the death rate decreases by 15 per 10,000 person-years as weekly exercise increases slightly.

Evaluating a function means calculating its values at specific points. This is central to understanding how changes in inputs affect the outputs of a function, such as the death rate for our example.

To illustrate the impact of exercise on the death rate, we calculate the function at \( x = 2 \) and \( x = 2.4 \). Here’s a step-by-step breakdown:

• First, evaluate \( R(2) = 5(2)^2 - 35(2) + 104 = 54 \). This tells us the death rate at 2000 kilocalories per week.
• Next, evaluate \( R(2.4) = 5(2.4)^2 - 35(2.4) + 104 = 48.8 \). At 2400 kilocalories, the death rate decreases to 48.8.

We use these values to determine \( \Delta R \), the actual change in the death rate, by calculating \( \Delta R = R(2.4) - R(2) = -5.2 \). So, the death rate decreases by 5.2 per 10,000 person-years when kilocalorie intake changes from 2000 to 2400. This function evaluation along with the derivative allows for a comprehensive understanding of the function’s behavior over the specified interval.