f(x)=x3−5x2+1
We distinguish the roots by Sturm's theorem:
f1(x)=f′(x)=3x2−10x
f:f1x3−5x2+1∣3x2−10x3x3−15x2+3x−53x3−10x2−5x2+3−15x2+9−15x2+50x−50x+9
f2(x)=50x−9
f1:f23x2−10x∣50x−9150x2−500x3x−50473150x2−27x−473x−473x+509⋅473−A,A>0
f3(x)=A,A>0
0123f+−−−f10−−−f2−+++f3++++2111
The smallest positive root of f(x):x∈(0,1) .
By Bisection method:
x∈(0,1)f(0)=1>0f(1)=−3<0x1=20+1=0.5f(x1)=f(0.5)=−0.125 ,which is negative.
Hence, the root lies in between 0 and 0.5
x2=20+0.5=0.25f(x2)=f(0.25)=0.703 , which is positive
Hence, the root lies in between 0.25 and 0.5
x3=20.25+0.5=0.375f(x3)=f(0.375)=0.349 ,which is positive
Hence, the root lies in between 0.375 and 0.5
x4=20.375+0.5=0.4375f(x4)=f(0.4375)=0.1267 ,which is positive
Hence, the root lies in between 0.4375 and 0.5
x5=20.4375+0.5=0.46875f(x5)=f(0.46875)=0.004
0.5-0.46875=0.03125<0.05
x∈(0.46875,0.5)
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