Textbook Question
Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.
f(x) = x³ + 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 7.5.30
Use l’Hôpital’s rule to find the limits in Exercises 7–52.
30. lim (θ → 0) ((1/2)^θ - 1) / θ
Verified step by step guidance1
Identify the limit expression: \(\lim_{\theta \to 0} \frac{(\frac{1}{2})^{\theta} - 1}{\theta}\).
Check the form of the limit by substituting \(\theta = 0\): the numerator becomes \((\frac{1}{2})^0 - 1 = 1 - 1 = 0\) and the denominator is \(0\), so the limit is of the indeterminate form \(\frac{0}{0}\).
Since the limit is of the form \(\frac{0}{0}\), apply l'Hôpital's Rule, which states that \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\) if the original limit is indeterminate.
Differentiate the numerator and denominator separately with respect to \(\theta\): the derivative of the numerator \(f(\theta) = (\frac{1}{2})^{\theta} - 1\) is \(f'(\theta) = (\frac{1}{2})^{\theta} \ln(\frac{1}{2})\), and the derivative of the denominator \(g(\theta) = \theta\) is \(g'(\theta) = 1\).
Rewrite the limit using these derivatives: \(\lim_{\theta \to 0} \frac{(\frac{1}{2})^{\theta} \ln(\frac{1}{2})}{1}\), then evaluate this limit by substituting \(\theta = 0\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits and Indeterminate Forms
Limits describe the behavior of a function as the input approaches a certain value. When direct substitution results in forms like 0/0 or ∞/∞, these are called indeterminate forms, requiring special techniques such as l’Hôpital’s rule to evaluate the limit.
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One-Sided Limits
l’Hôpital’s Rule
l’Hôpital’s rule is a method for evaluating limits that yield indeterminate forms 0/0 or ∞/∞ by differentiating the numerator and denominator separately and then taking the limit of their quotient. It simplifies complex limits involving functions that are difficult to evaluate directly.
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Exponential and Logarithmic Functions
Understanding the behavior and derivatives of exponential functions, such as (1/2)^θ, is essential. These functions can be rewritten using logarithms to facilitate differentiation, which is crucial when applying l’Hôpital’s rule to limits involving exponential expressions.
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Related Practice
Textbook Question
Use the results of Exercise 55 to show that the functions in Exercises 56–60 have inverses over their domains. Find a formula for df⁻¹/dx using Theorem 1.
f(x) = (1 − x)³
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Evaluate the integrals in Exercises 53–76.
53. ∫dx/√(9-x²)
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Evaluate the integrals in Exercises 97–110.
103. ∫₁⁴ (ln 2 · log₂x / x) dx
1
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Textbook Question
Evaluate the integrals in Exercises 33–54.
49. ∫ e^(sec πt) sec πt tan πt dt
Textbook Question
130. Use the identity arccot(u)=π/2 - arctan(u) to derive the formula for the derivative of arccot(u) in Table 7.4 from the formula for the derivative of arctan(u).