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Chapter 3: Problem 5

Find the linearization \(L(x)\) of \(f(x)\) at \(x=a\). $$f(x)=\tan x, \quad a=\pi$$

Short Answer

Expert verified
The linearization of \(f(x) = \tan x\) at \(x = \pi\) is \(L(x) = x - \pi\).

Step by step solution

01

Define Linearization Formula

The formula for the linearization of a function \(f(x)\) at \(x=a\) is given by \(L(x) = f(a) + f'(a)(x-a)\). We will use this to find the linearization of \(f(x) = \tan x\) at \(x = \pi\).
02

Evaluate \(f(a)\)

Substitute \(a = \pi\) into the function \(f(x) = \tan x\), so \(f(\pi) = \tan \pi = 0\).
03

Compute \(f'(x)\)

The derivative of \(f(x) = \tan x\) is \(f'(x) = \sec^2 x\).
04

Evaluate \(f'(a)\)

Substitute \(a = \pi\) into the derivative, so \(f'(\pi) = \sec^2 \pi = 1\) because \(\sec \pi = -1\).
05

Plug Values into Linearization Formula

Using the linearization formula \(L(x) = f(\pi) + f'(\pi)(x-\pi)\), substitute the values found: \(L(x) = 0 + 1(x - \pi)\). This simplifies to \(L(x) = x - \pi\).

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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, denoted as \(\tan(x)\), is one of the basic trigonometric functions. It is deeply connected to the structure of right triangles and circles. In a right triangle, the tangent of an angle \(\theta\) is the ratio of the opposite side to the adjacent side.
  • Mathematically, \(\tan(x) = \frac{\sin(x)}{\cos(x)}\).
  • The function is periodic with a period of \(\pi\), meaning it repeats every \(\pi\) units.
  • \(\tan(x)\) is undefined where \(\cos(x) = 0\), such as at \(x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots\)
Understanding the tangent function's properties helps us analyze its behavior around different points, which is crucial when considering linearization at a specific point like \(x = \pi\).
Derivative
The derivative of a function measures how the function changes as its input changes. In simpler terms, it tells us the slope of the tangent line to the curve of the function at any given point. For the tangent function \(f(x) = \tan(x)\), its derivative is \(f'(x) = \sec^2(x)\).
  • \(f'(x)\) is always positive where it is defined, indicating that the tangent function is increasing wherever it is defined.
  • The derivative helps us understand how sharply the function is changing at any given point.
At \(x = \pi\), \(\sec(\pi) = -1\), which means \(\sec^2(\pi) = 1\). Thus, the derivative \(f'(\pi) = 1\). This indicates the rate at which \(\tan(x)\) is changing at \(\pi\) is perfectly linear, aligned with its graph.
Linear Approximation
Linear approximation, or linearization, is a technique where we approximate a function by a line near a specific point \(x = a\). The formula for this approximation is given by:\[L(x) = f(a) + f'(a)(x-a)\]This expression forms the equation of the tangent line, giving the best linear representation of the original function around some point.
  • The process involves finding the function's value and derivative at that specific point.
  • At \(x = \pi\), for \(f(x) = \tan(x)\), we have \(f(\pi) = 0\) and \(f'(\pi) = 1\).
Substituting into the linearization formula, we get \(L(x) = 0 + 1 \times (x - \pi)\), simplifying to \(L(x) = x - \pi\). This linearization represents how \(\tan(x)\) behaves around \(x = \pi\) and can be useful for approximations in calculus and real-world applications.

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Most popular questions from this chapter

a. Find the linearization of \(f(x)=\log _{3} x\) at \(x=3 .\) Then round its coefficients to two decimal places. b. Graph the linearization and function together in the window \(0 \leq x \leq 8\) and \(2 \leq x \leq 4\)

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=e^{(\cos t+\ln t)}$$

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\sqrt[3]{\frac{x(x-2)}{x^{2}+1}}$$

Find \(d y / d x\). $$\ln x y=e^{x+y}$$

If \(f(x)=x^{n}, n \geq 1,\) show from the definition of the derivative that \(f^{\prime}(0)=0\).
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