Answer to Question #307439 in Differential Equations for Junior Pyakara

Question

The fourth-degree polynomial

f(x) =230x⁴ +18x³+9x² -221x-9

Has two real zeros, one in[-1,0] which is -0.0406593. Attempt to approximate this zero to within 10^-2 using the

a) secant method( using the endpoint of each Interval approximation)

b) Newton' method(use the midpoint of each Interval as the initial approximation

Answer 1

a. Let x₀=-1,x₁=0,e=0.01

f(0)=-9

f(-1)=230-18+9+221-9=433

$x_2=x_1-\frac{x_1-x_0}{f(x_1)-f(x_0)}=0-\frac{0-(-1)}{-9-433}(-9)=-0.02$

f(x₂)=0.0000368-0.000144+0.0036+4,42-9=-4.496

4.496>0.01

$x_3=-0.02-\frac{-0.02-0}{-4.496+9}(-4.496)=-0.04$

f(-0.04)=0.0006-0.001+0.0144+8.84-9=0.007

0.007<e

x=-0.04

b. x₀=-0.5, e=0.01

f'(x)=920x³+54x²+18x-221

f(-0.5)=14,375-2.25+2.25+110.5-9=115.875

f'(-0.5)=-115+13.5-9-221=-331.5

$x_1=x_0-f(x)/f'(x)=-0.5+115.875/331.5=-0.15$

x₀-x₁=-0.5+0.15=-0.35

0.35>e

f(-0.15)=0.116-0.06+0.2+33.15-9=24.406

f'(-0.15)=-3,105+1.215-2,7-221=-220.19

x₂=-0.15+24.406/220.19=-0.04

x₁-x₂=-0.15+0.04=-0.11>e

f(-0.04)=0.0006-0.001+0.0144+8.84-9=0.007

f'(-0.04)=-0.06+0.0864-0.72-221=-221.69

x₃=-0.04+0.007/221.69=-0.04

x₂- x₃=0.00003<e

x=-0.04