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)f(x) over the interval [a,b][a, b] is:

Average Rate of Change=f(b)−f(a)b−a=ΔyΔx\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))(a, f(a)) and (b,f(b))(b, f(b))
  • Physically: Average velocity (if f(x)f(x) represents position)
  • Generally: How fast ff changes on average between x=ax = a and x=bx = b

Example

For f(x)=x2f(x) = x^2 on [1,3][1, 3]: f(3)−f(1)3−1=9−12=82=4\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = \frac{8}{2} = 4

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

Instantaneous Rate of Change

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

lim⁡h→0f(a+h)−f(a)h\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}

Interpretation

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

Difference Quotient

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

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

Key Formulas

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

Instantaneous Rate of Change (at point x=ax = a): lim⁡h→0f(a+h)−f(a)h\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}

Alternative form (using xx instead of hh): lim⁡x→af(x)−f(a)x−a\lim_{x \to a} \frac{f(x) - f(a)}{x - a}

Real-World Applications

Position and Velocity

  • If s(t)s(t) = position at time tt
  • Average velocity = s(t2)−s(t1)t2−t1\frac{s(t_2) - s(t_1)}{t_2 - t_1}
  • Instantaneous velocity = lim⁡h→0s(t+h)−s(t)h\lim_{h \to 0} \frac{s(t + h) - s(t)}{h}

Cost and Marginal Cost

  • If C(x)C(x) = total cost to produce xx items
  • Average cost per item = C(x)x\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→0f(a+h)−f(a)hf'(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)=x2+3xf(x) = x^2 + 3x over the interval [1,4][1, 4].

💡 Show Solution

Use the average rate of change formula:

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

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

Step 1: Find f(1)f(1): f(1)=12+3(1)=1+3=4f(1) = 1^2 + 3(1) = 1 + 3 = 4

Step 2: Find f(4)f(4): f(4)=42+3(4)=16+12=28f(4) = 4^2 + 3(4) = 16 + 12 = 28

Step 3: Calculate the average rate of change: f(4)−f(1)4−1=28−43=243=8\frac{f(4) - f(1)}{4 - 1} = \frac{28 - 4}{3} = \frac{24}{3} = 8

Answer: The average rate of change is 88.

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

2Problem 2medium

❓ Question:

A ball is thrown upward. Its height (in feet) after tt seconds is given by h(t)=−16t2+64t+5h(t) = -16t^2 + 64t + 5. Find the average velocity of the ball between t=1t = 1 and t=3t = 3 seconds.

💡 Show Solution

Average velocity = average rate of change of position:

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

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

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

Step 3: Calculate average velocity: h(3)−h(1)3−1=53−532=02=0 ft/s\frac{h(3) - h(1)}{3 - 1} = \frac{53 - 53}{2} = \frac{0}{2} = 0 \text{ ft/s}

Answer: The average velocity is 00 ft/s.

Interpretation: The ball is at the same height at t=1t = 1 and t=3t = 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)=1xf(x) = \frac{1}{x} at x=2x = 2. Then estimate it using h=0.01h = 0.01.

💡 Show Solution

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

lim⁡h→0f(2+h)−f(2)h\lim_{h \to 0} \frac{f(2 + h) - f(2)}{h}

Step 1: Find f(2)f(2): f(2)=12f(2) = \frac{1}{2}

Step 2: Find f(2+h)f(2 + h): f(2+h)=12+hf(2 + h) = \frac{1}{2 + h}

Step 3: Write the difference quotient: f(2+h)−f(2)h=12+h−12h\frac{f(2 + h) - f(2)}{h} = \frac{\frac{1}{2+h} - \frac{1}{2}}{h}

Step 4: Simplify: =2−(2+h)2(2+h)h=−h2(2+h)h=−h2(2+h)⋅h=−12(2+h)= \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→0h \to 0): lim⁡h→0−12(2+h)=−12(2)=−14\lim_{h \to 0} \frac{-1}{2(2+h)} = \frac{-1}{2(2)} = -\frac{1}{4}

Estimate with h=0.01h = 0.01: −12(2+0.01)=−12(2.01)=−14.02≈−0.2488\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 −14=−0.25-\frac{1}{4} = -0.25 (exact). The estimate with h=0.01h = 0.01 gives approximately −0.2488-0.2488, which is very close!

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