Problem 6
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
Verified Answer
The formula is confirmed as the local linear approximation at \( x_0 = 0 \).
1Step 1: Identify the Function and the Point
The function we need to consider is \( f(x) = \frac{1}{\sqrt{1-x}} \) and the point at which we want to confirm the local linear approximation is \( x_0 = 0 \).
2Step 2: Calculate the Derivative at the Point
The first step in finding the linear approximation is to find the derivative of the function. The derivative is \( f'(x) = \frac{1}{2}(1-x)^{-3/2} \). Evaluate this derivative at the point \( x_0 = 0 \): \[ f'(0) = \frac{1}{2}(1-0)^{-3/2} = \frac{1}{2} \]
3Step 3: Write the Linear Approximation Formula
The linear approximation of a function \( f(x) \) at a point \( x_0 \) is given by:\[ f(x) \approx f(x_0) + f'(x_0)(x - x_0) \]Substitute \( x_0 = 0 \), \( f(x_0) = 1 \), and \( f'(x_0) = \frac{1}{2} \) into the formula:\[ f(x) \approx 1 + \frac{1}{2}(x - 0) = 1 + \frac{1}{2}x \]
4Step 4: Confirm the Given Formula
The given formula for the local linear approximation at \( x_0 = 0 \) is \( \frac{1}{\sqrt{1-x}} \approx 1 + \frac{1}{2}x \). Since this matches the formula we derived using the linear approximation method, the given formula is indeed the local linear approximation at \( x_0 = 0 \).
Key Concepts
Derivative CalculationLinear Approximation FormulaFunction Evaluation
Derivative Calculation
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:
- \( f'(x) = \frac{1}{2}(1-x)^{-3/2} \)
- \( f'(0) = \frac{1}{2} \)
Linear Approximation Formula
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:
- \( f(x) \approx f(x_0) + f'(x_0)(x - x_0) \)
- 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 \( f(x) \approx 1 + \frac{1}{2}x \).
Function Evaluation
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 \( f(0) = \frac{1}{\sqrt{1-0}} = 1 \).
Other exercises in this chapter
Problem 6
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Find \(d y / d x\). $$y=\ln \left|x^{3}-7 x^{2}-3\right|$$
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