Answer
See graph and explanations.
Work Step by Step
Step 1. Given the function
$f(x)=\begin{cases} \sqrt x, \ x\geq0 \\ \sqrt {-x}, \ x\lt0 \end{cases}$,
we have
$f'(x)=\begin{cases} \frac{1}{2\sqrt x}, \ x\geq0 \\ \frac{-1}{2\sqrt {-x}}, \ x\lt0 \end{cases}$
Step 2. Using Newton's Method, start with $x_0=h$; we have $x_1=x_0-\frac{f(x_0)}{f'(x_0)}=h-\frac{\sqrt h}{ \frac{1}{2\sqrt h}}=h-2h=-h$
Step 3. Starting with $x_0=-h$, we have $x_1=x_0-\frac{f(x_0)}{f'(x_0)}=-h-\frac{\sqrt {-(-h)}}{\frac{-1}{2\sqrt {-(-h)}}}=-h+2h=h$
Step 4. Graph the function as shown in the figure. The Newton's Method uses a tangent line to estimate the solutions. If we start at $x_0=h$ as shown in the figure, the tangent line will intersect with the x-axis at $x_1=-h$.
Step 5. Similarly, we can draw a tangent line when we start with $x_0=-h$ and get $x_1=h$ as shown in the figure.
