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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.1.25

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) = x⁵

Guida verificata passo dopo passo
1
Start with the given function: \(f(x) = x^5\). Our goal is to find the inverse function \(f^{-1}(x)\), which means we want to express \(x\) in terms of \(y\) where \(y = f(x)\).
Replace \(f(x)\) with \(y\): \(y = x^5\). To find the inverse, solve this equation for \(x\) in terms of \(y\).
Take the fifth root of both sides to isolate \(x\): \(x = \sqrt[5]{y} = y^{1/5}\). This gives the inverse function: \(f^{-1}(x) = x^{1/5}\).
Determine the domain and range of \(f^{-1}(x)\). Since \(f(x) = x^5\) is defined for all real numbers and is one-to-one, its inverse will also have domain and range as all real numbers.
Verify the inverse by checking the compositions: compute \(f(f^{-1}(x)) = (x^{1/5})^5\) and \(f^{-1}(f(x)) = (x^5)^{1/5}\), and confirm both simplify to \(x\).

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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 inputs, while the range is the set of all possible outputs. For inverse functions, the domain and range swap roles: the domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f.
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Power functions like f(x) = x⁵ are continuous and one-to-one over all real numbers, making them invertible. Since x⁵ is strictly increasing, its inverse is the fifth root function, which also has domain and range as all real numbers.
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