Average and Instantaneous Rates of Change

Understanding how functions change over intervals and at specific points

Average and Instantaneous Rates of Change

Average Rate of Change

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

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

Interpretation

Geometrically: Slope of the secant line connecting $(a, f(a))$ and $(b, f(b))$
Physically: Average velocity (if $f(x)$ represents position)
Generally: How fast $f$ changes on average between $x = a$ and $x = b$

Example

For $f(x) = x^2$ on $[1, 3]$ : $\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = \frac{8}{2} = 4$

The function increases by 4 units per unit of $x$ , on average.

Instantaneous Rate of Change

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

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

Interpretation

Geometrically: Slope of the tangent line at $(a, f(a))$
Physically: Instantaneous velocity at time $t = a$
Generally: How fast $f$ is changing right at $x = a$
In Calculus: This limit is called the derivative, written $f'(a)$

Difference Quotient

The expression $\frac{f(a + h) - f(a)}{h}$ is called the difference quotient.

As $h \to 0$ , the secant line approaches the tangent line.

Key Formulas

Average Rate of Change (over interval $[a, b]$ ): $\frac{f(b) - f(a)}{b - a}$

Instantaneous Rate of Change (at point $x = a$ ): $\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

Alternative form (using $x$ instead of $h$ ): $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

Real-World Applications

Position and Velocity

If $s(t)$ = position at time $t$
Average velocity = $\frac{s(t_2) - s(t_1)}{t_2 - t_1}$
Instantaneous velocity = $\lim_{h \to 0} \frac{s(t + h) - s(t)}{h}$

Cost and Marginal Cost

If $C(x)$ = total cost to produce $x$ items
Average cost per item = $\frac{C(x)}{x}$
Marginal cost (cost of one more item) ≈ instantaneous rate of change

Connection to Derivatives

In calculus, the instantaneous rate of change is the derivative: $f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

Precalculus focuses on:

Computing average rates of change
Understanding the concept of instantaneous rate of change
Setting up (but not necessarily evaluating) limit expressions

📚 Practice Problems

1Problem 1easy

❓ Question:

Find the average rate of change of $f(x) = x^2 + 3x$ over the interval $[1, 4]$ .

💡 Show Solution

Use the average rate of change formula:

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

where $a = 1$ and $b = 4$ .

Step 1: Find $f(1)$ : $f(1) = 1^2 + 3(1) = 1 + 3 = 4$

Step 2: Find $f(4)$ : $f(4) = 4^2 + 3(4) = 16 + 12 = 28$

Step 3: Calculate the average rate of change: $\frac{f(4) - f(1)}{4 - 1} = \frac{28 - 4}{3} = \frac{24}{3} = 8$

Answer: The average rate of change is $8$ .

Interpretation: On average, the function increases by 8 units for every 1 unit increase in $x$ over the interval $[1, 4]$ .

2Problem 2medium

❓ Question:

A ball is thrown upward. Its height (in feet) after $t$ seconds is given by $h(t) = -16t^2 + 64t + 5$ . Find the average velocity of the ball between $t = 1$ and $t = 3$ seconds.

💡 Show Solution

Average velocity = average rate of change of position:

$\text{Average Velocity} = \frac{h(3) - h(1)}{3 - 1}$

Step 1: Find $h(1)$ : $h(1) = -16(1)^2 + 64(1) + 5$ $= -16 + 64 + 5 = 53 \text{ feet}$

Step 2: Find $h(3)$ : $h(3) = -16(3)^2 + 64(3) + 5$ $= -16(9) + 192 + 5$ $= -144 + 192 + 5 = 53 \text{ feet}$

Step 3: Calculate average velocity: $\frac{h(3) - h(1)}{3 - 1} = \frac{53 - 53}{2} = \frac{0}{2} = 0 \text{ ft/s}$

Answer: The average velocity is $0$ ft/s.

Interpretation: The ball is at the same height at $t = 1$ and $t = 3$ . It went up and came back down to the same height, so the average velocity is zero (though it was moving the entire time!).

3Problem 3hard

❓ Question:

Write an expression for the instantaneous rate of change of $f(x) = \frac{1}{x}$ at $x = 2$ . Then estimate it using $h = 0.01$ .

💡 Show Solution

Expression for instantaneous rate of change at $x = 2$ :

$\lim_{h \to 0} \frac{f(2 + h) - f(2)}{h}$

Step 1: Find $f(2)$ : $f(2) = \frac{1}{2}$

Step 2: Find $f(2 + h)$ : $f(2 + h) = \frac{1}{2 + h}$

Step 3: Write the difference quotient: $\frac{f(2 + h) - f(2)}{h} = \frac{\frac{1}{2+h} - \frac{1}{2}}{h}$

Step 4: Simplify: $= \frac{\frac{2 - (2+h)}{2(2+h)}}{h} = \frac{\frac{-h}{2(2+h)}}{h} = \frac{-h}{2(2+h) \cdot h} = \frac{-1}{2(2+h)}$

Exact answer (taking $h \to 0$ ): $\lim_{h \to 0} \frac{-1}{2(2+h)} = \frac{-1}{2(2)} = -\frac{1}{4}$

Estimate with $h = 0.01$ : $\frac{-1}{2(2+0.01)} = \frac{-1}{2(2.01)} = \frac{-1}{4.02} \approx -0.2488$

Answer: The instantaneous rate of change is $-\frac{1}{4} = -0.25$ (exact). The estimate with $h = 0.01$ gives approximately $-0.2488$ , which is very close!

4Problem 4medium

❓ Question:

Find the average rate of change of f(x) = x² + 3x on the interval [1, 4].

💡 Show Solution

Step 1: Use the average rate of change formula: Average rate = [f(b) - f(a)] / (b - a)

Step 2: Identify a and b: a = 1, b = 4

Step 3: Calculate f(1): f(1) = 1² + 3(1) = 1 + 3 = 4

Step 4: Calculate f(4): f(4) = 4² + 3(4) = 16 + 12 = 28

Step 5: Find the average rate of change: Average rate = (28 - 4) / (4 - 1) = 24 / 3 = 8

Answer: 8

5Problem 5hard

❓ Question:

A particle moves along a line with position function s(t) = t³ - 6t² + 9t. Find the instantaneous velocity at t = 2.

💡 Show Solution

Step 1: Recall that instantaneous velocity is the derivative: v(t) = s'(t)

Step 2: Find the derivative of s(t) = t³ - 6t² + 9t: s'(t) = 3t² - 12t + 9

Step 3: Evaluate at t = 2: v(2) = 3(2)² - 12(2) + 9 = 3(4) - 24 + 9 = 12 - 24 + 9 = -3

Step 4: Interpret: The instantaneous velocity at t = 2 is -3 units/time The negative value indicates the particle is moving in the negative direction.

Answer: -3 units/time

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