Textbook Question
Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
1. a. arctan 1
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.3.145a
145. The linearization of eˣ at x = 0
a. Derive the linear approximation eˣ ≈ 1 + x at x = 0.
Verified step by step guidance1
Recall that the linearization (or linear approximation) of a function \( f(x) \) at a point \( a \) is given by the formula:
\[ L(x) = f(a) + f'(a)(x - a) \]
This represents the equation of the tangent line to the curve at \( x = a \).
Identify the function and the point of linearization: here, \( f(x) = e^x \) and \( a = 0 \).
Calculate \( f(a) = f(0) = e^0 \). Since \( e^0 = 1 \), this gives the value of the function at the point of linearization.
Find the derivative of the function: \( f'(x) = \frac{d}{dx} e^x = e^x \). Then evaluate it at \( x = 0 \), so \( f'(0) = e^0 = 1 \).
Substitute these values into the linearization formula:
\[ L(x) = f(0) + f'(0)(x - 0) = 1 + 1 \cdot x = 1 + x \]
This is the linear approximation of \( e^x \) near \( x = 0 \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Linearization of a Function
Linearization approximates a function near a point using the tangent line at that point. It replaces the function with a linear expression that is easier to work with, typically written as L(x) = f(a) + f'(a)(x - a), where a is the point of approximation.
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Linearization
Derivative of eˣ
The derivative of the exponential function eˣ is eˣ itself. At x = 0, this derivative equals 1, which is essential for finding the slope of the tangent line used in the linear approximation.
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Evaluating Functions and Derivatives at a Point
To find the linear approximation at x = 0, you must evaluate both the function and its derivative at that point. For eˣ, e⁰ = 1 and the derivative at 0 is also 1, leading to the linearization formula eˣ ≈ 1 + x.
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Related Practice
Textbook Question
In Exercises 1–4, solve for t.
1. a. e^(-0.3t) = 27
Textbook Question
153. The linearization of 2ˣ
a. Find the linearization of f(x) = 2ˣ at x = 0. Then round its coefficients to two decimal places.
Textbook Question
75. a. Find the open intervals on which the function is increasing and decreasing.
g(x) = x(ln x)²
Textbook Question
Evaluate the integrals in Exercises 67–74 in terms of
a. inverse hyperbolic functions.
73. ∫(from 0 to π)cos(x)dx/√(1+sin²x)
Textbook Question
[Technology Exercise] In Exercises 139–141, find the domain and range of each composite function. Then graph the compositions on separate screens. Do the graphs make sense in each case? Give reasons for your answers. Comment on any differences you see.
139. a. y=arctan(tan x)