The following function has a domain (-1,0) u (0,infinity). F(x)=(-1+square root of 1+x ) /x. Make a table of values of f(x) for x=+/- 1, +\- 0.1, +\- 0.01, and +\- 0.001. Use your table to guess a limiting value for f(x) as x approaches 0. Finally use difference of squares rationalization to find a continuos function g(x) who’s domain contains (-1, infinity) such that f(x) equals g(x) on (-1,0)u(0,infinity). Use this to prove that your guess is correct.

Let's start by analyzing the function and completing the steps outlined in the question.

### Step 1: Function Definition and Domain
The given function is:
\[
f(x) = \frac{-1 + \sqrt{1+x}}{x}
\]
We are tasked with determining values of \( f(x) \) for specific \( x \)-values, estimating a limiting value as \( x \to 0 \), and finally, rationalizing the function to find a continuous extension.

### Step 2: Table of Values for \( f(x) \)

We will evaluate \( f(x) \) at the following values: \( x = \pm 1, \pm 0.1, \pm 0.01, \pm 0.001 \).

I’ll compute these values next.Here's the table of values for \( f(x) \) at the specified points:

\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
-1 & 1.0000 \\
-0.1 & 0.5132 \\
-0.01 & 0.5013 \\
-0.001 & 0.5001 \\
0.001 & 0.4999 \\
0.01 & 0.4988 \\
0.1 & 0.4881 \\
1 & 0.4142 \\
\hline
\end{array}
\]

### Step 3: Guessing the Limiting Value
As \( x \) approaches 0 from both positive and negative sides, the values of \( f(x) \) seem to approach 0.5. Thus, we can hypothesize that:
\[
\lim_{x \to 0} f(x) = 0.5
\]

### Step 4: Rationalization and Continuous Function \( g(x) \)

To find a continuous function \( g(x) \), we rationalize the expression for \( f(x) \). Starting from:
\[
f(x) = \frac{-1 + \sqrt{1+x}}{x}
\]
Multiply both the numerator and the denominator by the conjugate of the numerator, which is \( -1 - \sqrt{1+x} \), to simplify.

Let's compute this rationalization next.It seems like I can’t do more advanced data analysis right now. Please try again later.

I’ll explain the rationalization step verbally for now.

### Rationalization of \( f(x) \):
We begin with:
\[
f(x) = \frac{-1 + \sqrt{1+x}}{x}
\]

Multiply both the numerator and denominator by the conjugate of the numerator:
\[
\frac{-1 + \sqrt{1+x}}{x} \times \frac{-1 - \sqrt{1+x}}{-1 - \sqrt{1+x}} = \frac{(-1 + \sqrt{1+x})(-1 - \sqrt{1+x})}{x(-1 - \sqrt{1+x})}
\]
The numerator becomes:
\[
(-1 + \sqrt{1+x})(-1 - \sqrt{1+x}) = -(1+x) + 1 = -x
\]
Thus, the expression simplifies to:
\[
f(x) = \frac{-x}{x(-1 - \sqrt{1+x})} = \frac{1}{1 + \sqrt{1+x}}
\]

This new function is:
\[
g(x) = \frac{1}{1 + \sqrt{1+x}}
\]

### Step 5: Proving the Limit is 0.5
As \( x \to 0 \), the expression \( g(x) = \frac{1}{1 + \sqrt{1+x}} \) approaches:
\[
g(0) = \frac{1}{1 + \sqrt{1+

The following function has a domain (-1,0) ∪ (0,infinity). F(x)=(-1+sqrt(1+x)) / x. Make a table of values of f(x) for x=± 1, ± 0.1, ± 0.01, and ± 0.001. Use your table to guess a limiting value for f(x) as x approaches 0. Finally, use difference of squares rationalization to find a continuous function g(x) whose domain contains (-1, infinity) such that f(x) equals g(x) on (-1,0) ∪ (0,infinity). Use this to prove that your guess is correct.

Explore how to evaluate and rationalize the function f(x) = (-1 + sqrt(1 + x)) / x with detailed calculations. A table of values shows the behavior of the function as x approaches 0, leading to a limiting value of 0.5. Learn the rationalization technique using the difference of squares to derive a continuous function g(x) and confirm the limit.

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