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Question Number 178867 by Spillover last updated on 22/Oct/22

Answered by Spillover last updated on 15/Jan/23

f(x)=x^4 −7x^3 +1 requare to show f(x_o )×f(x_1 )<0 given 0<x<1 f(x)=x^4 −7x^3 +1 f(0)=1 f(1)=−5 f(x_o )×f(x_1 )<0 1×−5<0 hence the root lies btn 0<x<1

f(x)=x4−7x3+1requaretoshowf(xo)×f(x1)<0given0<x<1f(x)=x4−7x3+1f(0)=1f(1)=−5f(xo)×f(x1)<01×−5<0hencetherootliesbtn0<x<1

Answered by Spillover last updated on 15/Jan/23

f(x)=x^4 −7x^4 +1 f ′(x)=4x^3 −21 from N−R method x_(n+1) =x_n −((f(x))/(f ′(x))) x_(n+1) =x_n −((x_n ^4 −7x_n ^3 +1)/(4x_n ^3 −21x_n ^2 )) x_(n+1) =((x_n ^4 −14x_n ^3 −1)/(4x_n ^3 −21x_n ^2 )) x_0 =1 1^(st ) Iteration n=0 x_1 =((x_0 ^4 −14x_0 ^3 −1)/(4x_0 ^3 −21x_0 ^2 )) =0.705882352 2^(nd) Iteration n=1 x_2 =0.571865018 3^(st ) Iteration n=2 x_3 =0.538828437 4^(st ) Iteration n=3 x_4 =0.536855534

f(x)=x4−7x4+1f′(x)=4x3−21fromN−Rmethodxn+1=xn−f(x)f′(x)xn+1=xn−xn4−7xn3+14xn3−21xn2xn+1=xn4−14xn3−14xn3−21xn2x0=11stIterationn=0x1=x04−14x03−14x03−21x02=0.7058823522ndIterationn=1x2=0.5718650183stIterationn=2x3=0.5388284374stIterationn=3x4=0.536855534

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