EXAMPLE 10 Show that there is a root of the equation 6x- 8x +3x-2 = 0 between 1 and 2. SOLUTION Let f(x) = 6x 8x + 3x- 2= 0. We are looking for a solution of the given equation, that is, a number c between 1 and 2 such that fc) - Therefore we take a = and N = in the Intermediate Value Theorem. We have (1) = 6- 8 + 3 - 2 = -1<0 and f(2) = 48 - 32+6-2 = 20 > 0. Thus (1) < 0 < f(2); that is N = 0 is a number between f(1) and (2). Now fis continuous since it is a polynomial, so that the Intermediate Value Theorem says there is a number c between and such that f(c) = In other words, the equation 6x - 8x + 3x - 2 = 0 has at least one root c in the open interval In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since (1.1) = -0.394 < 0 and (1.2) = 0.448 >0 %3! a root must lie between (smaller) and (larger). A calculator gives, by trial and error, (1.15) = -0.004750 < 0 and (1.16) = 0.080576 > 0. So a root lies in the open interval

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.5: Other Types Of Equations
Problem 65E
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ment-Responses/lastrdep=27131702#Q4
MY NOTES
PRACTICE ANOTHER
EXAMPLE 1O
Show that there is a root of the equation
6x - 8x2 + 3x - 2 = 0
between 1 and 2.
SOLUTION Let f(x) = 6x- 8x2 + 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) Therefore we take a =
b =
and N =
in the Intermediate Value Theorem. We have
f(1) = 6 - 8 + 3 - 2 = -1 <0
and
f(2) = 48 - 32 + 6 - 2 = 20 > 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
and
such that f(c) =
In other words, the equation 6x - 8x2 + 3x - 2 = 0 has at least one root c in
the open interval
In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since
f(1.1) = -0.394 < 0
and f(1.2) = 0.448 > 0
a root must lie between
(smaller) and
(larger). A calculator gives, by trial and error,
F(1.15) = -0.004750 < 0
and (1.16) = 0.080576 > 0.
So a root lies in the open interval
Transcribed Image Text:ment-Responses/lastrdep=27131702#Q4 MY NOTES PRACTICE ANOTHER EXAMPLE 1O Show that there is a root of the equation 6x - 8x2 + 3x - 2 = 0 between 1 and 2. SOLUTION Let f(x) = 6x- 8x2 + 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) Therefore we take a = b = and N = in the Intermediate Value Theorem. We have f(1) = 6 - 8 + 3 - 2 = -1 <0 and f(2) = 48 - 32 + 6 - 2 = 20 > 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 and such that f(c) = In other words, the equation 6x - 8x2 + 3x - 2 = 0 has at least one root c in the open interval In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f(1.1) = -0.394 < 0 and f(1.2) = 0.448 > 0 a root must lie between (smaller) and (larger). A calculator gives, by trial and error, F(1.15) = -0.004750 < 0 and (1.16) = 0.080576 > 0. So a root lies in the open interval
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