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Questions and Answers
What does Rolli's Theorem guarantee for a continuous and differentiable function on the interval [a,b], where f(a) = f(b)?
What does Rolli's Theorem guarantee for a continuous and differentiable function on the interval [a,b], where f(a) = f(b)?
- There exists a point c ∈ (a,b) such that f(c) = 0.
- There exists a point c ∈ (a,b) such that f’(c) is undefined.
- There exists a point c ∈ (a,b) such that f’(c) > 0.
- There exists a point c ∈ (a,b) such that f’(c) = 0. (correct)
In the context of Rolli's Theorem, what does it mean if f’(c) = 0 at point c?
In the context of Rolli's Theorem, what does it mean if f’(c) = 0 at point c?
- The tangent to the curve at point c is vertical.
- The tangent to the curve at point c is parallel to the x-axis. (correct)
- The function is increasing at point c.
- The function has a local maximum at point c.
Which equation follows from applying Rolli's Theorem to the function f(x) = x^3 - 3x^2 + 2x + 2 on [0,1]?
Which equation follows from applying Rolli's Theorem to the function f(x) = x^3 - 3x^2 + 2x + 2 on [0,1]?
- 3c^2 - 6c + 2 = 0 (correct)
- c^2 + 2c - 3 = 0
- c^2 - 2c + 1 = 0
- 3c^2 + 6c + 2 = 0
What is indicated by the statement that there exists a number c ∈ (a,b) such that f’(c) = (f(b) - f(a)) / (b - a) in the Mean Value Theorem?
What is indicated by the statement that there exists a number c ∈ (a,b) such that f’(c) = (f(b) - f(a)) / (b - a) in the Mean Value Theorem?
For the function f(x) = sin(x) in the interval [0, π], what is the derived value of c that satisfies the conclusion of Rolli's Theorem?
For the function f(x) = sin(x) in the interval [0, π], what is the derived value of c that satisfies the conclusion of Rolli's Theorem?
Flashcards
Rolle's Theorem
Rolle's Theorem
If a function is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and the function values at the endpoints are equal (f(a) = f(b)), then there exists a point c within the interval (a, b) where the derivative is zero (f'(c) = 0).
Mean Value Theorem
Mean Value Theorem
If a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c within the interval (a, b) where the derivative's value equals the average rate of change of the function over the interval.
Rolle's Theorem Condition 1
Rolle's Theorem Condition 1
The function must be continuous on the entire closed interval [a,b].
Rolle's Theorem Condition 2
Rolle's Theorem Condition 2
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Mean Value Theorem Condition
Mean Value Theorem Condition
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Study Notes
Rolle's Theorem
- Statement: If a function f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists a number c ∈ (a, b) such that f'(c) = 0.
- Geometric Interpretation: The tangent to the graph of f at x = c is parallel to the x-axis. This means the graph has a horizontal tangent at x = c.
Examples and Applications
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Example 1: Find a value of c satisfying the conclusion of Rolle's Theorem for f(x) = x³ - 3x² + 2x + 2 on [0, 1].
- f(0) = 2, f(1) = 2.
- f'(x) = 3x² - 6x + 2.
- Setting f'(c) = 0, we get 3c² - 6c + 2 = 0.
- Solving for c, c = (1 ± √3)/3
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Example 2: Find a value of c satisfying the conclusion of Rolle's Theorem for f(x) = sin x in [0, π].
- f(0) = 0, f(Ï€) = 0.
- f'(x) = cos x.
- Setting f'(c) = 0, we get cos c = 0.
- Solving for c, c = π/2.
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Example 3: Find a value of c for f(x) = x³ - x + 1 on [0, 1].
- f(0) = 1, f(1) = 1
- f'(x) = 3x² - 1
- Setting f'(c) = 0, we get 3c² - 1 = 0.
- Solving for c , c = ± 1/√3
Mean Value Theorem
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Statement: If a function f is continuous on [a, b] and differentiable on (a, b), then there exists a number c ∈ (a, b) such that:
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f'(c) = (f(b) - f(a)) / (b - a)*.
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Geometric Interpretation: There is a point on the graph of f between x = a and x = b where the tangent line is parallel to the secant line connecting the points (a, f(a)) and (b, f(b)).
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Example 1: Find the value of c for f(x) = x³ - x² + x on [0, 2].
- f(0) = 0, f(2) = 4
- f'(x) = 3x² - 2x + 1
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Setting f'(c) = (f(2) - f(0)) / (2 - 0), we get 3c² - 2c + 1 = 2.
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Solving for c: 3c² - 2c - 1 = 0, c = (1 ± √2)/3.
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