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The first step in confirming a local linear approximation involves calculating the derivative of the function. Differentiation helps us understand how a function changes as its input changes. In this exercise, we're looking at the function \( f(x) = \frac{1}{\sqrt{1-x}} \). To find its derivative, we use the chain rule, a fundamental technique in calculus. The formula for the derivative we obtain is:

• \( f'(x) = \frac{1}{2}(1-x)^{-3/2} \)

Once we've established this, we calculate the derivative at the specific point of interest, \( x_0 = 0 \). When substituting 0 into our derivative formula, it simplifies to:

• \( f'(0) = \frac{1}{2} \)

This tells us the slope of the tangent line to the curve \( f(x) \) at \( x_0 \). The slope is crucial for forming the line that approximates our function locally.

The idea of linear approximation is to use a line to estimate function values near a particular point. The linear approximation, also known as the tangent line approximation, uses the derivative and a given point to form a straight line. Remember this key formula for linear approximation:

• \( f(x) \approx f(x_0) + f'(x_0)(x - x_0) \)

In our specific exercise, we need \( x_0 = 0 \), which simplifies our calculations. Here's what we do step-by-step:

• Substitute into the formula: \( x_0 = 0 \), \( f(x_0) = 1 \) (since \( f(0) = \frac{1}{\sqrt{1-0}} = 1 \)), and \( f'(x_0) = \frac{1}{2} \).
• Plug these into the approximation formula: \( f(x) \approx 1 + \frac{1}{2}(x - 0) \).
• The simplified approximation is \( f(x) \approx 1 + \frac{1}{2}x \).

This line effectively estimates the function \( f(x) \) near \( x = 0 \). It relies on the concept that for values of \( x \) close to 0, the complex function \( f(x) \) behaves much like a straight line, making it easier to calculate approximations.

Evaluating a function at a specific point is an important task that provides a foundation for constructing local approximations. For our function \( f(x) = \frac{1}{\sqrt{1-x}} \), the evaluation at \( x_0 = 0 \) is straightforward.

• Substitution directly into the function gives \( f(0) = \frac{1}{\sqrt{1-0}} = 1 \).

This result is fundamental in forming the linear approximation, as it represents where the tangent or approximating line intersects the y-axis.Understanding where our function starts, allows us to build its linear approximation using the value of the function and the rate at which it changes, determined by the derivative. Evaluation, alongside the derivative, paints a fuller picture of how our function behaves around \( x_0 = 0 \), providing a clear starting point for further calculations or analyses.