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

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!).

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!