AP Calculus AB 3.2 Notes

AP Calculus AB 3.2 Rolle’s Theorem and the Mean Value Theorem

Key Concepts of Rolle’s Theorem

  • 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)=0f'(c) = 0

Example of Rolle’s Theorem

  1. Function: $f(x) = x^2 - 3x + 2$

    • Finding x-intercepts:

      • Set $f(x) = 0$:
        x2−3x+2=0x^2 - 3x + 2 = 0

      • Factoring:
        (x−1)(x−2)=0(x - 1)(x - 2) = 0

      • Solutions are x=1x = 1 and x=2x = 2.

    • Check for continuity and differentiability:

      • The polynomial function is continuous and differentiable everywhere.

    • Evaluate $f'(x)$:

      • f′(x)=2x−3f'(x) = 2x - 3

      • Set the derivative to zero to find critical points:
        2x−3=02x - 3 = 0
        2x=3<br>ightarrowx=rac322x = 3 <br>ightarrow x = rac{3}{2}

    • Conclusion: There exists a point at c=rac32c = rac{3}{2} where f′(c)=0f'(c) = 0, satisfying Rolle’s Theorem.

Mean Value 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)=racf(b)−f(a)b−af'(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 $(b, f(b))$.

Application of the Mean Value Theorem

  1. Function: $f(x) = x^4 - 2x^2$

    • Find values of $c$ in the interval $(-2, 2)$:

      • Derivative: f′(x)=4x3−4xf'(x) = 4x^3 - 4x

      • Set the derivative equal to zero:
        4x3−4x=04x^3 - 4x = 0
        4x(x2−1)=04x(x^2 - 1) = 0
        Point solutions are x=0x = 0, x=1x = 1, x=−1x = -1.

      • Conclusion: The values of cc for Mean Value Theorem in the interval $(-2, 2)$ are c=−1,0,1c = -1, 0, 1.

Further Application of the Mean Value Theorem

  1. Function: $f(x) = 5 - 4x$

    • Find values of $c$ in the open interval (1, 4):

      • Derivative: f′(x)=−4f'(x) = -4

    • Average rate of change over the interval [1, 4]:

      • Calculate $f(1)$ and $f(4)$:
        f(1)=5−4(1)=1f(1) = 5 - 4(1) = 1
        f(4)=5−4(4)=−11f(4) = 5 - 4(4) = -11

      • Average rate of change:
        racf(4)−f(1)4−1=rac−11−14−1=rac−123=−4rac{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 Application: Legal Case of Speeding

  1. 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 rac460rac{4}{60} hours.

        • Therefore, speed is rac5extmilesrac460exthours=5imes15=75extmi/hrrac{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.

Homework Assignments

  • 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