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Solve the Problem f(x)=xa,a>0f ( x ) = \sqrt [ a ] { x } , a > 0

Question 321

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Solve the problem.
-Apply Newton's method to f(x)=xa,a>0f ( x ) = \sqrt [ a ] { x } , a > 0 , and write an expression for xn+1x _ { n } + 1 . If the initial guess x0x _ { 0 } is greater than or equal to 1 , what happens to xn+1\left| x _ { n } + 1 \right| as nn \rightarrow \infty ?

Correct Answer:

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\[\begin{array}{l}
f ( x ) = \sqrt [ a ]...

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