Show that there is a root of the equation below between 1 and 2. 4x36x2 +...

 

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Show that there is a root of the equation below between 1 and 2. 4x36x2 + 3x-2=0 SOLUTION Let f(x) = 4x² - 6x² + 3x - 2. 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) =4-6 +3-2=-1 <0 f(2)= 32-24 +6-2=12>0 Thus f(1) < 0 <f(2); that is N = Ois a number between f(1) and f(2). Now f is 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 4x³ -6x² + 3x- 2 = 0 has at least one root c in the interval ( In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f(1.2) =-0.128 <0 f(1.3) = 0.548 >0 a root must lie between f(1.22) =-0.007008 <0 f(1.23) = 0.056068 >0 So a root lies in the interval (1.22, 1.23). and .A calculation gives, by trial and error, Show that there is a root of the equation below between 1 and 2. 4x36x2 + 3x-2=0 SOLUTION Let f(x) = 4x² - 6x² + 3x - 2. 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) =4-6 +3-2=-1 <0 f(2)= 32-24 +6-2=12>0 Thus f(1) < 0 <f(2); that is N = Ois a number between f(1) and f(2). Now f is 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 4x³ -6x² + 3x- 2 = 0 has at least one root c in the interval ( In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f(1.2) =-0.128 <0 f(1.3) = 0.548 >0 a root must lie between f(1.22) =-0.007008 <0 f(1.23) = 0.056068 >0 So a root lies in the interval (1.22, 1.23). and .A calculation gives, by trial and error,

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