2.1 - Defining Average and Instantaneous Rates of Change at a Point

Introduction to Rate of Change

Rate of change is something you’ve been working with since middle school. It’s just slope. In this lesson we’ll review average rate of change and introduce instantaneous rate of change, which is the foundation for derivatives.

Important: You’ll need a graphing calculator for this lesson, and make sure it’s set to RADIAN mode, not degrees. This matters any time you use sine, cosine, or tangent.

Average Rate of Change (Review)

Definition: Slope Between Two Points

The average rate of change is the slope of the line connecting two points on a function.

Slope formula:

\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta y}{\Delta x}

For a function on the interval $[a, b]$ :

\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}

Alternative form (the order doesn’t matter as long as you’re consistent):

\frac{y_1 - y_2}{x_1 - x_2}

Example 1: Finding Average Rate of Change Algebraically

Find the average rate of change of

f(x) = \ln(2x)

on the interval

[2, 5]

Step 1: Apply the formula

\frac{f(5) - f(2)}{5 - 2}

Step 2: Evaluate each function value

f(5) = \ln(2 \cdot 5) = \ln(10)

f(2) = \ln(2 \cdot 2) = \ln(4)

Step 3: Substitute

\frac{\ln(10) - \ln(4)}{5 - 2} = \frac{\ln(10) - \ln(4)}{3}

Step 4: Calculate with a calculator

\frac{2.3026\ldots - 1.3863\ldots}{3} = \frac{0.9163\ldots}{3} \approx 0.305

Answer:

0.305

(rounded to three decimal places)

Using Calculator Function Notation

For complicated functions, there’s an efficient calculator method that saves you from retyping expressions.

Calculator Steps (TI-84)

Step 1: Enter your function in Y1. Press Y= and enter the function, like ln(2X).

Step 2: Return to the home screen with 2nd → QUIT.

Step 3: Use function notation. Press VARS → scroll right to Y-VARS → select Function → choose Y1.

Step 4: Evaluate at specific values. Type Y1(5) and press ENTER to get

f(5)

, then Y1(2) for

f(2)

Step 5: Calculate the rate of change. Type (Y1(5) - Y1(2))/3 and press ENTER.

This is especially handy for long, complicated functions.

Critical Warning: Avoid Rounding Errors

The Problem with Early Rounding

Wrong approach:

\ln(10) \approx 2.30 \quad \text{(rounded early)}

\ln(4) \approx 1.39 \quad \text{(rounded early)}

\frac{2.30 - 1.39}{3} = \frac{0.91}{3} \approx 0.303

That gives an incorrect answer because of rounding error.

The Correct Approach: Round Only at the End

Keep full precision the whole way through. Only round (or truncate) to three decimal places at the very final answer.

Rule: Never round intermediate steps. Let your calculator hold onto all the digits until the end.

Example 2: Average Rate of Change from a Table

Given the table below, find the average rate of change of

g(x)

on the interval

[3, 25]

$x$	3	10	25	40
$g(x)$	8	15	-3	12

Identify the right values: We only need

x = 3

and

x = 25

, so

g(3) = 8

and

g(25) = -3

Calculate:

\frac{g(25) - g(3)}{25 - 3} = \frac{-3 - 8}{22} = \frac{-11}{22} = -\frac{1}{2}

Answer:

-\frac{1}{2}

, or

-0.5

Common mistake: Don’t assume the numbers in the first row go in the numerator. The

y

-values (function values) go on top and the

x

-values go on the bottom.

Introduction to Instantaneous Rate of Change

The Secant Line

The average rate of change is the slope of a secant line, which is a line connecting two points on a curve.

2.1 - Defining Average and Instantaneous Rates of Change at a Point

The secant line connects

(a, f(a))

and

(a + h, f(a + h))

Slope of the secant line:

\frac{f(a + h) - f(a)}{h}

The Tangent Line (Instantaneous Rate of Change)

So what happens as the two points get closer together? As

h \to 0

, the secant line approaches the tangent line.

The tangent line touches the curve at the point

(a, f(a))

and represents the instantaneous rate of change at that point. (It can cross the curve again somewhere else. What matters is that it matches the curve’s direction right at that point.)

Instantaneous rate of change at $x = a$ :

\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}

This limit gives you the slope of the tangent line at the point

(a, f(a))

Watching it happen with numbers. Take

f(x) = x^2

a = 3

. For any

h

, the secant slope simplifies to

\frac{(3 + h)^2 - 9}{h} = 6 + h

. Here’s what happens as

h

shrinks toward

0

from both sides:

$h$	$-0.1$	$-0.01$	$-0.001$	$0.001$	$0.01$	$0.1$
Secant slope	$5.9$	$5.99$	$5.999$	$6.001$	$6.01$	$6.1$

The secant slopes close in on

6

from both sides, so the instantaneous rate of change of

x^2

x = 3

6

Two Forms of the Definition

There are two equivalent ways to express instantaneous rate of change.

Form 1: Using $h$

\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}

What it’s saying: - Start at the point

(a, f(a))

- Move

h

units to the right, to

(a + h, f(a + h))

- Find the slope between those points - Let

h

approach zero

Form 2: Using $x$

\lim_{x \to a} \frac{f(x) - f(a)}{x - a}

What it’s saying: - Start at the point

(a, f(a))

- Consider another point

(x, f(x))

- Find the slope between those points - Let

x

approach

a

Both forms give you the same result.

Before you take the limit, both fractions are difference quotients, and each one is an average rate of change.

\frac{f(a + h) - f(a)}{h}

is the average rate of change of

f

on the interval from

a

a + h

, and

\frac{f(x) - f(a)}{x - a}

is the average rate of change on the interval from

a

x

. Taking the limit is what turns an average rate into an instantaneous one.

The value of either limit is called the derivative of $f$ at $x = a$ , written

f'(a)

and read “f prime of a.” This only works provided the limit exists. If the limit doesn’t exist (for example, the left side and the right side approach different values), then

f'(a)

doesn’t exist and there’s no instantaneous rate of change at

x = a

. Topic 2.4 shows when that happens.

Finding Instantaneous Rate of Change

Example 3: Using Form 1 (with $h$ )

Find the instantaneous rate of change of

f(x) = x - x^2

x = -1

Step 1: Identify

a

and

f(x)

. Here

a = -1

and

f(x) = x - x^2

Step 2: Set up the limit

\lim_{h \to 0} \frac{f(-1 + h) - f(-1)}{h}

Step 3: Evaluate

f(-1 + h)

f(-1 + h) = (h - 1) - (h - 1)^2

= (h - 1) - (h^2 - 2h + 1)

= h - 1 - h^2 + 2h - 1

= -h^2 + 3h - 2

Step 4: Evaluate

f(-1)

f(-1) = (-1) - (-1)^2 = -1 - 1 = -2

Step 5: Substitute and simplify

\lim_{h \to 0} \frac{(-h^2 + 3h - 2) - (-2)}{h} = \lim_{h \to 0} \frac{-h^2 + 3h}{h} = \lim_{h \to 0} \frac{h(-h + 3)}{h}

Step 6: Cancel

h

= \lim_{h \to 0} (-h + 3)

Step 7: Evaluate the limit

= 0 + 3 = 3

Answer: The instantaneous rate of change at

x = -1

is 3. That means the slope of the tangent line to the curve at

x = -1

3

Example 4: Using Form 2 (with $x$ )

Find the instantaneous rate of change of

f(x) = x - x^2

x = -1

, this time using the alternate form.

Step 1: Set up the limit

\lim_{x \to -1} \frac{f(x) - f(-1)}{x - (-1)}

Step 2: Substitute

\lim_{x \to -1} \frac{(x - x^2) - (-2)}{x + 1} = \lim_{x \to -1} \frac{x - x^2 + 2}{x + 1}

Step 3: Factor the numerator

= \lim_{x \to -1} \frac{-(x^2 - x - 2)}{x + 1} = \lim_{x \to -1} \frac{-(x - 2)(x + 1)}{x + 1}

Step 4: Cancel

(x + 1)

= \lim_{x \to -1} -(x - 2)

Step 5: Evaluate

= -(-1 - 2) = -(-3) = 3

Answer: 3, the same as Form 1.

Recognizing Derivative Expressions

On AP Calculus exams, you’ll need to identify the original function and the point from a limit expression.

Example 5: Identify from Form 1

Given:

\lim_{h \to 0} \frac{5\ln\left(\frac{2}{4 + h}\right) - 5\ln\left(\frac{1}{2}\right)}{h}

What’s the original function and the point?

Identify the pattern:

\frac{f(a + h) - f(a)}{h}

The function: Look at the first term,

5\ln\left(\frac{2}{x}\right)

, so

f(x) = 5\ln\left(\frac{2}{x}\right)

The point: Where is

h

being added? It’s added to make

4 + h

, so

a = 4

Answer:

f(x) = 5\ln\left(\frac{2}{x}\right)

x = 4

Example 6: Identify from Form 2

Given:

\lim_{x \to \frac{\pi}{3}} \frac{\sin(x) - \sin\left(\frac{\pi}{3}\right)}{x - \frac{\pi}{3}}

What’s the original function and the point?

Identify the pattern:

\frac{f(x) - f(a)}{x - a}

The function:

f(x) = \sin(x)

The point:

x = \frac{\pi}{3}

, the value

x

approaches.

Answer:

f(x) = \sin(x)

x = \frac{\pi}{3}

Practice Problems

Problem 1

Find the average rate of change of

h(x) = x^3 - 2x

[1, 3]

Answer:

\frac{h(3) - h(1)}{3 - 1} = \frac{(27 - 6) - (1 - 2)}{2} = \frac{21 - (-1)}{2} = \frac{22}{2} = 11

Problem 2

Find the instantaneous rate of change of

g(x) = x^2 + 3

x = 2

Answer:

\lim_{h \to 0} \frac{[(2 + h)^2 + 3] - [7]}{h} = \lim_{h \to 0} \frac{4 + 4h + h^2 + 3 - 7}{h}

= \lim_{h \to 0} \frac{4h + h^2}{h} = \lim_{h \to 0} (4 + h) = 4

Problem 3

Identify the function and the point:

\lim_{x \to 2} \frac{e^x - e^2}{x - 2}

Answer:

f(x) = e^x

x = 2

Key Takeaways

Average rate of change is the slope between two points: $\frac{f(b) - f(a)}{b - a}$

Use calculator function notation for complicated expressions

Never round early. Only round to three decimals at the final answer

A secant line connects two points, and its slope is the average rate of change

A tangent line touches at one point, and its slope is the instantaneous rate of change

Two forms for instantaneous rate of change:

Form 1: $\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$
Form 2: $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

Both forms give the same answer

The value of the limit is $f'(a)$ , the derivative at $x = a$ , provided the limit exists

Later you’ll learn shortcut rules for finding derivatives, but you have to understand this limit definition first

On exams, you’ll need to identify the original function and point from limit expressions

This concept is the foundation for derivatives

Back Next

2.1 - Defining Average and Instantaneous Rates of Change at a Point

Introduction to Rate of Change

Average Rate of Change (Review)

Definition: Slope Between Two Points

Example 1: Finding Average Rate of Change Algebraically

Using Calculator Function Notation

Calculator Steps (TI-84)

Critical Warning: Avoid Rounding Errors

The Problem with Early Rounding

The Correct Approach: Round Only at the End

Example 2: Average Rate of Change from a Table

Introduction to Instantaneous Rate of Change

The Secant Line

The Tangent Line (Instantaneous Rate of Change)

Two Forms of the Definition

Form 1: Using hhh

Form 2: Using xxx

Finding Instantaneous Rate of Change

Example 3: Using Form 1 (with hhh)

Example 4: Using Form 2 (with xxx)

Recognizing Derivative Expressions

Example 5: Identify from Form 1

Example 6: Identify from Form 2

Practice Problems

Problem 1

Problem 2

Problem 3

Key Takeaways

Form 1: Using $h$

Form 2: Using $x$

Example 3: Using Form 1 (with $h$ )

Example 4: Using Form 2 (with $x$ )