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Find the functions (a) \(f \circ g,(\) b) \(g \circ f,(\) c) \(f \circ f,\) and (d) \(g \circ g\) and their domains. \(f(x)=\sqrt{x}, \quad g(x)=\sqrt[3]{1-x}\)

Short Answer

Expert verified
(a) \(\sqrt{\sqrt[3]{1-x}}\), \(x \leq 1\); (b) \(\sqrt[3]{1-\sqrt{x}}\), \(0 \leq x \leq 1\); (c) \(\sqrt{\sqrt{x}}\), \(x \geq 0\); (d) \(\sqrt[3]{1-\sqrt[3]{1-x}}\), \(x \leq 1\).

Step by step solution

01

Analyze the Composition of Functions

To find the function compositions, we start by analyzing each one. We will substitute one function into the other as indicated by the composition notation. The compositions to find are: \(f \circ g = f(g(x))\), \(g \circ f = g(f(x))\), \(f \circ f = f(f(x))\), and \(g \circ g = g(g(x))\).
02

Calculate \(f \circ g\) and its Domain

Plug \(g(x) = \sqrt[3]{1-x}\) into \(f(x) = \sqrt{x}\). This gives us \(f(g(x)) = f(\sqrt[3]{1-x}) = \sqrt{\sqrt[3]{1-x}}\). The domain requires \(\sqrt[3]{1-x} \geq 0\). Since cube roots can take any real numbers, we need \(\sqrt[3]{1-x} \geq 0\). Solving \(1-x \geq 0\) gives \(x \leq 1\). Thus, the domain of \(f \circ g\) is \(x \leq 1\).
03

Calculate \(g \circ f\) and its Domain

Substitute \(f(x) = \sqrt{x}\) into \(g(x) = \sqrt[3]{1-x}\). This gives us \(g(f(x)) = g(\sqrt{x}) = \sqrt[3]{1-\sqrt{x}}\). The domain requires that \(1-\sqrt{x} \geq 0\), giving \(1 \geq \sqrt{x}\). Further, \(0 \leq x \leq 1\) because square roots must be non-negative. Thus, the domain of \(g \circ f\) is \(0 \leq x \leq 1\).
04

Calculate \(f \circ f\) and its Domain

Substitute \(f(x) = \sqrt{x}\) into itself to get \(f(f(x)) = f(\sqrt{x}) = \sqrt{\sqrt{x}}\). For \(\sqrt{x} \geq 0\), the domain is simply \(x \geq 0\) since \(x\) needs to be non-negative for the square root to exist. Hence, the domain of \(f \circ f\) is \(x \geq 0\).
05

Calculate \(g \circ g\) and its Domain

Substitute \(g(x) = \sqrt[3]{1-x}\) into itself to get \(g(g(x)) = g(\sqrt[3]{1-x}) = \sqrt[3]{1-\sqrt[3]{1-x}}\). There are no particular domain restrictions for cube roots. Thus, the domain of \(g \circ g\) is all real numbers \(x \leq 1\) determined initially for cube roots due to \(1-x \geq 0\).

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Key Concepts

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

Function Domains
Understanding the domain of a function is crucial. It determines the set of all possible input values for which the function is defined. Here, we look at two functions: a square root function, \( f(x) = \sqrt{x} \), and a cube root function, \( g(x) = \sqrt[3]{1-x} \).

For a square root function like \( f(x) \), the domain consists of all \( x \geq 0 \) since you can't take the square root of a negative number.
The cube root function, \( g(x) \), has a different approach. Cube roots are defined for all real numbers, but here, to keep it non-negative, we write:
\[ 1 - x \geq 0 \]\[ x \leq 1 \]
Thus, the domain is all real numbers \( x \leq 1 \). In function composition, understanding the domains of combined functions helps in finding the range for which the functions are valid.
Square Roots
Square roots are common in math, particularly when dealing with quadratic equations and geometry. The square root of a number \( x \) is a value \( y \) such that \( y^2 = x \).

In our example, \( f(x) = \sqrt{x} \), means finding the value that, when multiplied by itself, gives \( x \).
Since square roots can't handle negative numbers, the domain is limited to \( x \geq 0 \), ensuring all results are real numbers. Let's see this in practice:
  • For \( f(f(x)) \), or iterating the function on itself, we have \( \sqrt{\sqrt{x}} \). It's only defined for \( x \geq 0 \).

This shows the strict limitation of square roots to handle only non-negative inputs.
Cube Roots
Cube roots are related to finding a number which, when multiplied by itself twice, returns the original number. In formula \( g(x) = \sqrt[3]{1-x} \), we explore how cube roots work differently from square roots.

Cube roots can always find a real number output regardless of the input sign. They don't require the input to be non-negative. For composition, as seen in \( g(g(x)) = \sqrt[3]{1-\sqrt[3]{1-x}} \), there are no stricter domain restrictions, though our original form limits it to \( x \leq 1 \).
  • The general domain unrestricted by cube roots allows them to handle a wider range of functions compared to square roots.

This versatile nature makes cube roots vital for handling negative numbers, perfectly suiting them for many algebraic roles.
Inequalities in Functions
Inequalities play an essential role in determining where functions are valid (their domains) and resolving these inequalities is crucial in function compositions.

In our function compositions, like \( f \circ g \) or \( g \circ f \), we consider inequalities to find viable \( x \) values:
  • For \( f(g(x)) = \sqrt{\sqrt[3]{1-x}} \), we need \( \sqrt[3]{1-x} \geq 0 \).Solving \( 1-x\geq0 \) gives \( x \leq 1 \).
  • Conversely, \( g(f(x)) = \sqrt[3]{1-\sqrt{x}} \) demands \( 1-\sqrt{x} \geq 0 \) giving \( 0 \leq x \leq 1 \) on solving.

Our ability to handle such inequalities frames our understanding of how these functions interact, highlighting the combinations that yield valid results.

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