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The Value for Which Newton's Method Fails for the Function y=x3+3x2x+1y = - x ^ { 3 } + 3 x ^ { 2 } - x + 1

Question 127

Multiple Choice

The value for which Newton's method fails for the function below is shown in the graph. Give the reason why the method fails. y=x3+3x2x+1y = - x ^ { 3 } + 3 x ^ { 2 } - x + 1  The value for which Newton's method fails for the function below is shown in the graph. Give the reason why the method fails.  y = - x ^ { 3 } + 3 x ^ { 2 } - x + 1    A)   \lim _ { x \rightarrow \infty } \left\{ \begin{array} { l }  0 = x _ { 1 } = x _ { 3 } = \ldots \\ - 1 = x _ { 2 } = x _ { 4 } = \ldots \end{array} \right.  B)   f ^ { \prime } \left( x _ { 1 } \right)  = 0  C)   f ^ { \prime } \left( x _ { 2 } \right)  = 0  D)   f ^ { \prime \prime } \left( x _ { 2 } \right)  = 0  E)   \lim _ { x \rightarrow \infty } \left\{ \begin{array} { l }  1 = x _ { 1 } = x _ { 3 } = \ldots \\ 0 = x _ { 2 } = x _ { 4 } = \ldots \end{array} \right.


A) limx{0=x1=x3=1=x2=x4=\lim _ { x \rightarrow \infty } \left\{ \begin{array} { l } 0 = x _ { 1 } = x _ { 3 } = \ldots \\- 1 = x _ { 2 } = x _ { 4 } = \ldots\end{array} \right.
B) f(x1) =0f ^ { \prime } \left( x _ { 1 } \right) = 0
C) f(x2) =0f ^ { \prime } \left( x _ { 2 } \right) = 0
D) f(x2) =0f ^ { \prime \prime } \left( x _ { 2 } \right) = 0
E) limx{1=x1=x3=0=x2=x4=\lim _ { x \rightarrow \infty } \left\{ \begin{array} { l } 1 = x _ { 1 } = x _ { 3 } = \ldots \\0 = x _ { 2 } = x _ { 4 } = \ldots\end{array} \right.

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