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Lesson 1: Average and Instantaneous Rates of Change

Estimated time: 30-40 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Average Rate of Change

The average rate of change of a function f over the interval [a, b] measures how much the output changes per unit change in input, on average.

Average Rate of Change

The average rate of change of f on [a, b] is: [f(b) − f(a)] / (b − a)

Geometrically, this is the slope of the secant line connecting the points (a, f(a)) and (b, f(b)).

Example 1: Average Velocity

A ball is dropped from a tower. Its height after t seconds is h(t) = 100 − 16t² feet. Find the average velocity from t = 1 to t = 2.

Step 1: h(1) = 100 − 16(1) = 84 ft. h(2) = 100 − 16(4) = 36 ft.

Step 2: Average velocity = [h(2) − h(1)]/(2 − 1) = (36 − 84)/1 = −48 ft/s.

The negative sign indicates the ball is falling (height is decreasing).

Example 2: Average Rate from a Formula

Find the average rate of change of f(x) = x³ from x = 1 to x = 3.

Solution: [f(3) − f(1)]/(3 − 1) = (27 − 1)/2 = 26/2 = 13.

The Difference Quotient

To make the average rate of change more flexible, we use a general expression called the difference quotient.

Difference Quotient

For a function f and a small increment h: [f(a + h) − f(a)] / h

This gives the average rate of change of f over the interval [a, a + h].

Example 3: Computing a Difference Quotient

For f(x) = x², compute the difference quotient at x = a.

Step 1: f(a + h) = (a + h)² = a² + 2ah + h²

Step 2: f(a + h) − f(a) = a² + 2ah + h² − a² = 2ah + h²

Step 3: Divide by h: (2ah + h²)/h = 2a + h

From Average to Instantaneous

The average rate of change over [a, a+h] gives the slope of the secant line through two points. As h → 0, this secant line rotates toward the tangent line at x = a. The slope of the tangent line is the instantaneous rate of change.

Instantaneous Rate of Change

The instantaneous rate of change of f at x = a is:

limh→0 [f(a + h) − f(a)] / h

provided this limit exists. This is the slope of the tangent line to y = f(x) at the point (a, f(a)).

Example 4: Instantaneous Rate of Change

Find the instantaneous rate of change of f(x) = x² at x = 3.

Step 1: From Example 3, the difference quotient for f(x) = x² is 2a + h.

Step 2: At a = 3: difference quotient = 2(3) + h = 6 + h.

Step 3: Take the limit: limh→0 (6 + h) = 6.

The slope of the tangent line to y = x² at x = 3 is 6.

Example 5: Instantaneous Velocity

Using h(t) = 100 − 16t², find the instantaneous velocity at t = 1.

Step 1: [h(1+h) − h(1)]/h = [100 − 16(1+h)² − 84]/h

Step 2: = [100 − 16(1 + 2h + h²) − 84]/h = [100 − 16 − 32h − 16h² − 84]/h

Step 3: = (−32h − 16h²)/h = −32 − 16h

Step 4: limh→0 (−32 − 16h) = −32 ft/s.

The Tangent Line

Once you know the instantaneous rate of change (slope) at a point, you can write the equation of the tangent line using point-slope form.

Equation of the Tangent Line

The tangent line to y = f(x) at x = a is: y − f(a) = m(x − a)

where m = limh→0 [f(a+h) − f(a)]/h is the slope.

Example 6: Equation of a Tangent Line

Find the equation of the tangent line to f(x) = x² at x = 3.

Step 1: f(3) = 9, so the point is (3, 9).

Step 2: From Example 4, the slope m = 6.

Step 3: y − 9 = 6(x − 3), so y = 6x − 9.

Graphical Interpretation

The relationship between secant and tangent lines is fundamental:

Key Takeaways

Check Your Understanding

1. Find the average rate of change of f(x) = 2x² + 1 from x = 1 to x = 4.

[f(4) − f(1)]/(4 − 1) = (33 − 3)/3 = 30/3 = 10.

2. Compute the difference quotient for f(x) = 3x − x² at x = a.

f(a+h) = 3(a+h) − (a+h)² = 3a+3h − a² − 2ah − h². f(a+h) − f(a) = 3h − 2ah − h². Divide by h: 3 − 2a − h.

3. Find the instantaneous rate of change of f(x) = 3x − x² at x = 1.

limh→0 (3 − 2(1) − h) = 3 − 2 = 1.

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