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Chapter 3: Problem 4

Find the linearization of \(f(x)\) at \(x=a\). \(f(x)=\frac{1}{x^{2}}, a=0.5\)

Short Answer

Expert verified
The linearization of \( f(x) = \frac{1}{x^2} \) at \( x = 0.5 \) is \( L(x) = 12 - 16x \).

Step by step solution

01

- Understand Linearization

The linearization of a function at a given point is the first-order Taylor expansion around that point. It can be given as: \[ L(x) = f(a) + f'(a)(x - a) \]
02

- Evaluate the Function at x=a

Compute the value of the function at \(x = a\). \[ f(a) = f(0.5) = \frac{1}{(0.5)^2} = \frac{1}{0.25} = 4 \]
03

- Find the Derivative of the Function

Compute the derivative of the function \(f(x) = \frac{1}{x^2}\). Using the power rule: \[ f'(x) = -2x^{-3} = -\frac{2}{x^3} \]
04

- Evaluate the Derivative at x=a

Compute the value of the derivative at \(x = a\). \[ f'(0.5) = -\frac{2}{(0.5)^3} = -\frac{2}{0.125} = -16 \]
05

- Write the Linearization Formula

Insert the computed values into the linearization formula: \[ L(x) = f(a) + f'(a)(x - a) = 4 + (-16)(x - 0.5) \]
06

- Simplify the Expression

Simplify the linearization formula expression: \[ L(x) = 4 - 16(x - 0.5) = 4 - 16x + 8 = 12 - 16x \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor expansion
The Taylor expansion is a mathematical tool that approximates a function around a certain point using polynomials. The first-order Taylor expansion of a function, also known as linearization, helps us approximate the function with a straight line.
The general formula for a Taylor expansion around a point \(a\) is:
\[ f(x) \approx f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \ldots + \frac{f^n(a)}{n!}(x - a)^n \]
For linearization, we focus only on the first-order terms:
\[ L(x) = f(a) + f'(a)(x - a) \]
This essentially means we only use the function value and its first derivative at point \(a\) to approximate the function.
Function derivatives
To find the linear approximation of a function, we need to calculate its derivative. The derivative measures how a function changes as its input changes. It is fundamental in understanding the behavior of functions.
Given \( f(x) = \frac{1}{x^2} \), we find the derivative using rules like the power rule:
\[ f'(x) = -2x^{-3} = -\frac{2}{x^3} \]
The derivative, when evaluated at our point of interest \(a = 0.5\), is:
\[ f'(0.5) = -\frac{2}{(0.5)^3} = -\frac{2}{0.125} = -16 \]
These calculations are crucial for creating the linear approximation.
Power rule
The power rule is one of the basic rules for differentiation in calculus. It states that for a function of the form \( f(x) = x^n \), its derivative is \( f'(x) = nx^{n-1} \).
This rule helps simplify the differentiation process. In our problem, we have \( f(x) = \frac{1}{x^2} = x^{-2} \). Applying the power rule:
\[ f'(x) = -2x^{-3} \]
This simple rule allows us to quickly find the derivative, which is a key step in linearizing functions.
Knowing how to use the power rule effectively can greatly simplify many calculus problems you encounter.

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Most popular questions from this chapter

Differentiate implicity to find \(d y / d x\) and \(d^{2} y / d x^{2}\). $$ y^{2}-x y+x^{2}=5 $$

Differentiate implicily to find \(d y / d x\). $$ y^{5}=x^{3} $$

Graph the function. $$ f(x)=\left|\frac{1}{x}-2\right| $$

Use the indicated choice of \(x_{1}\) and Newton's method to solve the given equation. \(2 x-\sin x=\cos \left(x^{2}\right) ; x_{1}=\pi / 4\)

Then graph the tangent line to the graph at the point \((-0.8,0.384)\). $$ x^{3}=y^{2}(2-x) $$
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