Question #140052
Usin the method of false position. Find the root of equation x^6 - x^4 - x^3 - 1 = 0
1
Expert's answer
2020-10-25T19:15:35-0400
x6x4x31=x4(x21)(x3+1)=x^6-x^4-x^3-1=x^4(x^2-1)-(x^3+1)=

=x4(x1)(x+1)(x+1)(x2x+1)==x^4(x-1)(x+1)-(x+1)(x^2-x+1)=

=(x1)(x5+x4x2+x1)=(x-1)(x^5+x^4-x^2+x-1)

x1=1x1=1


Given f(x)=x6x4x31f(x)=x^6-x^4-x^3-1

Find points aa and bb  such that a<ba<b  and  f(a)f(b)<0f(a)f(b)<0

Let a=1,b=1.5a=1, b=1.5

f(a)=f(1)=1614131=2f(a)=f(1)=1^6-1^4-1^3-1=-2

f(b)=f(1.5)=1.561.541.531=1.953125f(b)=f(1.5)=1.5^6-1.5^4-1.5^3-1=1.953125

f(a)f(b)=f(1)f(1.5)<0f(a)f(b)=f(1)f(1.5)<0

Using iterative Regula-Falsi Formula, the first approximation,


x1=abaf(b)f(a)f(a)=x_1=a-\dfrac{b-a}{f(b)-f(a)}f(a)==11.511.953125+2(2)1.252964=1-\dfrac{1.5-1}{1.953125+2}(-2)\approx1.252964



f(1.252964)=1.562404f(1.252964)=-1.562404



x2=1.2529641.51.2529641.953125+1.562404(1.562404)x_2=1.252964-\dfrac{1.5-1.252964}{1.953125+1.562404}(-1.562404)\approx1.362754\approx1.362754


f(1.362754)=0.574795f(1.362754)=-0.574795



x3=1.3627541.51.3627541.953125+0.574795(0.574795)x_3=1.362754-\dfrac{1.5-1.362754}{1.953125+0.574795}(-0.574795)\approx1.393961\approx1.393961


f(1.393961)=0.147639f(1.393961)=-0.147639



x4=1.3939611.51.3939611.953125+0.147639(0.147639)x_4=1.393961-\dfrac{1.5-1.393961}{1.953125+0.147639}(-0.147639)\approx1.401413\approx1.401413


f(1.401413)=0.034199f(1.401413)=-0.034199



x5=1.4014131.51.4014131.953125+0.034199(0.034199)x_5=1.401413-\dfrac{1.5-1.401413}{1.953125+0.034199}(-0.034199)\approx1.403110\approx1.403110


f(1.403110)=0.007729f(1.403110)=-0.007729



x6=1.4031101.51.4031101.953125+0.007729(0.007729)x_6=1.403110-\dfrac{1.5-1.403110}{1.953125+0.007729}(-0.007729)\approx1.403492\approx1.403492



f(1.403492)=0.000390f(1.403492)=-0.000390


x7=1.4034921.51.4034921.953125+0.000390(0.000390)x_7=1.403492-\dfrac{1.5-1.403492}{1.953125+0.000390}(-0.000390)\approx1.403597\approx1.403597


f(1.403597)=0.000087f(1.403597)=-0.000087


x8=1.4035971.51.4035971.953125+0.000087(0.000087)x_8=1.403597-\dfrac{1.5-1.403597}{1.953125+0.000087}(-0.000087)\approx1.403601\approx1.403601


f(1.403601)=0.000020f(1.403601)=-0.000020



x8=1.4036011.51.4036011.953125+0.000020(0.000020)x_8=1.403601-\dfrac{1.5-1.403601}{1.953125+0.000020}(-0.000020)\approx1.403602\approx1.403602



f(1.403602)=0.000002f(1.403602)=-0.000002


If ε=0.0001,\varepsilon=0.0001, then f(x)<0.0001|f(x)|<0.0001

Root of f(x)f(x) is x=1.403602x=1.403602



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