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Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your answer to four decimal places.)

1/3x^3 + 1/2x^2 + 8 = 0, x1 = −3
I got -3.4808, but it's wrong. Help?

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Benny
2 answers

  1. It's ( -3.7103 )

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    ll

  2. 1/3x3+1/2x2+11=0,x1=−3

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    Anonymous

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