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 Problem: bisection method 2x^3-2x-5, a=1, b=4 [ Calculator, Method and examples ]Solution:Your problem -> bisection method 2x^3-2x-5, a=1, b=4Here 2x^3-2x-5=0Let f(x) = 2x^3-2x-51^(st) iteration :Here f(1) = -5 < 0 and f(4) = 115 > 0:. Now, Root lies between 1 and 4x_0=(1+4)/2=2.5f(x_0)=f(2.5)=2*(2.5)^3-2*(2.5)-5=21.25 > 02^(nd) iteration :Here f(1) = -5 < 0 and f(2.5) = 21.25 > 0:. Now, Root lies between 1 and 2.5x_1=(1+2.5)/2=1.75f(x_1)=f(1.75)=2*(1.75)^3-2*(1.75)-5=2.2188 > 03^(rd) iteration :Here f(1) = -5 < 0 and f(1.75) = 2.2188 > 0:. Now, Root lies between 1 and 1.75x_2=(1+1.75)/2=1.375f(x_2)=f(1.375)=2*(1.375)^3-2*(1.375)-5=-2.5508 < 04^(th) iteration :Here f(1.375) = -2.5508 < 0 and f(1.75) = 2.2188 > 0:. Now, Root lies between 1.375 and 1.75x_3=(1.375+1.75)/2=1.5625f(x_3)=f(1.5625)=2*(1.5625)^3-2*(1.5625)-5=-0.4956 < 05^(th) iteration :Here f(1.5625) = -0.4956 < 0 and f(1.75) = 2.2188 > 0:. Now, Root lies between 1.5625 and 1.75x_4=(1.5625+1.75)/2=1.6562f(x_4)=f(1.6562)=2*(1.6562)^3-2*(1.6562)-5=0.7742 > 06^(th) iteration :

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