Chapter 2 Section 7

Chapter 2 Section 7 - Chapter 2 Section 7 Intermediate...

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1 Chapter 2 Section 7 Intermediate Value Theorem
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2 Objectives: 1. Use the Intermediate Value Theorem to show that an equation has at least one root. 2. Approximate the solution of an equation using the Bisection Method.
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3 Intermediate Value Theorem Suppose f(x) is continuous on [a, b] and let M be any number between f(a) and f(b). Then there exists a number “c” in (a, b) such that f(c) = M. That is, a continuous function on an interval cannot skip values.
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4 Existence of zeros If f(x) is continuous on [a, b] and if f(a) and f(b) take opposite signs (one is positive and one is negative), then f(c) = 0 for some “c” in [a, b].
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5 Example: Show x 2 = √(x+1) has a root in the interval (1, 2) Rewrite x 2 = √(x+1) as x 2 - √(x+1) = 0. f(x) = x 2 - √(x+1) is continuous on its domain. If x 2 = √(x+1) has a root in (1, 2), then f(x) = 0 in (1, 2). f(1) = 1 2 - √(1+1) = 1 - √2 < 0 f(2) = 2 2 - √(2+1) = 4 - √3 > 0 therefore, by IVT, there is some number “c” in (1, 2) such that f(c) = 0 and x 2 = √(x+1) has a root in (1, 2).
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6 -4 -3 -2 -1 1 2 3 4 5 -4 -3 -2 -1 1 2 3 4
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This note was uploaded on 09/28/2009 for the course MATH 1550 taught by Professor Wei during the Spring '08 term at LSU.

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Chapter 2 Section 7 - Chapter 2 Section 7 Intermediate...

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