Problem 4 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(1-6,\) find the average rate of change of the function over each interval. \(f(x)=\ln x\) (a) \([1,4], \quad\) ( b) \([100,103]\)

Step-by-Step Solution

Answer

The average rate of change of the function over the interval [1,4] is \(\frac{\ln 4}{3}\) and over the interval [100,103] is \(\frac{\ln 103 - \ln 100}{3}\).

Step 1: Apply the formula for the first interval [1,4]

The average rate of change of the function over the interval [1,4] is calculated as \(\frac{f(4) - f(1)}{4 - 1}\). The function \(f(x)=\ln x\) is substituted into this formula. Therefore, it becomes \(\frac{\ln 4 - \ln 1}{4 - 1}\).

Step 2: Simplify the resulting expression

\(\ln 1 = 0\), so the expression simplifies to \(\frac{\ln 4}{3}\). This is the average rate of change of the function over the interval [1,4].

Step 3: Apply the formula for the second interval [100,103]

The average rate of change of the function over the interval [100,103] is calculated as \(\frac{f(103) - f(100)}{103 - 100}\). Substituting \(f(x)=\ln x\), it becomes \(\frac{\ln 103 - \ln 100}{103 - 100}\).

Step 4: Simplify the resulting expression

This expression can't be simplified further. Thus, the average rate of change of the function over the interval [100,103] is \(\frac{\ln 103 - \ln 100}{3}\).

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithmic function with a base of \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm function is essential in various fields, including mathematics, engineering, and the sciences, due to its ability to model exponential growth and decay processes.

Here's why the natural logarithm is unique:

• The inverse of the exponential function \( e^x \).
• The value of \( \ln(1) \) is 0, since \( e^0 = 1\).
• It's undefined for values \( x \leq 0 \).

For example, when you calculate \( \ln(4) \) or \( \ln(100) \), you're calculating the power to which \( e \) must be raised to obtain those numbers. Understanding the behavior of \( \ln(x) \) helps in analyzing growth patterns and computing average rates of change for functions, such as the one given.

Intervals

In the context of mathematical analysis, intervals are used to define a subset of the real number line. They can be finite or infinite, displaying portions of numbers over which functions can be assessed or analyzed.

The intervals in our problem, \([1, 4]\) and \([100, 103]\), are examples of closed intervals. This means:

• The endpoints 1 and 4 are included in the interval \([1, 4]\).
• Similarly, 100 and 103 are included in the interval \([100, 103]\).

Closed intervals are indicated using brackets \([a, b]\), and they allow us to consider the behavior of functions at their bounds.

When finding the average rate of change, these intervals help us determine the two specific points at which we evaluate the function, ultimately facilitating the calculation of changes within the range.

Function Analysis

Function analysis involves examining a function to understand its behavior, including its growth patterns, rates of change, and continuity over certain intervals. This particular problem focuses on calculating the average rate of change of the function \( f(x) = \ln(x) \) over two specified intervals.

To find the average rate of change between two points, like \( x_1 \) and \( x_2 \), on a function \( f(x) \), you can use the formula:\[\text{Average Rate of Change} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}\]For the intervals \([1, 4]\) and \([100, 103]\), this means calculating:

• \( \frac{\ln(4) - \ln(1)}{4 - 1} \) for the first interval, simplifying to \( \frac{\ln 4}{3} \) since \( \ln 1 = 0 \).
• \( \frac{\ln(103) - \ln(100)}{103 - 100} \) for the second interval, which doesn't simplify further.

This rate of change is a measure of how much the function \( \ln(x) \) increases or decreases on average between the two points in each interval. By understanding this analysis, you grasp the function’s consistency or variation across different ranges.