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Section 5.5 The Intermediate Value Theorem Rolle’s Theorem The Mean Value Theorem 3.6
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Intermediate Value Theorem (IVT) If f is continuous on [a, b] and N is a value between f(a) and f(b), then there is at least one point c between a and b where f takes on the value N. a f(a) b f(b) N c
![Intermediate Value Theorem (IVT) If f is continuous on [a, b] and N is a value between f(a) and f(b), then there is at least one point c between a and b where f takes on the value N.](http://images.slideplayer.com/27/8998796/slides/slide_2.jpg "a f(a) b f(b) N c.")
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Rolle’s Theorem If f is continuous on [a, b], if f(a) = 0, f(b) = 0, then there is at least one number c on (a, b) where f ‘ (c ) = 0 a b slope = 0 c f ‘ (c ) = 0
![Rolle’s Theorem If f is continuous on [a, b], if f(a) = 0, f(b) = 0, then there is at least one number c on (a, b) where f ‘ (c ) = 0 a b slope = 0 c f ‘ (c ) = 0](http://images.slideplayer.com/27/8998796/slides/slide_3.jpg "Rolle’s Theorem If f is continuous on [a, b], if f(a) = 0, f(b) = 0, then there is at least one number c on (a, b) where f ‘ (c ) = 0 a b slope = 0 c f ‘ (c ) = 0")
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Given the curve: x f(x) x+h f(x+h)
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The Mean Value Theorem (MVT) aka the ‘crooked’ Rolle’s Theorem If f is continuous on [a, b] and differentiable on (a, b) There is at least one number c on (a, b) at which a b f(a) f(b) c Conclusion: Slope of Secant Line Equals Slope of Tangent Line
![The Mean Value Theorem (MVT) aka the ‘crooked’ Rolle’s Theorem If f is continuous on [a, b] and differentiable on (a, b) There is at least one number c on (a, b) at which a b f(a) f(b) c Conclusion: Slope of Secant Line Equals Slope of Tangent Line](http://images.slideplayer.com/27/8998796/slides/slide_5.jpg "The Mean Value Theorem (MVT) aka the ‘crooked’ Rolle’s Theorem If f is continuous on [a, b] and differentiable on (a, b) There is at least one number c on (a, b) at which a b f(a) f(b) c Conclusion: Slope of Secant Line Equals Slope of Tangent Line")
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f(0) = -1 f(1) = 2
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Find the value(s) of c which satisfy Rolle’s Theorem for on the interval [0, 1]. Verify…..f(0) = 0 – 0 = 0 f(1) = 1 – 1 = 0 which is on [0, 1]
![Find the value(s) of c which satisfy Rolle’s Theorem for on the interval [0, 1].](http://images.slideplayer.com/27/8998796/slides/slide_7.jpg "Verify…..f(0) = 0 – 0 = 0 f(1) = 1 – 1 = 0 which is on [0, 1].")
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Find the value(s) of c that satisfy the Mean Value Theorem for
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Note: The Mean Value Theorem requires the function to be continuous on [-4, 4] and differentiable on (-4, 4). Therefore, since f(x) is discontinuous at x = 0 which is on [-4, 4], there may be no value of c which satisfies the Mean Value Theorem Since has no real solution, there is no value of c on [-4, 4] which satisfies the Mean Value Theorem
![Note: The Mean Value Theorem requires the function to be continuous on [-4, 4] and differentiable on (-4, 4).](http://images.slideplayer.com/27/8998796/slides/slide_9.jpg "Therefore, since f(x) is discontinuous at x = 0 which is on [-4, 4], there may be no value of c which satisfies the Mean Value Theorem Since has no real solution, there is no value of c on [-4, 4] which satisfies the Mean Value Theorem.")
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Given the graph of f(x) below, use the graph of f to estimate the numbers on [0, 3.5] which satisfy the conclusion of the Mean Value Theorem.
![Given the graph of f(x) below, use the graph of f to estimate the numbers on [0, 3.5] which satisfy the conclusion of the Mean Value Theorem.](http://images.slideplayer.com/27/8998796/slides/slide_10.jpg "Given the graph of f(x) below, use the graph of f to estimate the numbers on [0, 3.5] which satisfy the conclusion of the Mean Value Theorem.")
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f(x) is continuous and differentiable on [-2, 2] On the interval [-2, 2], c = 0 satisfies the conclusion of MVT
![f(x) is continuous and differentiable on [-2, 2] On the interval [-2, 2], c = 0 satisfies the conclusion of MVT](http://images.slideplayer.com/27/8998796/slides/slide_11.jpg "f(x) is continuous and differentiable on [-2, 2] On the interval [-2, 2], c = 0 satisfies the conclusion of MVT")
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f(x) is continuous and differentiable on [-2, 1] On the interval [-2, 1], c = 0 satisfies the conclusion of MVT
![f(x) is continuous and differentiable on [-2, 1] On the interval [-2, 1], c = 0 satisfies the conclusion of MVT](http://images.slideplayer.com/27/8998796/slides/slide_12.jpg "f(x) is continuous and differentiable on [-2, 1] On the interval [-2, 1], c = 0 satisfies the conclusion of MVT")
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Since f(x) is discontinuous at x = 2, which is part of the interval [0, 4], the Mean Value Theorem does not apply
![Since f(x) is discontinuous at x = 2, which is part of the interval [0, 4], the Mean Value Theorem does not apply](http://images.slideplayer.com/27/8998796/slides/slide_13.jpg "Since f(x) is discontinuous at x = 2, which is part of the interval [0, 4], the Mean Value Theorem does not apply")
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f(x) is continuous and differentiable on [-1, 2] c = 1 satisfies the conclusion of MVT
![f(x) is continuous and differentiable on [-1, 2] c = 1 satisfies the conclusion of MVT](http://images.slideplayer.com/27/8998796/slides/slide_14.jpg "f(x) is continuous and differentiable on [-1, 2] c = 1 satisfies the conclusion of MVT")
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f(3) = 39 f(-2) = 64 For how many value(s) of c is f ‘ (c ) = -5? If, how many numbers on [-2, 3] satisfy the conclusion of the Mean Value Theorem. A. 0 B. 1 C. 2 D. 3 E. 4 CALCULATOR REQUIRED X XX
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