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 − tan − 1 x . Differentiate the function f ( x ) = x 2 − tan − 1 x with respect to x . f ′ ( x ) = d f d x = d d x [ x 2 − tan − 1 x ] = 2 x − 1 1 + x 2 Substitute the values x 0 = 1 and x 0 = 1 2 in the function f ( x ) as follows. f ( 1 ) = ( 1 ) 2 − tan − 1 1 = 0.215 f ( 0.5 ) = ( 0.5 ) 2 − tan − 1 0.5 = − 0.213 Thus, consider the value of x 0 as 0.75 . The next approximation is x 1 = ( 0.75 ) − f ( 0.75 ) f ′ ( 0.75 ) = 0.75 − ( 0.75 ) 2 − tan − 1 ( 0.75 ) 2 ( 0.75 ) + 1 1 + ( 0.75 ) 2 = 0.844187336 Use the value x 1 = 0.844187336 in the relation as follows

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Chapter 27, Problem 68RE

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To solve: The equation x2=tan−1x by Newton’s Method and verify the solution with the help of graph.

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Chapter 27, Problem 67RE

Chapter 27, Problem 69RE

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Chapter 27 Solutions

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The equation x 2 = tan − 1 x by Newton’s Method and verify the solution with the help of graph.

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Chapter 27 Solutions