If f(x)= x^1/2 if x¡Ý 0 and -(-x)^1/2 if x>0

then the root of the equation f(x) = 0 is x = 0. Explain why Newton¡¯s method fails no matter which initial approximation p_0¡Ùo is used. Illustrate your explanation with a sketch.

To understand why Newton's method fails for the given equation f(x) = 0, let us first review how Newton's method works.

Newton's method is an iterative numerical method used to find the roots of a function. It involves iteratively improving an initial approximation by using the formula:

p_n+1 = p_n - f(p_n)/f'(p_n),

where p_n is the nth approximation and f'(p_n) is the derivative of the function evaluated at p_n.

In our case, let's calculate the derivative of the function f(x):

f'(x) = (1/2)x^(-1/2) if x ≥ 0, and
f'(x) = (1/2)(-x)^(-1/2) if x > 0.

Now, since we want to find the root of the equation f(x) = 0, we need to find a value of x for which f(x) = 0. Let's see what happens when we apply Newton's method to this equation.

Assume we start with an initial approximation p_0. On the first iteration, we calculate p_1 as:

p_1 = p_0 - f(p_0)/f'(p_0).

Now, let's consider two cases:
1. If p_0 is negative: Since f(x) is defined as x^1/2 for x ≥ 0, the function value f(p_0) will always be positive for any negative value of p_0. Similarly, the derivative f'(p_0) will also be positive because it depends on the positive square root of p_0. Therefore, we have p_1 = p_0 - positive/positive, which means p_1 will still be negative.

2. If p_0 is positive: In this case, f(x) is defined as -(-x)^1/2 for x > 0. So, f(p_0) will be negative because it involves the negative square root of p_0. The derivative f'(p_0) will be negative as well because it involves the negative square root of p_0. Thus, p_1 = p_0 - negative/negative, which means p_1 will still be positive.

We can repeat this process for subsequent iterations, but the same pattern persists. Regardless of the starting approximation p_0, the value will alternate between positive and negative at each iteration.

Now, let's illustrate this behavior graphically with a sketch:

On a graph, plot the function f(x) = x^1/2 for x ≥ 0 and f(x) = -(-x)^1/2 for x > 0. The graph will consist of two halves, each representing one of the two cases.

Starting with any initial approximation, draw a horizontal line at the height y = 0. The point(s) where this line intersects the graph will correspond to the roots of the equation f(x) = 0.

What you will observe is that on alternating iterations, your initial approximation will lead you to either the positive or negative half of the graph, but never to the root at x = 0. This behavior continues indefinitely, as Newton's method fails to converge to the root x = 0.

Hence, no matter which initial approximation p_0 you choose, Newton's method fails for the equation f(x) = 0, as it oscillates indefinitely between positive and negative values, never converging to the root at x = 0.