Chapter 3: Problem 8
Confirm that the stated formula is the local linear approximation at \(x_{0}=0 .\). $$ \frac{1}{1+x} \approx 1-x $$
Short Answer
Expert verified
The local linear approximation formula \( 1-x \) is confirmed correct at \( x_0 = 0 \).
Step by step solution
01
Identify the Function and Point of Approximation
The given function is \( f(x) = \frac{1}{1+x} \) and we are tasked to find its local linear approximation at \( x_0 = 0 \). The local linear approximation formula is \( f(x) \approx f(x_0) + f'(x_0)(x - x_0) \).
02
Evaluate the Function at the Point
Calculate \( f(x_0) = f(0) \). Since \( f(x) = \frac{1}{1+x} \), we have \( f(0) = \frac{1}{1+0} = 1 \).
03
Compute the Derivative
Find the first derivative \( f'(x) \) of \( f(x) = \frac{1}{1+x} \). This can be done using the power rule or quotient rule: \[ f'(x) = -\frac{1}{(1+x)^2} \]
04
Evaluate the Derivative at the Point
Calculate \( f'(x_0) = f'(0) \). Substituting \( x = 0 \) into the expression for the derivative, we get: \[ f'(0) = -\frac{1}{(1+0)^2} = -1 \]
05
Write the Linear Approximation Formula
Use the local linear approximation formula: \[ f(x) \approx f(0) + f'(0)(x - 0) \] Substitute the evaluated values: \[ f(x) \approx 1 + (-1)(x) = 1 - x \]
06
Confirm the Given Formula
The derived linear approximation \( 1 - x \) matches the given formula \( 1 - x \), confirming the approximation as accurate at \( x_0 = 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative Calculation
Understanding derivatives is essential in finding the local linear approximation of a function. A derivative represents how a function changes as its input changes. It gives us the slope of the function at a particular point.
For the function defined as \( f(x) = \frac{1}{1+x} \), we need to find the first derivative to calculate its slope at the point \( x_0 = 0 \). The derivative of \( f(x) \) is computed using the quotient rule which is suitable for functions of the form \( \frac{g(x)}{h(x)} \). Alternatively, you might use the power rule, but for this function, the quotient rule often makes more sense as it directly corresponds to \( f(x) = (1+x)^{-1} \).
Calculating the derivative, we obtain: \[ f'(x) = -\frac{1}{(1+x)^2} \]. This gives us an expression that tells how rapidly the function \( f(x) \) changes at any point \( x \). Evaluating this at \( x_0 = 0 \), \( f'(0) = -1 \), tells us that at this point, the function has a slope or rate of change of -1.
For the function defined as \( f(x) = \frac{1}{1+x} \), we need to find the first derivative to calculate its slope at the point \( x_0 = 0 \). The derivative of \( f(x) \) is computed using the quotient rule which is suitable for functions of the form \( \frac{g(x)}{h(x)} \). Alternatively, you might use the power rule, but for this function, the quotient rule often makes more sense as it directly corresponds to \( f(x) = (1+x)^{-1} \).
Calculating the derivative, we obtain: \[ f'(x) = -\frac{1}{(1+x)^2} \]. This gives us an expression that tells how rapidly the function \( f(x) \) changes at any point \( x \). Evaluating this at \( x_0 = 0 \), \( f'(0) = -1 \), tells us that at this point, the function has a slope or rate of change of -1.
Function Evaluation
Function evaluation is about determining the exact output of a function at a specific point. This is crucial in approximating functionalities since the first step involves establishing a baseline or initial value.
For our function \( f(x) = \frac{1}{1+x} \), we evaluate it at \( x_0 = 0 \) to find the point where our linear approximation is based. Plugging this value into the function, we get \( f(0) = \frac{1}{1+0} = 1 \). This means the function's initial output value is 1 at the starting point \( x_0 = 0 \).
Thus, in our local linear approximation formula, this initial value \( f(0) = 1 \) forms the constant part of the linearization, which will be adjusted according to the function's slope at this point.
For our function \( f(x) = \frac{1}{1+x} \), we evaluate it at \( x_0 = 0 \) to find the point where our linear approximation is based. Plugging this value into the function, we get \( f(0) = \frac{1}{1+0} = 1 \). This means the function's initial output value is 1 at the starting point \( x_0 = 0 \).
Thus, in our local linear approximation formula, this initial value \( f(0) = 1 \) forms the constant part of the linearization, which will be adjusted according to the function's slope at this point.
Linearization
Linearization is the process used to approximate a complex function with a simple linear function around a specific point. This is extremely useful for making calculations and predictions simpler and faster.
The local linear approximation formula is \( f(x) \approx f(x_0) + f'(x_0)(x - x_0) \). With the values calculated earlier, we substitute them into this formula:
Thus, the linear function \( 1-x \) approximates \( \frac{1}{1+x} \) around \( x_0=0 \). This simplification is essential in applications where a linear estimate can provide a close enough answer to complex equations near a chosen point.
The local linear approximation formula is \( f(x) \approx f(x_0) + f'(x_0)(x - x_0) \). With the values calculated earlier, we substitute them into this formula:
- The evaluated function value at the point \( x_0 \) is \( f(0)=1 \).
- The calculated derivative at this point is \( f'(0)=-1 \).
Thus, the linear function \( 1-x \) approximates \( \frac{1}{1+x} \) around \( x_0=0 \). This simplification is essential in applications where a linear estimate can provide a close enough answer to complex equations near a chosen point.