Change in Tandem | undefined Unit 1 Study Guide

The Core Idea: Change in Tandem

In precalculus, we are fundamentally interested in how the output of a function changes as its input changes. The concept of "Change in Tandem" introduces the most foundational tool for measuring this relationship over an interval: the average rate of change. This measurement quantifies the overall trend of a function between two distinct points. It answers the question, "On average, how much did the output value change for each one-unit change in the input value across this specific range?"

The average rate of change is expressed as a ratio. It is the total change in the function's output values, $f (b) - f (a)$ , divided by the total change in the input values, $b - a$ . This single value provides a powerful summary of the function's behavior over the interval $[a, b]$ . Whether the function's values generally increased, decreased, or experienced no net change is revealed by the sign of this calculated rate. Geometrically, this concept is visualized as the slope of the straight line connecting the two endpoints of the interval on the function's graph, known as the secant line.

Key Formulas & Definitions

The Average Rate of Change Formula

The average rate of change of a function $f (x)$ over a closed interval $[a, b]$ is given by the formula:

$Average Rate of Change = \frac{Change in Output}{Change in Input} = \frac{f ( b ) - f ( a )}{b - a}$

$f (b)$ is the value of the function at the end of the interval.
$f (a)$ is the value of the function at the beginning of the interval.
$b - a$ is the length of the interval.

The Secant Line

The average rate of change of a function $f (x)$ on the interval $[a, b]$ is geometrically equivalent to the slope of the secant line that passes through the two points $(a, f (a))$ and $(b, f (b))$ on the graph of $f (x)$ .

$m_{secant} = \frac{f ( b ) - f ( a )}{b - a}$

This means that the average rate of change provides a linear approximation of the function's change over that interval.

Understanding the "Average" in Average Rate of Change

A critical nuance of this topic is understanding that the average rate of change describes the overall or net behavior of a function across an entire interval, not its behavior at any specific point within that interval.

The sign of the average rate of change tells us about the net result from the starting point $a$ to the ending point $b$ :

Positive ( $> 0$ ): The function's value at the end of the interval is greater than the value at the beginning ( $f (b) > f (a)$ ). On average, the function increased.
**Negative ( $< 0`):** The function's value at the end of the interval is less than the value at the beginning ($f(b) < f(a)$ ). On average, the function decreased.
Zero (= 0): The function's value at the end of the interval is the same as the value at the beginning ( $f (b) = f (a)$ ). There was no net change in the function's value.

It is possible for a function to both increase and decrease within an interval, but still have a positive, negative, or zero average rate of change. For example, consider a stock price that starts at $50, rises to $70, and then falls to $55. Over this period, the stock both increased and decreased. However, the average rate of change over the entire interval would be positive because the final price ($55) is higher than the initial price ($50). The average rate of change smooths out these intermediate fluctuations to give a single measure of the overall trend. ## Core Concepts & Rules - **Definition:** The average rate of change is the ratio of the change in the function's output to the corresponding change in its input over a specified interval. - **Calculation:** The average rate of change for a function $f$ on $[a, b]$ is always calculated using the formula $\frac{f ( b ) - f ( a )}{b - a}$ .

Data Sources: This rate can be calculated when the function is defined by an algebraic formula or when it is represented by a table of values.
Geometric Interpretation: The average rate of change is precisely the slope of the secant line connecting the points $(a, f (a))$ and $(b, f (b))$ on the graph of the function.
Interpreting the Sign:
- A positive average rate of change signifies an average increase in function values over the interval.
- A negative average rate of change signifies an average decrease in function values over the interval.
- A zero average rate of change signifies that the function's values at the start and end of the interval are identical, resulting in no net change.

Step-by-Step Example 1: Calculating from a Function's Formula

Problem: Let $f (x) = 2 x^{2} - 3 x + 1$ . Calculate the average rate of change of $f (x)$ on the interval $[1, 4]$ and interpret its meaning.

Step 1: Identify the interval and the function.

The function is $f (x) = 2 x^{2} - 3 x + 1$ .

The interval is $[a, b] = [1, 4]$ , so $a = 1$ and $b = 4$ .

Step 2: Evaluate the function at the endpoints of the interval.

Calculate $f (a) = f (1)$ :

$f (1) = 2 (1)^{2} - 3 (1) + 1 = 2 (1) - 3 + 1 = 2 - 3 + 1 = 0$ .

So, the point on the graph is $(1, 0)$ .

Calculate $f (b) = f (4)$ :

$f (4) = 2 (4)^{2} - 3 (4) + 1 = 2 (16) - 12 + 1 = 32 - 12 + 1 = 21$ .

So, the other point on the graph is $(4, 21)$ .

Step 3: Substitute the values into the average rate of change formula.

$Average Rate of Change = \frac{f ( b ) - f ( a )}{b - a} = \frac{f ( 4 ) - f ( 1 )}{4 - 1}$

$= \frac{21 - 0}{4 - 1}$

Step 4: Compute the final value.

$= \frac{21}{3} = 7$

Step 5: Interpret the result.

The average rate of change of $f (x)$ on the interval $[1, 4]$ is $7$ . Because this value is positive, it indicates that, on average, the function's values increased by $7$ units for every $1$ -unit increase in $x$ across this interval. This value also represents the slope of the secant line connecting the points $(1, 0)$ and $(4, 21)$ .

Step-by-Step Example 2: Calculating from a Table of Values

Problem: The table below shows the depth of water, $D (t)$ , in feet, at a certain point in a harbor $t$ hours after midnight.

$t$ (hours)	0	3	5	9	12
$D (t)$ (feet)	8.5	12.1	10.3	4.1	8.5

(a) Find the average rate of change of the water depth from $t = 3$ to $t = 9$ .

(b) Find the average rate of change of the water depth from $t = 0$ to $t = 12$ . Interpret the meaning of your answer.

Part (a): Interval $[3, 9]$

Step 1: Identify the interval and corresponding function values from the table.

The interval is $[a, b] = [3, 9]$ .

From the table, $D (a) = D (3) = 12.1$ .

From the table, $D (b) = D (9) = 4.1$ .

Step 2: Substitute the values into the average rate of change formula.

$Average Rate of Change = \frac{D ( 9 ) - D ( 3 )}{9 - 3} = \frac{4.1 - 12.1}{9 - 3}$

Step 3: Compute the final value.

$= \frac{- 8.0}{6} = - \frac{8}{6} = - \frac{4}{3} \approx - 1.33$

The average rate of change is $- 4/3$ feet per hour. The negative sign indicates that, on average, the water depth was decreasing between 3 and 9 hours after midnight.

Part (b): Interval $[0, 12]$

Step 1: Identify the interval and corresponding function values from the table.

The interval is $[a, b] = [0, 12]$ .

From the table, $D (a) = D (0) = 8.5$ .

From the table, $D (b) = D (12) = 8.5$ .

Step 2: Substitute the values into the average rate of change formula.

$Average Rate of Change = \frac{D ( 12 ) - D ( 0 )}{12 - 0} = \frac{8.5 - 8.5}{12 - 0}$

Step 3: Compute the final value.

$= \frac{0}{12} = 0$

Step 4: Interpret the result.

The average rate of change of the water depth from $t = 0$ to $t = 12$ is $0$ feet per hour. This means there was no net change in the water depth between midnight ( $t = 0$ ) and noon ( $t = 12$ ). Although the table shows the depth changed during the interval (it rose to 12.1 ft and fell to 4.1 ft), the depth at the end of the 12-hour period was the same as it was at the beginning.

Using Your Calculator

While the average rate of change is a conceptual and algebraic calculation, a graphing calculator (like a TI-84) can be used to ensure accuracy and speed, especially with complex functions.

Scenario: Find the average rate of change for $f (x) = \frac{x ^{3} + 5}{x}$ on the interval $[2, 5]$ .

Step 1: Define the Function

Press the [Y=] button.
In Y1, enter the function. Use parentheses to ensure correct order of operations: Y1 = (√(X^3 + 5)) / X

Step 2: Calculate on the Home Screen

Press [2nd] $t h e n$ [MODE] $t o$ [QUIT] $an d re t u r n t o t h e h o m escree n . - T h e f or m u l ai s$ (f(5) - f(2)) / (5 - 2) $. W ec an t y p e t hi s u s in g t h e ‘ Y 1‘ v a r iab l e . - T o a ccess ‘ Y 1‘, p ress ‘ [V A RS]$ , go to the $Y - V A RS$ menu, select $1 : F u n c t i o n ...$ , and then 1:Y1`.
Type the full expression using parentheses for the numerator and denominator:
(Y1(5) - Y1(2)) / (5 - 2)
Press [ENTER].

Step 3: Read the Result

The calculator will display the result, which is approximately $0.379$ . This method avoids potential rounding errors and manual calculation mistakes. The calculator's primary role here is for accurate evaluation of $f (a)$ and $f (b)$ and performing the final division.

AP Exam Quick Hit

Common Question Types

Calculation from a Function: You will be given a function, $g (x)$ , and an interval, $[a, b]$ , and asked to compute the average rate of change.
- Example: "What is the average rate of change of $f (x) = sin (π x) + x$ on the interval $[0, 2]$ ?"
Calculation from a Table: You will be given a table of values for a function, $h (t)$ , and asked to find the average rate of change between two input values in the table.
- Example: "Using the table of values for the function $h (t)$ provided, find the average rate of change on the interval $[5, 9]$ ."
Interpretation: You will be given an average rate of change and asked to explain its meaning in the context of a problem, or you will be asked to determine the sign of the average rate of change based on a graph.
- Example: "The average rate of change of the temperature of a cup of coffee, $C (t)$ , from $t = 0$ minutes to $t = 10$ minutes is $- 8.2$ degrees Celsius per minute. What does the sign of this value imply about the temperature of the coffee?"

Common Mistakes

Flipping the Formula: A very common error is to incorrectly calculate $\frac{b - a}{f ( b ) - f ( a )}$ instead of the correct $\frac{f ( b ) - f ( a )}{b - a}$ . Always remember it's "change in y" over "change in x".
Forgetting to Divide: Students sometimes calculate only the change in the function's value, $f (b) - f (a)$ , and forget to divide by the length of the interval, $b - a$ .
Mixing up Order: Calculating $\frac{f ( a ) - f ( b )}{b - a}$ or $\frac{f ( b ) - f ( a )}{a - b}$ . The order must be consistent in the numerator and denominator: $(y_{2} - y_{1}) / (x_{2} - x_{1})$ .
Misinterpreting "Average": Believing that a positive average rate of change on $[a, b]$ means the function was increasing at every point in that interval. A function can have intervals of decrease within $[a, b] ‘ an d s t i ll ha v e a p os i t i v e a v er a g er a t eo f c han g e . - * * C a l c u l a t or P a re n t h eses E rror : * * Wh e n u s in g a c a l c u l a t or, t y p in g ‘ Y 1 (b) - Y 1 (a) / (b - a) ‘ w i t h o u tp a re n t h eses a ro u n d t h e n u m er a t or . T h ec a l c u l a t o r^{'} sor d ero f o p er a t i o n s w i ll co m p u t e ‘ Y 1 (a) / (b - a)$ first, leading to an incorrect answer. The correct entry is $(Y1(b) - Y1(a)) / (b - a)`.

Change in Tandem - AP PreCalculus Study Guide