The solution of the equation x 2 − 4 x + 4 − ln x = 0 , 1 ≤ x ≤ 2 and for 2 ≤ x ≤ 4 within 10 − 7 , using newton’s method, secant method and method of false position.

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Answer 1

Question

Chapter 2.3, Problem 12ES

a.

To determine

The solution of the equation x2−4x+4−lnx=0,1≤x≤2 and for 2≤x≤4 within 10−7, using newton’s method, secant method and method of false position.

b.

To determine

The solution of the equation x+1−2sinπx=0,0≤x≤12 and for 12≤x≤1 within 10−7, using newton’s method, secant method and method of false position.

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Answer 2

Question

Chapter 2.3, Problem 12ES

a.

To determine

The solution of the equation x2−4x+4−lnx=0,1≤x≤2 and for 2≤x≤4 within 10−7, using newton’s method, secant method and method of false position.

b.

To determine

The solution of the equation x+1−2sinπx=0,0≤x≤12 and for 12≤x≤1 within 10−7, using newton’s method, secant method and method of false position.

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Answer 3

Question

Chapter 2.3, Problem 12ES

a.

To determine

The solution of the equation x2−4x+4−lnx=0,1≤x≤2 and for 2≤x≤4 within 10−7, using newton’s method, secant method and method of false position.

b.

To determine

The solution of the equation x+1−2sinπx=0,0≤x≤12 and for 12≤x≤1 within 10−7, using newton’s method, secant method and method of false position.

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Answer 4

Question

Chapter 2.3, Problem 12ES

a.

To determine

The solution of the equation x2−4x+4−lnx=0,1≤x≤2 and for 2≤x≤4 within 10−7, using newton’s method, secant method and method of false position.

b.

To determine

The solution of the equation x+1−2sinπx=0,0≤x≤12 and for 12≤x≤1 within 10−7, using newton’s method, secant method and method of false position.

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Answer 5

Chapter 2.3, Problem 12ES

a.

To determine

The solution of the equation x2−4x+4−lnx=0,1≤x≤2 and for 2≤x≤4 within 10−7, using newton’s method, secant method and method of false position.

b.

To determine

The solution of the equation x+1−2sinπx=0,0≤x≤12 and for 12≤x≤1 within 10−7, using newton’s method, secant method and method of false position.

Answer 6

Chapter 2.3, Problem 12ES

a.

To determine

The solution of the equation x2−4x+4−lnx=0,1≤x≤2 and for 2≤x≤4 within 10−7, using newton’s method, secant method and method of false position.

Answer 7

Chapter 2.3, Problem 12ES

a.

To determine

The solution of the equation x2−4x+4−lnx=0,1≤x≤2 and for 2≤x≤4 within 10−7, using newton’s method, secant method and method of false position.

Answer 8

b.

To determine

The solution of the equation x+1−2sinπx=0,0≤x≤12 and for 12≤x≤1 within 10−7, using newton’s method, secant method and method of false position.

Answer 9

b.

To determine

The solution of the equation x+1−2sinπx=0,0≤x≤12 and for 12≤x≤1 within 10−7, using newton’s method, secant method and method of false position.

Answer 10

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Answer 11

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Answer 12

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Answer 13

Ch. 2.1 - Use the Bisection method to find p3 for f(x)=xcosx...Ch. 2.1 - Let f(x) = 3(x +1)(x 12)(x 1) = 0. Use the...Ch. 2.1 - Use the Bisection method to find solutions...Ch. 2.1 - Use the Bisection method to find solutions...Ch. 2.1 - Use the Bisection method to find solutions...Ch. 2.1 - Prob. 7ES Ch. 2.1 - Prob. 8ES Ch. 2.1 - Prob. 9ES Ch. 2.1 - Prob. 10ES Ch. 2.1 - Prob. 11ES

The solution of the equation x 2 − 4 x + 4 − ln x = 0 , 1 ≤ x ≤ 2 and for 2 ≤ x ≤ 4 within 10 − 7 , using newton’s method, secant method and method of false position.

Solution Summary: The author explains Newton's method, secant method and method of false position.

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The solution of the equation x2−4x+4−lnx=0,1≤x≤2 and for 2≤x≤4 within 10−7, using newton’s method, secant method and method of false position.

The solution of the equation x+1−2sinπx=0,0≤x≤12 and for 12≤x≤1 within 10−7, using newton’s method, secant method and method of false position.

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