Problem 20 - Algebra and Trigonometry

Question

\(17-28\) A function is given. Determine the average rate of change of the function between the given values of the variable. $$ f(z)=1-3 z^{2} ; \quad z=-2, z=0 $$

Step-by-Step Solution

Answer

The average rate of change is 6.

Step 1: Identify the Function and Interval

The function given is \(f(z) = 1 - 3z^2\). We need to find the average rate of change over the interval \([-2, 0]\).

Step 2: Calculate Function Values

Evaluate the function at the endpoints of the interval. For \(z = -2\), \(f(-2) = 1 - 3(-2)^2 = 1 - 12 = -11\). For \(z = 0\), \(f(0) = 1 - 3(0)^2 = 1\).

Step 3: Apply the Average Rate of Change Formula

The average rate of change is given by the formula \[\frac{f(b) - f(a)}{b - a}\]. Here, \(a = -2\), \(b = 0\), \(f(b) = 1\), and \(f(a) = -11\). Substitute these values into the formula: \[\frac{1 - (-11)}{0 - (-2)} = \frac{1 + 11}{2} = \frac{12}{2} = 6.\]

Step 4: Interpret the Result

The average rate of change of the function \(f(z) = 1 - 3z^2\) between \(z = -2\) and \(z = 0\) is 6. This means that, on average, the function increases by 6 units for each unit increase in \(z\) over this interval.

Key Concepts

Function Evaluation

Function evaluation is a fundamental concept in mathematics that involves computing the output of a function for a given input. In the context of the given exercise, we evaluate the function at two specific points within a given interval:

• For the point \( z = -2 \), we substitute in the function \( f(z) = 1 - 3z^2 \) to find the value \( f(-2) = 1 - 3(-2)^2 \).
• The calculation yields \( f(-2) = 1 - 12 = -11 \).

The second evaluation is:

• For \( z = 0 \), substituting into the function gives \( f(0) = 1 - 3(0)^2 = 1 \).

Function evaluation helps us determine the behavior of the function at specific points, enabling us to analyze changes in its context, such as finding the average rate of change.

Interval Notation

Interval notation is a way of representing a subset of numbers on the real number line. It succinctly communicates the start and end points of an interval. For the given problem, we have the interval \([-2, 0]\), which includes:

• The numbers starting from \(-2\) to \(0\).
• The square brackets \([ \text{and} ]\) indicating that both \(-2\) and \(0\) are included in the interval.

This notation is useful in calculus and applied mathematics for defining the domain over which we evaluate functions or calculate changes. It tells us exactly where to consider the function and ensures precision in interpretation.

Quadratic Functions

Quadratic functions are polynomial functions of degree two with a general form of \( ax^2 + bx + c \). In this exercise, the function given is \( f(z) = 1 - 3z^2 \), which is a quadratic function with:

• A leading coefficient of \(-3\), which means the parabola opens downward.
• No linear term, indicating symmetry about the y-axis.

Quadratic functions are characterized by their curved, parabolic shape. Understanding their form helps in determining various properties, such as vertex, axis of symmetry, and concavity. In this context, the key aspect is determining how this shape affects the rate of change over specified intervals.

Calculus Basics

Calculus basics often revolve around understanding rates of changes and behavior of functions. The average rate of change is a fundamental concept, akin to calculating the slope in algebra for linear functions. For the quadratic function \( f(z) = 1 - 3z^2 \), the average rate of change gives us a snapshot of how the function behaves over an interval:

• The formula \( \frac{f(b) - f(a)}{b - a} \) measures how much \( f(z) \) changes per unit change in \( z \).
• In our solution, \( f(-2) = -11 \) and \( f(0) = 1 \), thus the average rate of change over the interval \([-2, 0]\) is \( \frac{12}{2} = 6 \).

This concept helps in understanding broader applications in physics, engineering, and economics where rates of change are critical. It also prepares the foundation for more advanced calculus concepts such as derivatives.