Use Newton's Method to approximate the root of 2.2x5−4.4x3+1.3x2−0.9x−4.0=0 in the interval [−2,−1] correct to six decimal places.
We apply Newton's Method with
f(x)=2.2x5−4.4x3+1.3x2−0.9x−4 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)=11x4−13.2x2+2.6x−0.9
Based from the graph, we choose x=1.4
Using Approximation Formula
xn+1=xn−f(xn)f′(xn)x2=x1−2.2x51−4.4x31+1.3x21−0.9x1−411x41−13.2x21+2.6x1−0.9x2=−1.4−2.2(−1.4)5−4.4(−1.4)3+1.3(−1.4)2−0.9(−1.4)−411(−1.4)4−13.2(−1.4)2+2.6(−1.4)−0.9x2≈−1.4x3=−1.404176−f(−1.404176)f′(−1.404176)x3≈−1.404118x4=−1.404118−f(−1.404118)f′(−1.404118)x4≈−1.404118
Since x3 and x4 agree to six decimal decimal places, therefore x≈−1.404118
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