Mean Value Theorem - AP Calculus BC

  1. Identify the condition that ensures Mean Value Theorem's conclusion holds.
    Continuity and differentiability on the interval. Both properties must hold for the theorem to apply.
  2. Find cc for f(x)=x2f(x) = x^2 over [1,3][1, 3] using the Mean Value Theorem.
    c=2c = 2. Set f′(c)=2c=9−13−1=4f'(c) = 2c = \frac{9-1}{3-1} = 4, so c=2c = 2.
  3. For f(x)=x4f(x) = x^4, determine cc on [−1,1][-1, 1] using the Mean Value Theorem.
    c=0c = 0. Set f′(c)=4c3=0f'(c) = 4c^3 = 0, so c=0c = 0.
  4. Determine if f(x)=∣x∣f(x) = |x| satisfies the Mean Value Theorem on [−1,1][-1, 1].
    No, f(x)=∣x∣f(x) = |x| is not differentiable at x=0x = 0. The function has a sharp corner at x=0x = 0.
  5. What conditions must a function satisfy to apply the Mean Value Theorem?
    Continuous on [a,b][a, b] and differentiable on (a,b)(a, b). These ensure the function has no breaks or sharp corners.
  6. What is the primary conclusion of the Mean Value Theorem?
    There exists a c in (a,b)c \text{ in } (a, b) where f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b)-f(a)}{b-a}. States the existence of a point with equal rates.
  7. What theorem guarantees an instant rate of change equals average rate over an interval?
    Mean Value Theorem. Establishes when instantaneous equals average rate of change.
  8. What key property of f(x)f(x) ensures Mean Value Theorem's applicability on [a,b][a, b]?
    Continuity on [a,b][a, b] and differentiability on (a,b)(a, b). These are the fundamental requirements for the theorem.
  9. State the Mean Value Theorem for derivatives.
    If f(x)f(x) is continuous on [a,b][a, b] and differentiable on (a,b)(a, b), then f(b)−f(a)b−a=f′(c)\frac{f(b)-f(a)}{b-a} = f'(c) for some c in (a,b)c \text{ in } (a, b). Guarantees a point where instantaneous rate equals average rate.
  10. For f(x)=x3f(x) = x^3, determine if the Mean Value Theorem applies on [−1,1][-1, 1].
    Yes, f(x)=x3f(x) = x^3 is continuous and differentiable on [−1,1][-1, 1]. Polynomials are continuous and differentiable everywhere.
  11. Is f(x)=sin(x)f(x) = \text{sin}(x) on [0,π][0, \text{π}] suitable for the Mean Value Theorem?
    Yes, f(x)=sin(x)f(x) = \text{sin}(x) is continuous and differentiable on [0,π][0, \text{π}]. Sine function is smooth everywhere.
  12. Verify Mean Value Theorem applicability for f(x)=ln(x)f(x) = \text{ln}(x) on [1,3][1, 3].
    Yes, f(x)=ln(x)f(x) = \text{ln}(x) is continuous and differentiable on [1,3][1, 3]. Logarithm is continuous and differentiable for x>0x > 0.
  13. Which theorem ensures a point cc exists such that f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b)-f(a)}{b-a}?
    Mean Value Theorem. This theorem establishes the existence of such a point.
  14. What is the geometric meaning of the Mean Value Theorem?
    A tangent to the curve at cc is parallel to the secant line through (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)). The tangent line at cc has the same slope as the secant line.
  15. Evaluate Mean Value Theorem applicability for f(x)=x3−6x2f(x) = x^3 - 6x^2 on [0,3][0, 3].
    Yes, f(x)=x3−6x2f(x) = x^3 - 6x^2 is continuous and differentiable on [0,3][0, 3]. Polynomial functions are always continuous and differentiable.
  16. Does the function f(x)=x1/3f(x) = x^{1/3} satisfy the Mean Value Theorem on [−1,1][-1, 1]?
    No, f(x)=x1/3f(x) = x^{1/3} is not differentiable at x=0x = 0. The derivative is undefined at x=0x = 0.
  17. Determine if f(x)=1xf(x) = \frac{1}{x} satisfies Mean Value Theorem on [2,6][2, 6].
    Yes, f(x)=1xf(x) = \frac{1}{x} is continuous and differentiable on [2,6][2, 6]. No discontinuities or undefined derivatives in this interval.
  18. Determine if f(x)=x3−3x2f(x) = x^3 - 3x^2 satisfies Mean Value Theorem on [0,3][0, 3].
    Yes, f(x)=x3−3x2f(x) = x^3 - 3x^2 is continuous and differentiable on [0,3][0, 3]. Polynomials satisfy all required conditions.
  19. For f(x)=exf(x) = e^x, verify if the Mean Value Theorem applies on [0,1][0, 1].
    Yes, f(x)=exf(x) = e^x is continuous and differentiable on [0,1][0, 1]. Exponential functions are smooth everywhere.
  20. Verify if f(x)=1ln(x)f(x) = \frac{1}{\text{ln}(x)} meets Mean Value Theorem conditions on [2,5][2, 5].
    Yes, f(x)=1ln(x)f(x) = \frac{1}{\text{ln}(x)} is continuous and differentiable on [2,5][2, 5]. Logarithm is positive and smooth on this interval.
  21. Given f(x)=x2−4xf(x) = x^2 - 4x, find cc on [0,4][0, 4] using the Mean Value Theorem.
    c=2c = 2. Set f′(c)=2c−4=0f'(c) = 2c - 4 = 0, so c=2c = 2.
  22. Identify the function requirement that ensures differentiability.
    The function must be continuous. Differentiability requires continuity as a prerequisite.
  23. Does f(x)=1xf(x) = \frac{1}{x} meet Mean Value Theorem prerequisites on [1,4][1, 4]?
    Yes, f(x)=1xf(x) = \frac{1}{x} is continuous and differentiable on [1,4][1, 4]. The function has no discontinuities or non-differentiable points.
  24. Determine if f(x)=x3f(x) = x^3 fits Mean Value Theorem criteria on [−2,1][-2, 1].
    Yes, f(x)=x3f(x) = x^3 is continuous and differentiable on [−2,1][-2, 1]. Cubic functions are continuous and differentiable everywhere.

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