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EXAMPLE 10 Show that there is a root of the equation 6x - 9x2 + 3x - 2 = 0 between 1 and 2. SOLUTION Let f(x) = 6x³ – 9x² + 3x – 2 = 0. We are looking for a solution of the given equation, that is, a number c between 1 and 2 such that f(c) = [0 take a = 1 . Therefore we in the Intermediate Value v, b = 2 , and N = 0 Theorem. We have f(1) = 6 - 9 + 3 - 2 = -2 < o and f(2) = 48 – 36 + 6 – 2 = 16 > 0. Thus f(1) < 0 < f(2); that is N = 0 is a number between f(1) and f(2). Now fis continuous since it is a polynomial, so that the Intermediate Value Theorem says there is a number c between 1 and 2 such that f(c) = 0 . In other words, the equation 6x – 9x² + 3x – 2 = 0 has at least one root c in the open interval (-2,16) In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f(1.3) = -0.128 < 0 and f(1.4) = 1.024 > 0 (smaller) and 1.4 a root must lie between 1.3 by trial and error, (larger). A calculator gives, f(1.31) = -0.026354 < 0 and f(1.32) = 0.078208 > 0. So a root lies in the open interval (-0.128,1.024)

EXAMPLE 10 Show that there is a root of the equation 6x - 9x2 + 3x - 2 = 0 between 1 and 2. SOLUTION Let f(x) = 6x³ – 9x² + 3x – 2 = 0. We are looking for a solution of the given equation, that is, a number c between 1 and 2 such that f(c) = [0 take a = 1 . Therefore we in the Intermediate Value v, b = 2 , and N = 0 Theorem. We have f(1) = 6 - 9 + 3 - 2 = -2 < o and f(2) = 48 – 36 + 6 – 2 = 16 > 0. Thus f(1) < 0 < f(2); that is N = 0 is a number between f(1) and f(2). Now fis continuous since it is a polynomial, so that the Intermediate Value Theorem says there is a number c between 1 and 2 such that f(c) = 0 . In other words, the equation 6x – 9x² + 3x – 2 = 0 has at least one root c in the open interval (-2,16) In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f(1.3) = -0.128 < 0 and f(1.4) = 1.024 > 0 (smaller) and 1.4 a root must lie between 1.3 by trial and error, (larger). A calculator gives, f(1.31) = -0.026354 < 0 and f(1.32) = 0.078208 > 0. So a root lies in the open interval (-0.128,1.024)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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EXAMPLE 10 Show that there is a root of the equation 6x - 9x2 + 3x - 2 = 0 between 1 and 2. SOLUTION Let f(x) = 6x³ – 9x² + 3x – 2 = 0. We are looking for a solution of the given equation, that is, a number c between 1 and 2 such that f(c) = [0 take a = 1 . Therefore we in the Intermediate Value v, b = 2 , and N = 0 Theorem. We have f(1) = 6 - 9 + 3 - 2 = -2 < o and f(2) = 48 – 36 + 6 – 2 = 16 > 0. Thus f(1) < 0 < f(2); that is N = 0 is a number between f(1) and f(2). Now fis continuous since it is a polynomial, so that the Intermediate Value Theorem says there is a number c between 1 and 2 such that f(c) = 0 . In other words, the equation 6x – 9x² + 3x – 2 = 0 has at least one root c in the open interval (-2,16) In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f(1.3) = -0.128 < 0 and f(1.4) = 1.024 > 0 (smaller) and 1.4 a root must lie between 1.3 by trial and error, (larger). A calculator gives, f(1.31) = -0.026354 < 0 and f(1.32) = 0.078208 > 0. So a root lies in the open interval (-0.128,1.024)
Transcribed Image Text:EXAMPLE 10 Show that there is a root of the equation 6x - 9x2 + 3x - 2 = 0 between 1 and 2. SOLUTION Let f(x) = 6x³ – 9x² + 3x – 2 = 0. We are looking for a solution of the given equation, that is, a number c between 1 and 2 such that f(c) = [0 take a = 1 . Therefore we in the Intermediate Value v, b = 2 , and N = 0 Theorem. We have f(1) = 6 - 9 + 3 - 2 = -2 < o and f(2) = 48 – 36 + 6 – 2 = 16 > 0. Thus f(1) < 0 < f(2); that is N = 0 is a number between f(1) and f(2). Now fis continuous since it is a polynomial, so that the Intermediate Value Theorem says there is a number c between 1 and 2 such that f(c) = 0 . In other words, the equation 6x – 9x² + 3x – 2 = 0 has at least one root c in the open interval (-2,16) In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f(1.3) = -0.128 < 0 and f(1.4) = 1.024 > 0 (smaller) and 1.4 a root must lie between 1.3 by trial and error, (larger). A calculator gives, f(1.31) = -0.026354 < 0 and f(1.32) = 0.078208 > 0. So a root lies in the open interval (-0.128,1.024)
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