Derivatives - Rolle Theorem (Continued)
Recap: Rolle Theorem
- The Rolle Theorem states that if a function 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 c in (a, b) such that f’(c) = 0.
Application of the Rolle Theorem
- The Rolle Theorem can be used to prove the existence of solutions to certain types of equations.
- It is also used to prove the Mean Value Theorem.
Today’s Topic
- In today’s lecture, we will continue our discussion on the Rolle Theorem.
- We will explore some properties and applications of the theorem.
Properties of Rolle Theorem
- The function must be continuous on the closed interval [a, b].
- The function must be differentiable on the open interval (a, b).
- The function must satisfy f(a) = f(b).
Note
- The condition f(a) = f(b) is necessary for the theorem to hold.
- If this condition is not satisfied, the Rolle Theorem cannot be used.
Example 1
Given the function f(x) = x^2 - 4x + 3 on the interval [1, 3], prove that there exists at least one solution c in (1, 3) such that f’(c) = 0.
Solution:
- First, we need to check if the function satisfies the conditions of the Rolle Theorem.
- The function is continuous on the closed interval [1, 3] and differentiable on the open interval (1, 3).
- Next, we need to check if f(1) = f(3).
- f(1) = (1)^2 - 4(1) + 3 = 0
- f(3) = (3)^2 - 4(3) + 3 = 0
- Therefore, f(1) = f(3)
- As all the conditions are satisfied, we can conclude that there exists at least one solution c in (1, 3) such that f’(c) = 0.
Application of the Rolle Theorem
- The Rolle Theorem is often used in calculus to prove other theorems and results.
Example 2
Prove that the function f(x) = sin(x) - x has at most one root.
Solution:
- First, we need to show that f(x) is continuous on the closed interval [0, π] and differentiable on the open interval (0, π).
- Second, we need to show that f(0) = f(π) = 0.
- f(0) = sin(0) - 0 = 0
- f(π) = sin(π) - π = 0
- Therefore, f(0) = f(π) = 0
- As all the conditions are satisfied, we can conclude that there exists at least one solution c in (0, π) such that f’(c) = 0.
- Since f(x) is a smooth function and the derivative f’(x) = cos(x) - 1 never equals zero on the interval (0, π), there can be at most one root for the equation f(x) = 0.
Limitations of the Rolle Theorem
- The Rolle Theorem is valid only if the function satisfies f(a) = f(b).
- If the function does not satisfy this condition, we cannot apply the Rolle Theorem.
Example 3
Consider the function f(x) = x^2 on the interval [1, 4].
- f(1) = (1)^2 = 1
- f(4) = (4)^2 = 16
- f(1) ≠ f(4)
- Therefore, the conditions of the Rolle Theorem are not satisfied.
- We cannot apply the Rolle Theorem to this function.
Connection to the Mean Value Theorem
- The Rolle Theorem is a special case of the Mean Value Theorem.
- If a function satisfies the conditions of the Rolle Theorem, it also satisfies the conditions of the Mean Value Theorem.
- Let f(x) be continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in (a, b) such that:
f’(c) = (f(b) - f(a))/(b - a)
Note
- The Mean Value Theorem is a more general theorem as compared to the Rolle Theorem.
Example 4
Prove that the function f(x) = x^3 - 3x is continuous on the closed interval [-1, 2] and differentiable on the open interval (-1, 2).
Solution:
- The function is a polynomial function, and all polynomial functions are continuous and differentiable for all real numbers.
- Therefore, the function f(x) = x^3 - 3x is continuous on the closed interval [-1, 2] and differentiable on the open interval (-1, 2).
Example 5
Given the function f(x) = x^2 + x - 2 on the interval [-2, 1], show that there exists at least one solution c in (-2, 1) such that f’(c) = 0.
Solution:
- First, we need to check if the function satisfies the conditions of the Rolle Theorem.
- The function is continuous on the closed interval [-2, 1] and differentiable on the open interval (-2, 1).
- Next, we need to check if f(-2) = f(1).
- f(-2) = (-2)^2 - 2 + 1 = 7
- f(1) = (1)^2 + 1 - 2 = 0
- Therefore, f(-2) ≠ f(1)
- As the conditions are not satisfied, we cannot apply the Rolle Theorem to this function.
Summary
- The Rolle Theorem states that if a function 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 c in (a, b) such that f’(c) = 0.
- The Rolle Theorem is often used to prove the existence of solutions to equations.
- It is a special case of the more general Mean Value Theorem.
- Be cautious of the limitations of the Rolle Theorem, where f(a) ≠ f(b).