The Intermediate Value Theorem

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Presentation transcript:

The Intermediate Value Theorem Section 5.5 The Intermediate Value Theorem Rolle’s Theorem The Mean Value Theorem 3.6

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. b f(b) N c a f(a)

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 slope = 0 c f ‘ (c ) = 0 a b

Given the curve: f(x+h) x+h x f(x)

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 Conclusion: Slope of Secant Line Equals Slope of Tangent Line a b f(a) f(b) c

Show that has exactly one solution on [0, 1]. f(0) = -1 and f(1) = 2 By IVT, there is AT LEAST ONE solution on [0, 1] Since there is no max/min on (0, 1), there is EXACTLY One solution on [0, 1].

f(0) = -1 f(1) = 2

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 that satisfy the Mean Value Theorem for

Find the value(s) of c that satisfy the Mean Value Theorem for 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

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.

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, 1] On the interval [-2, 1], c = 0 satisfies the conclusion of MVT

Since f(x) is discontinuous at x = 2, which is part of the interval [0, 4], the Mean Value Theorem does not apply

f(x) is continuous and differentiable on [-1, 2] c = 1 satisfies the conclusion of MVT

CALCULATOR REQUIRED 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 f(3) = 39 f(-2) = 64 For how many value(s) of c is f ‘ (c ) = -5? X X X