Tuesday, November 24, 2015

Single Variable Calculus, Chapter 4, 4.8, Section 4.8, Problem 14

Use Newton's Method to approximate the root of 2.2x54.4x3+1.3x20.9x4.0=0 in the interval [2,1] correct to six decimal places.







We apply Newton's Method with


f(x)=2.2x54.4x3+1.3x20.9x4 and f(x)=2.2ddx(x5)4.4ddx(x3)+1.3ddx(x2)0.9ddx(x)ddx(4)f(x)=(2.2)(5x4)(4.4)(3x2)+(1.3)(2x)0.9(1)0f(x)=11x413.2x2+2.6x0.9


Based from the graph, we choose x=1.4

Using Approximation Formula


xn+1=xnf(xn)f(xn)x2=x12.2x514.4x31+1.3x210.9x1411x4113.2x21+2.6x10.9x2=1.42.2(1.4)54.4(1.4)3+1.3(1.4)20.9(1.4)411(1.4)413.2(1.4)2+2.6(1.4)0.9x21.4x3=1.404176f(1.404176)f(1.404176)x31.404118x4=1.404118f(1.404118)f(1.404118)x41.404118


Since x3 and x4 agree to six decimal decimal places, therefore x1.404118

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