AP Calculus AB 3.2 Notes

Rolle's Theorem states that:
- If a function $f(x)$ is continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, and if $f(a) = f(b)$, then there exists at least one value of $c$ in the interval $(a, b)$ such that
  f'(c) = 0

Function: $f(x) = x^2 - 3x + 2$
- Finding x-intercepts:
  - Set $f(x) = 0$:
    x^2 - 3x + 2 = 0
  - Factoring:
    (x - 1)(x - 2) = 0
  - Solutions are x = 1 and x = 2.
- Check for continuity and differentiability:
  - The polynomial function is continuous and differentiable everywhere.
- Evaluate $f'(x)$:
  - f'(x) = 2x - 3
  - Set the derivative to zero to find critical points:
    2x - 3 = 0
    2x = 3
    ightarrow x = rac{3}{2}
- Conclusion: There exists a point at c = rac{3}{2} where f'(c) = 0, satisfying Rolle’s Theorem.

Mean Value Theorem states that:
- If a function $f(x)$ is continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one value of $c$ in the interval $(a, b)$ such that
  f'(c) = rac{f(b) - f(a)}{b - a}
- Geometric interpretation: The theorem guarantees that there is at least one tangent line to the curve that is parallel to the secant line between points $(a, f(a))$ and $(b, f(b))$.

Function: $f(x) = x^4 - 2x^2$
- Find values of $c$ in the interval $(-2, 2)$:
  - Derivative: f'(x) = 4x^3 - 4x
  - Set the derivative equal to zero:
    4x^3 - 4x = 0
    4x(x^2 - 1) = 0
    Point solutions are x = 0, x = 1, x = -1.
  - Conclusion: The values of c for Mean Value Theorem in the interval $(-2, 2)$ are c = -1, 0, 1.

Function: $f(x) = 5 - 4x$
- Find values of $c$ in the open interval (1, 4):
  - Derivative: f'(x) = -4
- Average rate of change over the interval [1, 4]:
  - Calculate $f(1)$ and $f(4)$:
    f(1) = 5 - 4(1) = 1
    f(4) = 5 - 4(4) = -11
  - Average rate of change:
    rac{f(4) - f(1)}{4 - 1} = rac{-11 - 1}{4 - 1} = rac{-12}{3} = -4
- Conclusion: Since the derivative $f'(x) = -4$, which matches the average rate of change, the Mean Value Theorem holds true in (1, 4).

Scenario: Scenario involving police cars and speeds.
- You pass a stationary police car in a speed zone of 55 mi/hr and are clocked at 55 mi/hr.
- Four minutes later, you pass another police car located 5 miles down the road, clocked at 50 mi/hr.
- Police Concern: Even though you weren’t speeding at either instance, the police argue you were speeding between the two points based on calculus principles.
- Teacher's Proof:
  - Use the Mean Value Theorem:
  - Average speed between the two points (from the first police car to the second) can be calculated as:
    - Time taken is 4 minutes, which is rac{4}{60} hours.
    - Therefore, speed is rac{5 ext{ miles}}{ rac{4}{60} ext{ hours}} = 5 imes 15 = 75 ext{ mi/hr}.
  - Since the constant speeds recorded are less than 75 mi/hr, and given the mean speed must exist between the points, this suggests the vehicle was speeding.

Assignments from page 178:
- Questions: 3, 4, 9, 12, 15, 16, 17, 21, 29, 37, 42, 43, 46, 48, 49, 50, 54, 61, 62, 70-72