Question: Suppose you attempt to use Newton's method to approximate a solution of the equation x 3 1 2 x + 7 = 0 3 -

Suppose you attempt to use Newton's method to approximate a solution of the equation
x312x+7=03-12+7=0.
Letx0=20=2be the initial approximation, and then calculatex11andx22. If the value is undefined, enterDNE.
x1=1=
x2=2=
What do you notice?
Newton's method fails because f'(x0)=0(0)=0 but f(x0)0(0)0.
Newton's method fails because the approximations are getting larger in absolute value.
Newton's method fails because the approximations are alternating between two numbers.
Newton's method is successful at approaching a number so far.

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