Unit 2.1 – Defining Average and Instantaneous Rates of Change at a Point

Problem: Find the average rate of change of \(f(x) = x^2\) on the interval \([1, 4]\)

Solution:

1. Identify the interval: \(a = 1\), \(b = 4\)
2. Evaluate function at endpoints:
   • \(f(1) = 1^2 = 1\)
   • \(f(4) = 4^2 = 16\)
3. Apply the formula:
   \[ \frac{f(b) - f(a)}{b - a} = \frac{f(4) - f(1)}{4 - 1} = \frac{16 - 1}{3} = \frac{15}{3} = 5 \]

Answer: The average rate of change is 5

Interpretation: On average, \(f(x)\) increases by 5 units for every 1 unit increase in x over the interval \([1, 4]\)

Problem: A car's position (in miles) after t hours is given by \(s(t) = 60t + 10\). Find the average velocity from \(t = 1\) to \(t = 3\) hours.

Solution:

1. Evaluate at endpoints:
   • \(s(1) = 60(1) + 10 = 70\) miles
   • \(s(3) = 60(3) + 10 = 190\) miles
2. Calculate average rate:
   \[ \frac{s(3) - s(1)}{3 - 1} = \frac{190 - 70}{2} = \frac{120}{2} = 60 \text{ mph} \]

Answer: Average velocity = 60 mph

Interpretation: The car traveled an average of 60 miles per hour during the 2-hour trip

Problem: Find the instantaneous rate of change of \(f(x) = x^2\) at \(x = 3\) using the limit definition

Solution:

1. Set up the limit:
   \[ f'(3) = \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h} \]
2. Evaluate \(f(3 + h)\) and \(f(3)\):
   • \(f(3 + h) = (3 + h)^2 = 9 + 6h + h^2\)
   • \(f(3) = 9\)
3. Substitute into the limit:
   \[ f'(3) = \lim_{h \to 0} \frac{(9 + 6h + h^2) - 9}{h} = \lim_{h \to 0} \frac{6h + h^2}{h} \]
4. Factor and simplify:
   \[ = \lim_{h \to 0} \frac{h(6 + h)}{h} = \lim_{h \to 0} (6 + h) \]
5. Evaluate the limit:
   \[ = 6 + 0 = 6 \]

Answer: \(f'(3) = 6\)

Interpretation: At \(x = 3\), the function \(f(x) = x^2\) is changing at a rate of 6 units per unit increase in x

Problem: Find \(f'(2)\) for \(f(x) = x^3\) using \(\lim_{x \to a} \frac{f(x) - f(a)}{x - a}\)

Solution:

1. Set up the limit with \(a = 2\):
   \[ f'(2) = \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2} = \lim_{x \to 2} \frac{x^3 - 8}{x - 2} \]
2. Factor the numerator:
   • \(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\)
3. Cancel common factor:
   \[ f'(2) = \lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{x - 2} = \lim_{x \to 2} (x^2 + 2x + 4) \]
4. Evaluate the limit:
   \[ = 2^2 + 2(2) + 4 = 4 + 4 + 4 = 12 \]

Answer: \(f'(2) = 12\)

Problem: For \(g(x) = x^2 + 2x\), find:
(a) Average rate of change on \([1, 3]\)
(b) Instantaneous rate at \(x = 2\)
(c) Compare the results

Solution:

(a) Average rate on [1, 3]:

• \(g(1) = 1^2 + 2(1) = 3\)
• \(g(3) = 3^2 + 2(3) = 15\)
• Average rate = \(\frac{15 - 3}{3 - 1} = \frac{12}{2} = 6\)

(b) Instantaneous rate at x = 2:

• \(g'(2) = \lim_{h \to 0} \frac{g(2+h) - g(2)}{h}\)
• \(g(2+h) = (2+h)^2 + 2(2+h) = 4 + 4h + h^2 + 4 + 2h = 8 + 6h + h^2\)
• \(g(2) = 8\)
• \(g'(2) = \lim_{h \to 0} \frac{6h + h^2}{h} = \lim_{h \to 0} (6 + h) = 6\)

(c) Comparison:

Interestingly, the average rate over \([1, 3]\) equals the instantaneous rate at \(x = 2\) (the midpoint)! This happens because \(g(x)\) is a quadratic function, which has a symmetric shape. The tangent at the middle equals the secant slope across the interval.

Try these problems to test your understanding:

1. Find the average rate of change of \(f(x) = 3x^2 - 2x\) on \([0, 2]\)
2. Find \(f'(1)\) for \(f(x) = x^2 + 3x\) using the limit definition
3. A ball's height (in feet) after t seconds is \(h(t) = -16t^2 + 64t\). Find the average velocity from \(t = 1\) to \(t = 2\)
4. For \(g(x) = \frac{1}{x}\), find \(g'(2)\) using the limit definition

Answers:

1. 4: \(\frac{f(2) - f(0)}{2 - 0} = \frac{8 - 0}{2} = 4\)
2. 5: \(f'(1) = \lim_{h \to 0} \frac{(1+h)^2 + 3(1+h) - 4}{h} = 5\)
3. 16 ft/s: \(\frac{h(2) - h(1)}{2 - 1} = \frac{64 - 48}{1} = 16\) feet per second
4. -1/4: \(g'(2) = \lim_{h \to 0} \frac{\frac{1}{2+h} - \frac{1}{2}}{h} = -\frac{1}{4}\)