Question #110544
Find the interval of unit length that contains the smallest positive root of the
equation f(x) = x
3 −5x
2 +1 = 0. Starting with this interval, find an interval of
length 0.05 or less that contains the root, by Bisection method.
1
Expert's answer
2020-04-20T18:45:30-0400

f(x)=x35x2+1f(x)=x^3-5x^2+1

We distinguish the roots by Sturm's theorem:

f1(x)=f(x)=3x210xf_1(x)=f'(x)=3x^2-10x

f:f1x35x2+13x210x3x315x2+3x53x310x25x2+315x2+915x2+50x50x+9f:f_1\\ x^3-5x^2+1\quad|3x^2-10x\\ 3x^3-15x^2+3\quad x-5\\ 3x^3-10x^2\\ -5x^2+3\\ -15x^2+9\\ -15x^2+50x\\ -50x+9

f2(x)=50x9f_2(x)=50x-9

f1:f23x210x50x9150x2500x3x47350150x227x473x473x+947350A,A>0f_1:f_2\\ 3x^2-10x\quad\quad|50x-9\\ 150x^2-500x\quad3x-\frac{473}{50}\\ 150x^2-27x\\ -473x\\ -473x+\frac{9\cdot473}{50}\\ -A, A>0

f3(x)=A,A>0f_3(x)=A, A>0

ff1f2f30+0+21++12++13++1\begin{matrix} & f&f_1&f_2&f_3 &\\ 0 & +&0&-&+&2\\ 1&-&-&+&+&1\\ 2&-&-&+&+&1\\ 3&-&-&+&+&1 \end{matrix}

 The smallest positive root of f(x):x(0,1)f(x):x\in(0,1) .


By Bisection method:

x(0,1)f(0)=1>0f(1)=3<0x1=0+12=0.5f(x1)=f(0.5)=0.125x\in(0,1)\\ f(0)=1>0\\ f(1)=-3<0\\ x_1=\frac{0+1}{2}=0.5\\ f(x_1)=f(0.5)=-0.125  ,which is negative.  

Hence, the root lies in between 0 and 0.5

x2=0+0.52=0.25f(x2)=f(0.25)=0.703x_2=\frac{0+0.5}{2}=0.25\\ f(x_2)=f(0.25)=0.703 , which is positive

Hence, the root lies in between 0.25 and 0.5

x3=0.25+0.52=0.375f(x3)=f(0.375)=0.349x_3=\frac{0.25+0.5}{2}=0.375\\ f(x_3)=f(0.375)=0.349 ,which is positive

Hence, the root lies in between 0.375 and 0.5

x4=0.375+0.52=0.4375f(x4)=f(0.4375)=0.1267x_4=\frac{0.375+0.5}{2}=0.4375\\ f(x_4)=f(0.4375)=0.1267 ,which is positive

Hence, the root lies in between 0.4375 and 0.5

x5=0.4375+0.52=0.46875f(x5)=f(0.46875)=0.004x_5=\frac{0.4375+0.5}{2}=0.46875\\ f(x_5)=f(0.46875)=0.004

0.5-0.46875=0.03125<0.05

x(0.46875,0.5)x\in(0.46875,0.5)



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