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In Exercises \(11-20,\) for the given functions \(f\) and \(g,\) find the composite functions \(f \circ g\) and \(g \circ f,\) and state their domains. $$\begin{aligned} &f(x)=\sqrt[3]{x^{2}-1}\\\ &g(x)=x^{2 / 3}+1 \end{aligned}$$

Short Answer

Expert verified
The composite functions are \( f \circ g(x) = \sqrt[3]{x^{4/3} + 2x^{2/3}} \) and \( g \circ f(x) = (x^2 - 1)^{2/9} + 1 \), both with domain \( x \in \mathbb{R} \).

Step by step solution

01

Understand the functions

We are given two functions: \( f(x) = \sqrt[3]{x^2 - 1} \) and \( g(x) = x^{2/3} + 1 \). Our task is to find the compositions \( f \circ g \) and \( g \circ f \) and determine their domains.
02

Find \( f \circ g(x) \)

To find \( f \circ g(x) \), we substitute \( g(x) \) into \( f(x) \). So, \( f(g(x)) = f\left(x^{2/3} + 1\right) = \sqrt[3]{(x^{2/3} + 1)^2 - 1} \). Simplifying, we have \( f(g(x)) = \sqrt[3]{x^{4/3} + 2x^{2/3}} \).
03

Determine the domain of \( f \circ g \)

The domain of \( f \circ g \) requires that \( x^{4/3} + 2x^{2/3} \) is a valid input for the cube root function, which allows any real number. Since there is no restriction from the function itself, \( f \circ g \) is defined for all real numbers, \( x \in \mathbb{R} \).
04

Find \( g \circ f(x) \)

To find \( g \circ f(x) \), substitute \( f(x) \) into \( g(x) \). So, \( g(f(x)) = g(\sqrt[3]{x^2 - 1}) = \left(\sqrt[3]{x^2 - 1}\right)^{2/3} + 1 = (x^2 - 1)^{2/9} + 1 \).
05

Determine the domain of \( g \circ f \)

For \( g(f(x)) = (x^2 - 1)^{2/9} + 1 \) to be defined, \( x^2 - 1 \) must be real, which is always true, so there are no restrictions here. The domain of \( g \circ f \) is all real numbers, \( x \in \mathbb{R} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domains of Functions
When evaluating functions, it's important to understand where they "live" or in simpler terms, what values they can accept. The set of all input values that a function can handle without causing any mathematical mishaps is called the domain. Consider the functions: - For a polynomial like \(g(x) = x^{2/3} + 1\), it's generally defined for all real numbers, as raising numbers to a fraction typically has no issues. - The function \(f(x) = \sqrt[3]{x^2 - 1}\) is also well-defined for any real number because the cube root can work with positives, negatives, and zero. Thus, both have domains of all real numbers, denoted as \(x \in \mathbb{R}\).In the world of composite functions, finding the domains means ensuring that the output of the first function fits within the input range of the second function. Thankfully, since both functions are generous with their domains, the composite functions \(f \circ g\) and \(g \circ f\) are defined for all real input values. This means they, too, have a domain of \(x \in \mathbb{R}\).
Function Composition
Function composition involves creating a new function by "plugging" one function into another. Imagine function \(f(x)\) as a machine that takes an input, processes it, and spits out a result. Function \(g(x)\) also takes an input and gives an output. By composing, we chain the output of \(g(x)\) directly into \(f(x)\). To compute \(f \circ g(x)\): - Replace the \(x\) in \(f(x)\) with \(g(x)\) getting \(f(g(x))\). - This becomes \(f\left(x^{2/3} + 1\right) = \sqrt[3]{(x^{2/3} + 1)^2 - 1}\). - After simplifying, it arrives at \(\sqrt[3]{x^{4/3} + 2x^{2/3}}\).For \(g \circ f(x)\): - Similarly, replace \(x\) in \(g(x)\) with the expression \(f(x)\) yielding \(g(f(x))\). - Now, \(g(\sqrt[3]{x^2 - 1}) = (x^2 - 1)^{2/9} + 1\).Function composition is a powerful tool enabling complex transformations, useful in areas ranging from basic algebra to advanced calculus.
Cube Root Functions
Cube root functions have a unique charm. Unlike square roots, they don't discriminate against negative values. A cube root, given by the symbol \(\sqrt[3]{x}\), finds a number \(y\) such that \(y^3 = x\). Key features of cube root functions include: - The domain is all real numbers \(x \in \mathbb{R}\). - These functions have a more gradual curve than square roots, smoothly passing from negative to positive values.To see this in action: - The function \(f(x) = \sqrt[3]{x^2 - 1}\) adapts any input \(x\) and modifies it by computing the cube root of \(x^2 - 1\). Because cubes encompass both positive and negative numbers, this function is free from the co-domain restrictions seen with square root counterparts.Understanding cube roots is essential for grasping complex functions in mathematics since they often reappear in advanced mathematical techniques and solutions.

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