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Chapter 4: Problem 52

Find the linear approximation \(L(x)\) to \(y=f(x)\) near \(x=a\) for the function. \(f(x)=\tan x, a=\frac{\pi}{4}\)

Short Answer

Expert verified
The linear approximation is \( L(x) = 2x + 1 - \frac{\pi}{2} \) near \( x = \frac{\pi}{4} \).

Step by step solution

01

Find the Value of the Function at the Point

To find the linear approximation of the function near a point, we first need to evaluate the function at that point. Here, the function is \( f(x) = \tan x \) and we need to evaluate it at \( x = \frac{\pi}{4} \). Since \( \tan \left( \frac{\pi}{4} \right) = 1 \), we find that \( f \left( \frac{\pi}{4} \right) = 1 \).
02

Calculate the Derivative of the Function

The next step is to find the derivative of the function, \( f'(x) \). For \( f(x) = \tan x \), the derivative \( f'(x) = \sec^2 x \).
03

Evaluate the Derivative at the Given Point

Now, we need to evaluate the derivative at the point \( x = \frac{\pi}{4} \). We know that \( \sec \left( \frac{\pi}{4} \right) = \sqrt{2} \), so \( \sec^2 \left( \frac{\pi}{4} \right) = 2 \). Thus, \( f' \left( \frac{\pi}{4} \right) = 2 \).
04

Formulate the Linear Approximation Formula

The linear approximation, \( L(x) \), is given by the formula \( L(x) = f(a) + f'(a)(x - a) \). Plugging in our values: \( a = \frac{\pi}{4} \), \( f(a) = 1 \), and \( f'(a) = 2 \), we have \( L(x) = 1 + 2(x - \frac{\pi}{4}) \).
05

Simplify the Linear Approximation Expression

Simplify the linear approximation expression: \( L(x) = 1 + 2x - \frac{\pi}{2} \). Therefore, the linear approximation near \( x = \frac{\pi}{4} \) is \( L(x) = 2x + 1 - \frac{\pi}{2} \).

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Key Concepts

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

Tangent Function
The tangent function, often represented as \( \tan x \), is a periodic function vital in trigonometry and calculus. This function takes an angle as input and returns the ratio of the sine and cosine of that angle, \( \tan x = \frac{\sin x}{\cos x} \). It is important to note:
  • The tangent function has a period of \( \pi \). This means that every \( \pi \) units along the \( x \)-axis, the function repeats its values.
  • The function is undefined at odd multiples of \( \frac{\pi}{2} \), as the cosine in the denominator equals zero at these points.
  • Tangent is often used in various fields such as engineering, physics, and computer science for modeling cyclical phenomena.
In this exercise, we are asked to find the linear approximation of \( \tan x \) near \( x = \frac{\pi}{4} \). Understanding the behavior of the tangent function around this point aids in calculating small changes in function values effectively.
Derivative Evaluation
Derivative evaluation is a fundamental concept in calculus that involves determining the instantaneous rate of change of a function with respect to one of its variables. Mathematically, if \( f(x) \) is our function, the derivative, \( f'(x) \), measures how \( f(x) \) changes as \( x \) changes.
  • For the tangent function, \( f(x) = \tan x \), its derivative is \( f'(x) = \sec^2 x \), a significant result in calculus.
  • The symbol \( \sec x \) represents the secant function, given by \( \sec x = \frac{1}{\cos x} \).
In this exercise, evaluating \( f'(x) \) at \( x = \frac{\pi}{4} \) gives \( \sec^2 \left( \frac{\pi}{4} \right) = 2 \), as \( \sec \left( \frac{\pi}{4} \right) = \sqrt{2} \). Knowing how to compute derivatives correctly provides a deeper insight into function behavior and helps form the basis for linear approximations.
Linearization Formula
The linearization formula is a powerful tool in calculus used to approximate the value of a function near a given point. It is derived from the tangent line equation and provides an easier way to estimate function values. The general linearization formula is given by:
  • \( L(x) = f(a) + f'(a)(x-a) \)
where:
  • \( f(a) \) is the function's value at the point \( a \).
  • \( f'(a) \) is the derivative at \( a \).
  • \( x-a \) represents a small change from the point \( a \).
In this exercise, for the function \( f(x) = \tan x \) at \( x = \frac{\pi}{4}\), we use \( f(a) = 1 \) and \( f'(a) = 2 \) to find \( L(x) = 1 + 2(x - \frac{\pi}{4}) \). Linearization simplifies complex functions into manageable linear forms, aiding computation and understanding.
Calculus
Calculus is a core area of mathematics centered around change and motion. It encompasses two main branches: differential calculus and integral calculus. Each provides unique methods for analyzing and understanding various phenomena.
  • Differential calculus is the part dealing with the concept of a derivative. It focuses on understanding and calculating the rate of change of quantities.
  • Integral calculus focuses on accumulation and areas under curves, putting together pieces to form a complete picture.
The problem addressed here involves differential calculus, particularly the use of derivatives to find the linear approximation of a function, in this case, \( \tan x \). By leveraging calculus concepts, students can solve complex real-world problems and understand the principles governing the world around them.

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