Explanation Method used: Newton’ method: If x n is an approximation a solution of f ( x ) = 0 and f ′ ( x n ) ≠ 0 the next approximation is given by x n + 1 = x n − f ( x n ) f ′ ( x n ) . Calculation: Consider the function f ( x ) = x 2 − 3 + ln 4 x . Differentiate the function f ( x ) = x 2 − 3 + ln 4 x with respect to x . f ′ ( x ) = d f d x = d d x [ x 2 − 3 + ln 4 x ] = 2 x − 0 + 4 4 x = 2 x + 1 x Substitute the values x 1 = 1 and x 1 = 2 in the function f ( x ) as follows. f ( 1 ) = ( 1 ) 2 − 3 + ln 4 ( 1 ) = − 0.6137 f ( 2 ) = ( 2 ) 2 − 3 + ln 4 ( 2 ) = 4 − 3 + ln 8 = 3.079 Thus, consider the value of x 1 as 1.5. The next approximation is x 2 = ( 1.5 ) − f ( 1.5 ) f ′ ( 1.5 ) = 1.5 − ( 1.5 ) 2 − 3 + ln 4 ( 1.5 ) 2 ( 1.5 ) + 1 1.5 = 1.5 − 1.0417594 3.6666667 ≈ 1.2158838 Use the value x 2 = 1

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Chapter 27.8, Problem 21E

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To solve: The equation x2−3+ln4x=0 by Newton’s Method and verify the solution with the help of graph.

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Chapter 27.8, Problem 20E

Chapter 27.8, Problem 22E

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The equation x 2 − 3 + ln 4 x = 0 by Newton’s Method and verify the solution with the help of graph.

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