Textbook Question
Evaluate the integrals in Exercises 67–74 in terms of
a. inverse hyperbolic functions.
67. ∫(from 0 to 2√3)dx/√(4+x²)
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 7.2.75a
75. a. Find the open intervals on which the function is increasing and decreasing.
g(x) = x(ln x)²
Verified step by step guidance1
Identify the domain of the function \(g(x) = x(\ln x)^2\). Since \(\ln x\) is defined only for \(x > 0\), the domain is \((0, \infty)\).
Find the first derivative \(g'(x)\) using the product rule. Let \(u = x\) and \(v = (\ln x)^2\). Then, \(g'(x) = u'v + uv'\). Compute \(u' = 1\) and \(v' = 2 \ln x \cdot \frac{1}{x} = \frac{2 \ln x}{x}\).
Substitute these into the derivative: \(g'(x) = 1 \cdot (\ln x)^2 + x \cdot \frac{2 \ln x}{x} = (\ln x)^2 + 2 \ln x\).
Factor the derivative to simplify: \(g'(x) = \ln x \left( \ln x + 2 \right)\).
Determine the critical points by setting \(g'(x) = 0\), which gives \(\ln x = 0\) or \(\ln x = -2\). Solve these to find \(x = 1\) and \(x = e^{-2}\). Use these points to test intervals in the domain \((0, \infty)\) to find where \(g'(x)\) is positive (increasing) or negative (decreasing).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Domain of the Function
Before analyzing increasing or decreasing behavior, determine the domain where the function is defined. For g(x) = x(ln x)², the natural logarithm ln x requires x > 0, so the domain is (0, ∞). Understanding the domain ensures correct interval analysis.
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Finding the Domain and Range of a Graph
First Derivative and Critical Points
The first derivative g'(x) indicates the slope of the function. Finding g'(x) and solving g'(x) = 0 identifies critical points where the function may change from increasing to decreasing or vice versa. These points help partition the domain into intervals for testing.
Recommended video:
04:50Critical Points
Increasing and Decreasing Intervals
A function is increasing where its derivative is positive and decreasing where its derivative is negative. By testing values in intervals determined by critical points, one can classify each interval as increasing or decreasing, providing a complete description of the function's behavior.
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07:32Determining Where a Function is Increasing & Decreasing
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
145. The linearization of eˣ at x = 0
a. Derive the linear approximation eˣ ≈ 1 + x at x = 0.
Textbook Question
78. Which one is correct, and which one is wrong? Give reasons for your answers.
a. lim (x → 0) (x² - 2x) / (x² - sin x) = lim (x → 0) (2x - 2) / (2x - cos x) = lim (x → 0) 2 / (2 + sin x) = 2 / (2 + 0) = 1
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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)