x6−x4−x3−1=x4(x2−1)−(x3+1)=
=x4(x−1)(x+1)−(x+1)(x2−x+1)=
=(x−1)(x5+x4−x2+x−1) x1=1
Given f(x)=x6−x4−x3−1
Find points a and b such that a<b and f(a)f(b)<0
Let a=1,b=1.5
f(a)=f(1)=16−14−13−1=−2
f(b)=f(1.5)=1.56−1.54−1.53−1=1.953125
f(a)f(b)=f(1)f(1.5)<0 Using iterative Regula-Falsi Formula, the first approximation,
x1=a−f(b)−f(a)b−af(a)==1−1.953125+21.5−1(−2)≈1.252964
f(1.252964)=−1.562404
x2=1.252964−1.953125+1.5624041.5−1.252964(−1.562404)≈≈1.362754
f(1.362754)=−0.574795
x3=1.362754−1.953125+0.5747951.5−1.362754(−0.574795)≈≈1.393961
f(1.393961)=−0.147639
x4=1.393961−1.953125+0.1476391.5−1.393961(−0.147639)≈≈1.401413
f(1.401413)=−0.034199
x5=1.401413−1.953125+0.0341991.5−1.401413(−0.034199)≈≈1.403110
f(1.403110)=−0.007729
x6=1.403110−1.953125+0.0077291.5−1.403110(−0.007729)≈≈1.403492
f(1.403492)=−0.000390
x7=1.403492−1.953125+0.0003901.5−1.403492(−0.000390)≈≈1.403597
f(1.403597)=−0.000087
x8=1.403597−1.953125+0.0000871.5−1.403597(−0.000087)≈≈1.403601
f(1.403601)=−0.000020
x8=1.403601−1.953125+0.0000201.5−1.403601(−0.000020)≈≈1.403602
f(1.403602)=−0.000002
If ε=0.0001, then ∣f(x)∣<0.0001
Root of f(x) is x=1.403602
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