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Ch. 1 - Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 1, Problem 1.89

Finding all inverses Find all the inverses associated with the following functions, and state their domains.


ƒ(x) = 2 / ( x² + 2)

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1
The given function is \( f(x) = \frac{2}{x^2 + 2} \). This is a rational function where the numerator is a constant and the denominator is a quadratic expression.
To find the inverse, we need to solve for \( x \) in terms of \( y \). Start by setting \( y = \frac{2}{x^2 + 2} \).
Rearrange the equation to express \( x^2 + 2 \) in terms of \( y \): \( y(x^2 + 2) = 2 \). Then, solve for \( x^2 \): \( x^2 = \frac{2}{y} - 2 \).
Take the square root of both sides to solve for \( x \): \( x = \pm \sqrt{\frac{2}{y} - 2} \).
The expression \( \frac{2}{y} - 2 \) must be non-negative for the square root to be defined, so \( \frac{2}{y} \geq 2 \), which implies \( y \leq 1 \). Therefore, the domain of the inverse function is \( y \in (-\infty, 1] \).

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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 essentially reverses the effect of the original function. If a function f takes an input x and produces an output y, the inverse function f⁻¹ takes y back to x. For a function to have an inverse, it must be one-to-one, meaning each output is produced by exactly one input.
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Domain and Range

The domain of a function is the set of all possible input values (x-values) that the function can accept, while the range is the set of all possible output values (y-values) that the function can produce. When finding the inverse of a function, the domain of the original function becomes the range of the inverse, and vice versa.
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Finding Inverses Algebraically

To find the inverse of a function algebraically, you typically start by replacing f(x) with y, then swap x and y in the equation. After that, you solve for y to express it in terms of x. This new expression represents the inverse function, and it is important to also determine its domain based on the original function.
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