Problem 6 - Calculus Early Transcendentals

Question

Confirm that the stated formula is the local linear approximation at \(x_{0}=0\) $$ \frac{1}{\sqrt{1-x}} \approx 1+\frac{1}{2} x $$

Step-by-Step Solution

Answer

The given formula is the local linear approximation at \( x_0 = 0 \).

Step 1: Recall the formula for local linear approximation

The local linear approximation of a function \( f(x) \) at a point \( x_0 \) is given by \( f(x) \approx f(x_0) + f'(x_0)(x - x_0) \). This formula is used to approximate the function around \( x_0 \).

Step 2: Define the function

In this problem, the function \( f(x) \) is \( f(x) = \frac{1}{\sqrt{1-x}} \). We are asked to approximate this function around \( x_0 = 0 \).

Step 3: Evaluate the function at the point

Calculate \( f(0) = \frac{1}{\sqrt{1-0}} = 1 \).

Step 4: Find the derivative of the function

First, recall that \( f(x) = (1-x)^{-1/2} \). Using the chain rule, the derivative \( f'(x) \) is \( f'(x) = -\frac{1}{2} (1-x)^{-3/2}(-1) = \frac{1}{2}(1-x)^{-3/2} \).

Step 5: Evaluate the derivative at the point

Calculate \( f'(0) = \frac{1}{2}(1-0)^{-3/2} = \frac{1}{2} \).

Step 6: Write the linear approximation formula

Substitute into the local linear approximation formula: \( f(x) \approx 1 + \frac{1}{2}(x - 0) = 1 + \frac{1}{2}x \).

Step 7: Confirm the given formula

The given formula \( \frac{1}{\sqrt{1-x}} \approx 1 + \frac{1}{2}x \) matches the derived local linear approximation formula at \( x_0 = 0 \). Thus, the formula is confirmed.

Key Concepts

Function Approximation

Function approximation is all about finding simpler ways to represent complex functions. The goal is to approximate a complicated function with a simpler one, making calculations easier or faster.
Local linear approximation is one such technique. It involves using a tangent line to approximate a function near a specific point. This allows us to predict the value of the function in a small neighborhood around the point of interest.
This technique is particularly helpful because:

• It simplifies the computation near the point.
• Provides a good estimation of the function's value.
• Gives insights into the function's behavior around that point.

In the context of the exercise, we used this method to approximate the function \( f(x) = \frac{1}{\sqrt{1-x}} \) at \( x_0 = 0 \). By using the local linear approximation formula, we derived an easier form: \( 1 + \frac{1}{2}x \). This new expression makes it easy to calculate values of \( f(x) \) when \( x \) is near 0.

Derivatives

Derivatives tell us how a function changes as its input changes. They are essential in understanding the rate of change and the slope of the function at any given point.
To find the local linear approximation, knowing the derivative of the function at the given point is crucial. This derivative helps determine the slope of the tangent line, which forms the basis for the approximation.
In the problem, we computed the derivative of \( f(x) = \frac{1}{\sqrt{1-x}} \). We recognized it as \((1-x)^{-1/2} \) and used the power rule to find:

• The derivative is \( f'(x) = \frac{1}{2}(1-x)^{-3/2} \).
• This derivative measures how quickly the function changes as \( x \) changes.

Ultimately, we evaluated this derivative at \( x_0 = 0 \) to find \( f'(0) = \frac{1}{2} \). This value indicates the slope of the tangent line, which tells us the rate of change of \( f(x) \) near 0.

Chain Rule

The chain rule is a technique from calculus used for differentiating composite functions. It's like a "rule of thumb" for breaking down the process of finding derivatives of complex nested functions.
When we have a function inside another function, the chain rule steps in and helps us determine the overall derivative.
Here's what you need to know about how the chain rule was applied in the problem:

• We looked at \( f(x) = \frac{1}{\sqrt{1-x}} \) as a composite function: the outer function is \( u^{-1/2} \) and the inner function is \( u = 1-x \).
• The chain rule tells us to take the derivative of the outer function and multiply it by the derivative of the inner function.
• Using this method, we found \( f'(x) = \frac{1}{2}(1-x)^{-3/2} \) involves combining both derivatives correctly.

By using the chain rule effectively, we acquired the derivative needed for applying our local linear approximation, confirming our understanding of how these concepts are intertwined.