Solve the problem. -Show that if $h 0$, applying Newton's method to $$f ( x ) = \left\{ \begin{array} { l l } \sqrt { x - 8 } , & x \geq 8 \ - \sqrt { 8 - x } , & x

Question

Solve the problem.
-Show that if $h > 0$, applying Newton's method to
$$f ( x ) = \left\{ \begin{array} { l l } 
\sqrt { x - 8 } , & x \geq 8 \
- \sqrt { 8 - x } , & x < 8
\end{array} \right.$$
leads to $x _ { 2 } = h$ if $x _ { 0 } = h$ and to $x _ { 2 } = - h$ if $x _ { 0 } = - h$ when $0 < 8 < h$.

Quizplus · Accepted Answer

The Answer of Solve the problem.
-Show that if $h > 0$, applying Newton's method to
$$f ( x ) = \left\{ \begin{array} { l l } 
\sqrt { x - 8 } , & x \geq 8 \
- \sqrt { 8 - x } , & x < 8
\end{array} \right.$$
leads to $x _ { 2 } = h$ if $x _ { 0 } = h$ and to $x _ { 2 } = - h$ if $x _ { 0 } = - h$ when $0 < 8 < h$.

Solve the Problem h>0h > 0h>0 , Applying Newton's Method To f(x)={x−8,x≥8−8−x,x<8f ( x ) = \left\{ \begin{array} { l l } \sqrt { x - 8 } , & x \geq 8 \\- \sqrt { 8 - x } , & x < 8\end{array} \right.f(x)={x−8​,−8−x​,​x≥8x<8​

Solve the Problem $h > 0$ , Applying Newton's Method To $f ( x ) = \left\{ \begin{array} { l l } \sqrt { x - 8 } , & x \geq 8 \\- \sqrt { 8 - x } , & x < 8\end{array} \right.$