Answer to Question #110544 in Quantitative Methods for Anju Jayachandran

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)=x^3-5x^2+1"

We distinguish the roots by Sturm's theorem:

"f_1(x)=f'(x)=3x^2-10x"

"f:f_1\\\\\nx^3-5x^2+1\\quad|3x^2-10x\\\\\n3x^3-15x^2+3\\quad x-5\\\\\n3x^3-10x^2\\\\\n-5x^2+3\\\\\n-15x^2+9\\\\\n-15x^2+50x\\\\\n-50x+9"

"f_2(x)=50x-9"

"f_1:f_2\\\\\n3x^2-10x\\quad\\quad|50x-9\\\\\n150x^2-500x\\quad3x-\\frac{473}{50}\\\\\n150x^2-27x\\\\\n-473x\\\\\n-473x+\\frac{9\\cdot473}{50}\\\\\n-A, A>0"

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

"\\begin{matrix}\n & f&f_1&f_2&f_3 &\\\\\n 0 & +&0&-&+&2\\\\\n1&-&-&+&+&1\\\\\n2&-&-&+&+&1\\\\\n3&-&-&+&+&1\n\\end{matrix}"

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


By Bisection method:

"x\\in(0,1)\\\\\nf(0)=1>0\\\\\nf(1)=-3<0\\\\\nx_1=\\frac{0+1}{2}=0.5\\\\\nf(x_1)=f(0.5)=-0.125"  ,which is negative.  

Hence, the root lies in between 0 and 0.5

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

Hence, the root lies in between 0.25 and 0.5

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

Hence, the root lies in between 0.375 and 0.5

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

Hence, the root lies in between 0.4375 and 0.5

"x_5=\\frac{0.4375+0.5}{2}=0.46875\\\\\nf(x_5)=f(0.46875)=0.004"

0.5-0.46875=0.03125<0.05

"x\\in(0.46875,0.5)"



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