Derivatives - Rolle Theorem

  • The Rolle’s theorem is a special case of the mean value theorem.

  • It is applicable when a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b).

  • The theorem states that if a function f(x) satisfies the following conditions:

    1. f(x) is continuous on the closed interval [a, b].
    2. f(x) is differentiable on the open interval (a, b).
    3. f(a) = f(b).

  • Then, there exists at least one value c in the open interval (a, b) such that f’(c) = 0.

Example:

Let’s consider a function f(x) = x^2 on the interval [0,2].

  • Since f(x) is a continuous function on [0,2] and differentiable on (0,2).
  • We can find the derivative of f(x) as f’(x) = 2x.
  • Now, let’s check the conditions of Rolle’s theorem.
    • f(x) is continuous on [0,2].
    • f(x) is differentiable on (0,2).
    • f(0) = 0 and f(2) = 4.
  • Therefore, based on the theorem, there exists at least one value c in the interval (0,2) such that f’(c) = 0.
  • In this case, we can see that f’(c) = 2c = 0.
  • Solving this equation, we get c = 0.
  • Hence, the function f(x) = x^2 satisfies the conditions of Rolle’s theorem, and there exists a value c in the interval (0,2) such that f’(c) = 0.
  1. Rolle’s Theorem - Proof
    • Let f(x) be a function that satisfies the conditions of Rolle’s theorem on the interval [a, b].
    • Since f(x) is continuous on [a, b], it must be bounded on [a, b].
    • Let M be the maximum value of f(x) on [a, b] and m be the minimum value of f(x) on [a, b].
    • If f(x) is a constant function, then f’(x) = 0 for all values of x and we are done.
    • Otherwise, assume that f(x) is not constant and f(a) ≠ f(b).
    • Since f(x) is continuous on [a, b], it must attain its maximum and minimum values on [a, b].
    • Therefore, there exist points c and d in (a, b) such that f(c) = M and f(d) = m.
    • By the Extreme Value Theorem, f(x) must have an absolute maximum and minimum on [a, b].
    • Since f(c) = M = f(d) = m, we have f’(c) = f’(d) = 0.
    • Hence, there exists at least one value c in (a, b) such that f’(c) = 0.
  1. Importance of Rolle’s Theorem
    • Rolle’s theorem is an important tool in calculus and analysis.
    • It provides a theoretical foundation for the behavior of differentiable functions.
    • It helps in proving the existence of critical points and stationary points of a function.
    • It is used in proving results related to the mean value theorem and other theorems in calculus.
    • It is used in optimization problems and finding intervals of increase and decrease.
  1. Application of Rolle’s Theorem - Example 1
    • Consider the function f(x) = x^3 - 2x^2 - 1 on the interval [-1, 2].
    • Verify if the function satisfies the conditions of Rolle’s theorem.
    • f(x) is continuous on [-1, 2] and differentiable on (-1, 2).
    • f(-1) = (-1)^3 - 2(-1)^2 - 1 = -2 and f(2) = (2)^3 - 2(2)^2 - 1 = -5.
    • Since f(-1) ≠ f(2), we can apply Rolle’s theorem.
    • Therefore, there exists at least one value c in the interval (-1, 2) such that f’(c) = 0.
  1. Application of Rolle’s Theorem - Example 2
    • Consider the function f(x) = sin(x) on the interval [0, π/2].
    • Verify if the function satisfies the conditions of Rolle’s theorem.
    • f(x) is continuous on [0, π/2] and differentiable on (0, π/2).
    • f(0) = sin(0) = 0 and f(π/2) = sin(π/2) = 1.
    • Since f(0) = f(π/2), we can apply Rolle’s theorem.
    • Therefore, there exists at least one value c in the interval (0, π/2) such that f’(c) = 0.
  1. Rolle’s Theorem vs Mean Value Theorem
    • Rolle’s theorem is a special case of the mean value theorem.
    • Both theorems involve the existence of a value in an interval.
    • Rolle’s theorem deals with a function having equal values at the endpoints of an interval.
    • Mean value theorem deals with average rates of change of a function.
    • Mean value theorem includes Rolle’s theorem as a special case.
  1. Summary
    • Rolle’s theorem states that if a function is continuous on a closed interval and differentiable on the open interval, and if the function has equal values at the endpoints of the interval, then there exists at least one value in the open interval at which the derivative is zero.
    • Rolle’s theorem is a special case of the mean value theorem.
    • It is an important tool in calculus and analysis.
    • It helps in proving the existence of critical points and stationary points.
    • It is used in optimization problems and finding intervals of increase and decrease.
  1. Practice Questions
    1. Verify if the function f(x) = x^2 - 4x + 3 satisfies the conditions of Rolle’s theorem on the interval [1, 3].
    2. Find the value of c that satisfies the conclusion of Rolle’s theorem for the function f(x) = x^3 - 6x^2 + 9x - 4 on the interval [2, 4].
    3. Prove that the function f(x) = x^2 is not applicable for Rolle’s theorem on the interval [-1, 1].
    4. Use Rolle’s theorem to prove that the function f(x) = 4x^3 - 3x - 1 has a root in the interval [-1, 1].

  1. Answer Key - Practice Questions

    1. f(x) = x^2 - 4x + 3 is continuous on [1, 3] and differentiable on (1, 3).
    • f(1) = (1)^2 - 4(1) + 3 = 0 and f(3) = (3)^2 - 4(3) + 3 = 0.
    • Since f(1) = f(3) = 0, we can apply Rolle’s theorem.
    • Therefore, there exists at least one value c in the interval (1, 3) such that f’(c) = 0.
    1. f(x) = x^3 - 6x^2 + 9x - 4 is continuous on [2, 4] and differentiable on (2, 4).
    • f(2) = (2)^3 - 6(2)^2 + 9(2) - 4 = 0 and f(4) = (4)^3 - 6(4)^2 + 9(4) - 4 = 0.
    • Since f(2) = f(4) = 0, we can apply Rolle’s theorem.
    • Therefore, there exists at least one value c in the interval (2, 4) such that f’(c) = 0.

    1. f(x) = x^2 is not applicable for Rolle’s theorem as it does not satisfy the condition f(a) = f(b).

    2. f(x) = 4x^3 - 3x - 1 is continuous on [-1, 1] and differentiable on (-1, 1).

    • f(-1) = 4(-1)^3 - 3(-1) - 1 = -6 and f(1) = 4(1)^3 - 3(1) - 1 = 0.
    • Since f(-1) ≠ f(1), we can apply Rolle’s theorem.
    • Therefore, there exists at least one value c in the interval (-1, 1) such that f’(c) = 0.
  1. Key Takeaways
    • Rolle’s theorem involves the existence of a value in an interval where the derivative of a function is zero.
    • It is applicable when a function is continuous on a closed interval and differentiable on the open interval, with equal values at the endpoints.
    • Rolle’s theorem is a special case of the mean value theorem.
    • It has various applications in calculus and analysis, including the study of critical points, stationary points, and optimization problems.
  1. References
    • Stewart, James. Calculus: Early Transcendentals. Cengage Learning, 2016.
    • Larson, Ron, et al. Calculus. Cengage Learning, 2017.
    • Anton, Howard, et al. Calculus: Early Transcendentals. John Wiley & Sons, 2015.
  1. Application of Rolle’s Theorem - Example 3
    • Consider the function f(x) = x^2 - 2x on the interval [0, 2].
    • Verify if the function satisfies the conditions of Rolle’s theorem.
    • f(x) is continuous on [0, 2] and differentiable on (0, 2).
    • f(0) = (0)^2 - 2(0) = 0 and f(2) = (2)^2 - 2(2) = 0.
    • Since f(0) = f(2) = 0, we can apply Rolle’s theorem.
    • Therefore, there exists at least one value c in the interval (0, 2) such that f’(c) = 0.
  1. Derivatives of Trigonometric Functions
    • The derivative of sin(x) is cos(x), i.e., d/dx(sin(x)) = cos(x).
    • The derivative of cos(x) is -sin(x), i.e., d/dx(cos(x)) = -sin(x).
    • The derivative of tan(x) is sec^2(x), i.e., d/dx(tan(x)) = sec^2(x).
    • The derivative of cot(x) is -csc^2(x), i.e., d/dx(cot(x)) = -csc^2(x).
    • The derivative of sec(x) is sec(x)tan(x), i.e., d/dx(sec(x)) = sec(x)tan(x).
    • The derivative of csc(x) is -csc(x)cot(x), i.e., d/dx(csc(x)) = -csc(x)cot(x).
  1. Derivatives of Exponential Functions
    • The derivative of e^x is e^x, i.e., d/dx(e^x) = e^x.
    • The derivative of a^x (a > 0, a ≠ 1) is a^x ln(a), i.e., d/dx(a^x) = a^x ln(a).
    • The derivative of ln(x) is 1/x, i.e., d/dx(ln(x)) = 1/x.
    • The derivative of log_a(x) is (1/x)(1/ln(a)), i.e., d/dx(log_a(x)) = (1/x)(1/ln(a)).
    • The derivative of log_e(x) = ln(x).
  1. Derivatives of Hyperbolic Functions
    • The derivative of sinh(x) is cosh(x), i.e., d/dx(sinh(x)) = cosh(x).
    • The derivative of cosh(x) is sinh(x), i.e., d/dx(cosh(x)) = sinh(x).
    • The derivative of tanh(x) is sech^2(x), i.e., d/dx(tanh(x)) = sech^2(x).
    • The derivative of coth(x) is -csch^2(x), i.e., d/dx(coth(x)) = -csch^2(x).
    • The derivative of sech(x) is -sech(x) tanh(x), i.e., d/dx(sech(x)) = -sech(x) tanh(x).
    • The derivative of csch(x) is -csch(x) coth(x), i.e., d/dx(csch(x)) = -csch(x) coth(x).
  1. Differentiation of Composite Functions
    • When differentiating a composite function, we use the chain rule.
    • If y = f(g(x)), then dy/dx = (df/dg)(dg/dx).
    • Example: Let y = sin(3x^2). Using the chain rule, dy/dx = cos(3x^2)(6x).
  1. Higher Order Derivatives
    • The second derivative of a function f(x) is denoted as d²f/dx².
    • To find the second derivative, we differentiate the first derivative.
    • Example: Let f(x) = x^3 + 2x^2 + 3x. The first derivative is f’(x) = 3x^2 + 4x + 3. The second derivative is f’’(x) = 6x + 4.
  1. Differentiation of Implicit Functions
    • An implicit function is defined implicitly in terms of x and y.
    • To differentiate an implicit function, we use the implicit differentiation technique.
    • Example: Let x^2 + y^2 = 4. Differentiating both sides with respect to x, we get 2x + 2yy’ = 0. Solving for y’, we get y’ = -x/y.
  1. Applications of Derivatives in Optimization
    • Derivatives are used to optimize functions, finding maximum and minimum values.
    • To find the maximum/minimum points, we equate the derivative to zero and solve for x.
    • Example: Find the point on the parabola y = x^2 - 4x + 3 that is closest to the point (2, 1). We need to minimize the distance function d(x) = √((x-2)^2 + (x^2-4x+3-1)^2).
  1. Related Rates
    • Related rates problems involve finding how the rate of change of one quantity affects the rate of change of another related quantity.
    • To solve related rates problems, we set up an equation involving the rates of change and differentiate both sides implicitly.
    • Example: A ladder is sliding down a wall at a rate of 2 m/s. When the top of the ladder is 5 m above the ground, it is moving away from the wall at a rate of 4 m/s. Find the rate at which the base of the ladder is moving away from the wall.
  1. Summary
    • Rolle’s theorem states the existence of at least one value where a function’s derivative is zero.
    • Differentiation of trigonometric, exponential, and hyperbolic functions follows specific rules.
    • The chain rule is used to differentiate composite functions.
    • Higher order derivatives are obtained by differentiating the first derivative.
    • Implicit differentiation is used for differentiating implicit functions.
    • Derivatives have applications in optimization, related rates, and various other areas of mathematics and science.