Question

1. Determine the root of function f(x)= x+2x-2r-1 by using Newton's method with x=0.8 and error, e=0.005. 2. Use Newton's method to approximate the root for f(x) = -x-1. Do calculation in 4 decimal points. Letx=1 and error, E=0.005. 3. Given 7x)=x-2x2+x-3 Use Newton's method to estimate the root at 4 decimal points. Take initial value, Xo4. 4. Find the root of f(x)=x2-9x+1 accurate to 3 decimal points. Use Newton's method with initial value, X=2

Answer 1

solution: given f(x)= x3 - 2x²+x-3., do=4 t '(x)= 3X2-4x+1 for mula newton's method flan) f'(x98) Inti & dign- iteration: 8,- do - $(x) au- flu To f'(o ) pode of '(U) . 3 5 4- 33 33 33 - 3 aecond iteration: = 3- x a «,- 3 (X) fo(ai) $(3) + (3) 3- q 2.4375 third iteration: sa a, (22) 2.4375-7 (24375) f'(x2) t(2.4375) 2.4375- 2.0368 9.07421 =

Serie 130 Touth e sa tion:. . . 200130 - $ (2-21.30 L f'(x3) '(2-2130) 2.2.130 - 0.25613 6.84010 to E.17755 fifth iteration :- $(ay) +(2-1755) 55. Xy- 2-12SS f'(ou) f'(2-1755) 2:1755- 0.00610 6.4964 2.1745. the root is 2017454

w) gioun f(x)= x2-9X+1, Xo =2.. f'(x)= 32-9 formula newton's method. Anti in- $(In) filan) first iteration: to- $(x) = fi (70) 2- $12) 7' (2) -9) - 5 second iteration : x ,- 5 5 F(xi) fi(81) (5) f'(s) S: 3. 772. 66 38d iteration : goo . &z= -f(ag) = 3.772- +(3.772) f'(&2). f'(3.772) - 3.77C 20.7199 33-6839 3.156.

yth iteration: ५- ४१- (43) 1 (3.156) 6 (3.156)- - (४3) 5.( 3.156) . : 3.156 - 4.0308 20.881 __ : 2.96 2. 5th iteration : ४६ = ४५-६(५) . .१62- 5 (2.962) 5' (2.962) (५५) _ 2.962- ०.3289 17. 3203 .. 2.१५ 3 6th iteration : ४८ - १.१५२ - (१.१५3) 66 (४६) - 5' (s) 7 (2943) - 2.१५3- 0.00 30 55. 16 • 9837 2.१५ 28.

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