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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.1.29
Chapter 7, Problem 7.1.29

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) = 1/x², x > 0

Verified step by step guidance
1
Start by writing the function given: \(f(x) = \frac{1}{x^{2}}\) with the domain \(x > 0\).
To find the inverse function \(f^{-1}(x)\), set \(y = \frac{1}{x^{2}}\) and then interchange \(x\) and \(y\) to get \(x = \frac{1}{y^{2}}\).
Solve the equation \(x = \frac{1}{y^{2}}\) for \(y\): multiply both sides by \(y^{2}\) to get \(x y^{2} = 1\), then divide by \(x\) to get \(y^{2} = \frac{1}{x}\).
Take the square root of both sides to solve for \(y\): \(y = \pm \sqrt{\frac{1}{x}} = \pm \frac{1}{\sqrt{x}}\). Since the original domain of \(f\) is \(x > 0\) and \(f(x) > 0\), choose the positive root to maintain the inverse function's domain and range consistency, so \(f^{-1}(x) = \frac{1}{\sqrt{x}}\).
Identify the domain and range of \(f^{-1}\): the domain of \(f^{-1}\) is the range of \(f\), which is \(y > 0\), so \(x > 0\) for \(f^{-1}\). The range of \(f^{-1}\) is the domain of \(f\), which is \(y > 0\). Finally, verify the inverse by checking that \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).

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Key Concepts

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

Inverse Functions

An inverse function reverses the effect of the original function, swapping inputs and outputs. For a function f(x), its inverse f⁻¹(x) satisfies f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. Finding the inverse involves solving y = f(x) for x in terms of y.
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Domain and Range of Functions and Their Inverses

The domain of a function is the set of all possible input values, while the range is the set of possible outputs. For an inverse function, the domain and range swap roles compared to the original function. Identifying these sets ensures the inverse is well-defined.
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Restrictions on the Domain for Invertibility

A function must be one-to-one (injective) to have an inverse. Restricting the domain, such as x > 0 for f(x) = 1/x², ensures the function is one-to-one and invertible. Without this, the inverse would not be a proper function.
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Related Practice
Textbook Question

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)

Textbook Question

Evaluate the integrals in Exercises 97–110.

109. ∫ (dx / (x log₁₀x))

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Textbook Question

Evaluate the integrals in Exercises 39–56.

41. ∫2y dy/(y²-25)

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Textbook Question

In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.

y = ln(e^(θ)/(1+e^θ))

Textbook Question

Evaluate the integrals in Exercises 53–76.

69. ∫dx/((2x-1)√((2x-1)²-4))

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Textbook Question

In Exercises 25–36, find the derivative of y with respect to the appropriate variable.

31. y = cos⁻¹(x) - x sech⁻¹(x)