Problem 44 - College Algebra with Modeling and Visualization

Question

Exercises \(39-44\) : Average Rate of Change Find the average rate of change of \(f\) from \(-2\) to \(2 .\) What is the average rate of change of \(f\) from \(x_{1}\) to \(x_{2}\), where \(x_{1} \neq x_{2} ?\) $$ f(x)=5 x+1 $$

Step-by-Step Solution

Answer

The average rate of change is 5.

Step 1: Understand the Average Rate of Change Formula

The average rate of change of a function \( f(x) \) from \( x_1 \) to \( x_2 \) is given by the formula: \[ \text{Average Rate of Change} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \] This represents how much the function changes, on average, per unit increase in \( x \).

Step 2: Identify Values for x1 and x2

For this exercise, we want to find the average rate of change from \( x_1 = -2 \) to \( x_2 = 2 \).

Step 3: Evaluate Function at x1 and x2

Substitute \( x_1 = -2 \) and \( x_2 = 2 \) into the function \( f(x) = 5x + 1 \). Calculate \( f(-2) \) and \( f(2) \). \[ f(-2) = 5(-2) + 1 = -10 + 1 = -9 \] \[ f(2) = 5(2) + 1 = 10 + 1 = 11 \]

Step 4: Calculate the Difference in Function Values

Find the difference \( f(x_2) - f(x_1) \) using the values from Step 3:\[ f(2) - f(-2) = 11 - (-9) = 11 + 9 = 20 \]

Step 5: Compute the Average Rate of Change

Apply the values obtained in Steps 2 and 4 into the formula:\[ \text{Average Rate of Change} = \frac{20}{2 - (-2)} = \frac{20}{2 + 2} = \frac{20}{4} = 5 \]

Step 6: Conclusion

The average rate of change of the function \( f(x) = 5x + 1 \) from \( x = -2 \) to \( x = 2 \) is 5.

Key Concepts

Linear Functions

Linear functions are one of the simplest forms of functions in mathematics. They are represented by the formula \( f(x) = ax + b \), where \( a \) and \( b \) are constants. In linear functions, the variable \( x \) expresses a direct relationship that is graphically represented as a straight line on a coordinate plane.
Here are key characteristics of linear functions:

• They display a constant rate of change. The graph of a linear function will always be a straight line, where the slope \( a \) remains uniform across the graph.
• Linear functions are defined everywhere, meaning you can substitute any real number in for \( x \) to find \( f(x) \).
• The \( b \) term in the function represents the y-intercept, or the point at which the line crosses the y-axis.

In the provided exercise, our function is \( f(x) = 5x + 1 \). This simple linear function tells us a few things:

• The slope \( a \) is \( 5 \), indicating that for every one-unit increase in \( x \), \( f(x) \) increases by 5 units.
• The y-intercept is \( 1 \), which means when \( x = 0 \), \( f(x) = 1 \).

Function Evaluation

Function evaluation is the process of finding the output of a function for specific input values. When you have a function equation, you can determine the corresponding function value by substituting the input value into the function formula. In terms of linear functions like \( f(x) = 5x + 1 \), this involves replacing \( x \) with a given number.
In the step-by-step solution for our exercise:

• To find \( f(-2) \), substitute \( -2 \) into the function: \( f(-2) = 5(-2) + 1 = -10 + 1 = -9 \).
• Similarly, for \( f(2) \), replace \( x \) with \( 2 \): \( f(2) = 5(2) + 1 = 10 + 1 = 11 \).

By evaluating a function, you're essentially finding points on the line described by the function. This simple substitution approach reveals outputs that correspond with the inputs used, which helps in solving for behaviors such as the average rate of change.

Difference Quotient

The difference quotient is a fundamental concept in calculus, used to measure the average rate of change of a function over a specific interval. It is represented by the formula:\[\text{Difference Quotient} = \frac{f(x_2) - f(x_1)}{x_2 - x_1} \]This computation involves finding the difference in function values at two points \( x_1 \) and \( x_2 \) and dividing by the difference in the input values \( x_2 - x_1 \). This essentially gives us a slope of the secant line between the points on a curve of a function.
For a linear function, like \( f(x) = 5x + 1 \), the difference quotient simplifies things significantly because:

• Calculating \( f(x_2) - f(x_1) \) gives the total change in function value between \( x_1 \) and \( x_2 \).
• The difference \( x_2 - x_1 \) is simply the change in input values.
• The resulting quotient is the constant slope of a linear function. For \( x_1 = -2 \) and \( x_2 = 2 \), the difference in function values is 20, and the difference in \( x \) values is 4, yielding an average rate of change of 5. This matches the slope \( a \) of the function.

The difference quotient, therefore, gives us a powerful tool for understanding how this function behaves over an interval, with its value being identical to the slope for linear functions.