Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.7.63

Chapter 7, Problem 7.7.63

Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms, as shown in the following table.
sinh⁻¹x = ln(x + √(x² + 1)), -∞ < x < ∞
cosh⁻¹x = ln(x + √(x² - 1)), x ≥ 1
tanh⁻¹x = (1/2)ln((1+x)/(1-x)), |x| < 1
sech⁻¹x = ln((1+√(1-x²))/x), 0 < x ≤ 1
csch⁻¹x = ln(1/x + √(1+x²)/|x|), x ≠ 1
coth⁻¹x = (1/2)ln((x+1)/(x-1)), |x| > 1
Use these formulas to express the numbers in Exercises 61–66 in terms of natural logarithms.
63. tanh⁻¹(-1/2)

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Identify the formula for the inverse hyperbolic tangent function given in the problem: \(\tanh^{-1} x = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right)\), valid for \(|x| < 1\).

Substitute the given value \(x = -\frac{1}{2}\) into the formula: \(\tanh^{-1} \left(-\frac{1}{2}\right) = \frac{1}{2} \ln \left( \frac{1 + (-\frac{1}{2})}{1 - (-\frac{1}{2})} \right)\).

Simplify the numerator and denominator inside the logarithm: numerator becomes \(1 - \frac{1}{2}\) and denominator becomes \(1 + \frac{1}{2}\).

Rewrite the expression inside the logarithm as a single fraction: \(\frac{1 - \frac{1}{2}}{1 + \frac{1}{2}}\).

Express the entire result as \(\tanh^{-1} \left(-\frac{1}{2}\right) = \frac{1}{2} \ln \left( \text{simplified fraction} \right)\), completing the expression in terms of natural logarithms.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, such as sinh⁻¹, cosh⁻¹, and tanh⁻¹, are the inverses of hyperbolic functions. They return the value whose hyperbolic function equals a given number. Understanding their domain and range is essential for correctly applying their logarithmic expressions.

Logarithmic Expressions of Inverse Hyperbolic Functions

Inverse hyperbolic functions can be expressed using natural logarithms, which simplifies evaluation and analysis. For example, tanh⁻¹x = (1/2)ln((1+x)/(1−x)) for |x| < 1. This relationship leverages properties of exponentials and logarithms to rewrite inverse hyperbolic functions in a more manageable form.

Domain Restrictions and Validity

Each inverse hyperbolic function has specific domain restrictions to ensure the logarithmic expressions are valid and real-valued. For tanh⁻¹x, the domain is |x| < 1, which must be checked before substitution to avoid undefined or complex results.

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