/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Find the linearization \(L(x)\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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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

State the formula for linearization

To find the linearization of a function at a point, we use the formula: \[L(x) = f(a) + f'(a)(x-a)\]where \(f(a)\) is the function evaluated at the point \(a\), and \(f'(a)\) is the derivative of the function evaluated at \(a\).
02

Evaluate the function at the given point

Substitute \(x = \pi\) into the function \(f(x) = \tan x\):\[f(\pi) = \tan \pi = 0\]Thus, \(f(\pi) = 0\).
03

Calculate the derivative of the function

The derivative of \(f(x) = \tan x\) is \(f'(x) = \sec^2 x\). Standard differentiation rules for trigonometric functions give this result.
04

Evaluate the derivative at the given point

Now, substitute \(x = \pi\) into the derivative:\[f'(\pi) = \sec^2 \pi = \frac{1}{\cos^2 \pi} = \frac{1}{(-1)^2} = 1\]Thus, \(f'(\pi) = 1\).
05

Construct the linearization function

Using the values calculated, substitute into the linearization formula:\[L(x) = f(\pi) + f'(\pi)(x-\pi) = 0 + 1\cdot(x-\pi) = x - \pi\]Thus, the linearization of \(f(x)\) at \(x = \pi\) is \(L(x) = x - \pi\).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which tells us how the function's output changes with respect to changes in its input. In simpler terms, it measures the rate at which a quantity changes. Differentiation is based on the concept of limits and results in a new function that gives the slope or tangent of the original function at any particular point.

To differentiate a function, we often follow established rules, such as the power rule, product rule, quotient rule, and chain rule. These rules simplify the process and help us determine derivatives efficiently. Once we have the derivative, it can be used for various purposes like finding the rate of change, solving optimization problems, or determining the behavior of physical systems.

In our original exercise, the derivative of the function \(f(x) = \tan x\) is \(f'(x) = \sec^2 x\). This trigonometric differentiation example showcases a standard result from calculus that helps us understand the tangent function's gradient changes.
Trigonometric Functions
Trigonometric functions are mathematical functions related to the angles and sides of triangles. There are six primary trigonometric functions: sine, cosine, tangent, cosecant, secant, and cotangent. Each function has unique properties and is used to solve various problems involving periodic phenomena in physics, engineering, and many other fields.

In the given exercise, the function we are working with is the tangent function, \(f(x) = \tan x\). Tangent is defined as the ratio of the sine of an angle to the cosine of that angle: \(\tan x = \frac{\sin x}{\cos x}\). At certain points, the tangent function has specific values, such as \(\tan(0) = 0\) and \(\tan(\pi) = 0\), which are important when evaluating or manipulating expressions.

Using trigonometric differentiation, such as determining the derivative \(f'(x) = \sec^2 x\), can yield insights into how these functions behave as they change. Understanding these properties is crucial for applications in wave motion, signal processing, and even in simple geometric transformations.
Derivative Evaluation
Once we have the derivative of a function, the next step is derivative evaluation at specific points. This evaluation helps us understand the function's behavior locally around those points, especially in terms of its increase or decrease.

In the original exercise, after finding the derivative \(f'(x) = \sec^2 x\), it's evaluated at the point \(x = \pi\) to get \(f'(\pi) = \sec^2 \pi = 1\). This process tells us the slope of the tangent line to the function at the particular point \(x = \pi\).

Evaluating derivatives is necessary for creating approximate linear models of functions at specific points, a process called linearization. The linearization provides a simple linear function that closely matches the behavior of the original function near that point, useful in making approximations and predictions. In our exercise, this leads to the linearization \(L(x) = x - \pi\), a linear expression that provides a convenient way to approximate \(\tan x\) around \(x = \pi\).

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Most popular questions from this chapter

What is special about the functions $$f(x)=\sin ^{-1} \frac{1}{\sqrt{x^{2}+1}} \text { and } g(x)=\tan ^{-1} \frac{1}{x} ?$$ Explain.

Find the derivative of \(y\) with respect to the given independent variable. $$y=\log _{2}\left(8 t^{\ln 2}\right)$$

Chain Rule Suppose that \(f(x)=x^{2}\) and \(g(x)=|x| .\) Then the compositions$$(f \circ g)(x)=|x|^{2}=x^{2} \quad \text { and } \quad(g \circ f)(x)=\left|x^{2}\right|=x^{2}$$ are both differentiable at \(x=0\) even though \(g\) itself is not differentiable at \(x=0 .\) Does this contradict the Chain Rule? Explain.

In Exercises \(69-74,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\). c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can that satisfies \(|x-a|<\delta \Rightarrow|f(x)-L(x)|<\varepsilon\) for \(\varepsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see whether your \(\delta\) -estimate holds true. $$f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2$$

Find the domain and range of each composite function. Then graph the composition of the two functions on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see. a. \(y=\cos ^{-1}(\cos x)\) b. \(y=\cos \left(\cos ^{-1} x\right)\)

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