Explain why Newton 's method doesn 't work for finding the root of the equation x 3 3 x 9 0 if the initial approximation is chosen to be x 1 1 f ( x ) x 3 3 x 9 f ' ( x ) If x 1 1 , then f ' ( x 1 ) and the tangent line used for approximating x 2 is Select horizontal vertical Attempting to find x 2 results in trying to Select multiply divide by zero

Question

Explain why Newton 's method doesn 't work for finding the root of the equation x 3 3 x   9   0 if the initial approximation is chosen to be x 1   1   f ( x )   x 3 3 x   9 f  ' ( x )     If x 1   1 , then f  ' ( x 1 )   and the tangent line used for approximating x 2 is       Select       horizontal vertical   Attempting to find x 2 results in trying to       Select       multiply divide by zero

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Explain why Newton's method doesn't work for finding the root of the equation x 3 3 x + 9 = 0 if the initial approximation

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