Problem 13 - Brief Applied Calculus

Question

Find the average rate of change of the given function between the following pairs of \(x\) -values. [Hint: See pages 95-96.] a. \(x=3\) and \(x=5\) b. \(x=3\) and \(x=4\) c. \(x=3\) and \(x=3.5\) d. \(x=3\) and \(x=3.1\) e. \(x=3\) and \(x=3.01\) f. What number do your answers seem to be approaching? $$ f(x)=5 x+1 $$

Step-by-Step Solution

Answer

The average rate of change approaches 5, which is consistent for a linear function.

Step 1: Understanding the Problem

We need to find the average rate of change of the function \( f(x) = 5x + 1 \) between different pairs of \(x\)-values. The average rate of change represents the slope of the secant line between the two points determined by these \(x\)-values on the graph of the function.

Step 2: Using the Average Rate of Change Formula

The average rate of change of a function \( f(x) \) between \( x = a \) and \( x = b \) is given by the formula \[ \frac{f(b) - f(a)}{b - a}. \] This formula calculates the change in the function value divided by the change in \(x\).

Step 3: Solving Part a: x=3 and x=5

Calculate \( f(3) = 5(3) + 1 = 16 \) and \( f(5) = 5(5) + 1 = 26 \). The average rate of change is \[ \frac{26 - 16}{5 - 3} = \frac{10}{2} = 5. \]

Step 4: Solving Part b: x=3 and x=4

Calculate \( f(3) = 16 \) and \( f(4) = 5(4) + 1 = 21 \). The average rate of change is \[ \frac{21 - 16}{4 - 3} = \frac{5}{1} = 5. \]

Step 5: Solving Part c: x=3 and x=3.5

Calculate \( f(3) = 16 \) and \( f(3.5) = 5(3.5) + 1 = 18.5 \). The average rate of change is \[ \frac{18.5 - 16}{3.5 - 3} = \frac{2.5}{0.5} = 5. \]

Step 6: Solving Part d: x=3 and x=3.1

Calculate \( f(3) = 16 \) and \( f(3.1) = 5(3.1) + 1 = 16.5 \). The average rate of change is \[ \frac{16.5 - 16}{3.1 - 3} = \frac{0.5}{0.1} = 5. \]

Step 7: Solving Part e: x=3 and x=3.01

Calculate \( f(3) = 16 \) and \( f(3.01) = 5(3.01) + 1 = 16.05 \). The average rate of change is \[ \frac{16.05 - 16}{3.01 - 3} = \frac{0.05}{0.01} = 5. \]

Step 8: Observing the Limit of the Average Rate of Change

Notice that in all calculations, as the second \(x\)-value approaches 3, the average rate of change consistently results in 5. This suggests that the derivative of the function \( f(x) = 5x + 1 \) at \( x = 3 \) is 5. For any linear function \( ax + b \), the rate of change is constant and equals \( a \), the coefficient of \( x \).

Key Concepts

Secant Line

In mathematics, a secant line is a straight line that intersects a curve at two or more points. It represents the average rate of change of the function over an interval

• The secant line is particularly useful when trying to find the average rate of change over a specific interval.
• It can be visualized as connecting two points on the curve defined by the function.

In our exercise, the secant line can be determined for each pair of given x-values. For example, between the points where \( x = 3 \) and \( x = 5 \), the secant line represents the average rate of change between these two points.
The formula to determine the average rate of change, and thus the slope of the secant line, is \[ \frac{f(b) - f(a)}{b - a} \]
This formula helps calculate the slope of the secant line, by evaluating the differences in the function's values for the chosen x-values.

Linear Function

A linear function is a function whose graph is a straight line. This type of function is expressed in the form of \( f(x) = ax + b \), where \( a \) and \( b \) are constants. For example, the function given in the exercise, \( f(x) = 5x + 1 \), is a linear function.

• The main characteristic of a linear function is its constant rate of change. This is represented by the coefficient of \( x \), which in our example is 5.
• Linear functions have a simple graph - a straight line - which makes them easy to analyze.

One key feature of linear functions is that the slope or rate of change is the same regardless of the \( x \)-values chosen. In the context of this exercise, that means the average rate of change remains constant at 5 for any interval. This characteristic simplifies the calculation of the average rate of change, as it will equal the slope of the entire line.

Derivative

The derivative of a function at a point is a measure of how the function's output value changes as its input value changes. It is essentially the rate of change or the slope of the tangent line at that point.

• For linear functions, such as \( f(x) = 5x + 1 \), the derivative is a constant. This means that the rate of change at any point is the same.
• In more complex functions, the derivative can vary at different points.

When finding the limit of the average rate of change as the interval approaches a specific point, you essentially find the derivative at that point.
In the provided exercise, as the intervals get smaller and closer to \( x = 3 \), the average rate consistently equals 5. This number is the derivative of the function at \( x = 3 \). For the function \( f(x) = 5x + 1 \), the derivative at any point is 5, reflecting the constant change due to its linearity.