Problem 28 - A Graphical Approach to College Algebra

Question

For each pair of functions, (a) find \((f+g)(x),(f-g)(x),\) and \((f g)(x)\) (b) give the domains of the functions in part (a); (c) find \(\frac{f}{8}\) and give its domain; (d) find \(f \circ g\) and give its domain; and (e) find \(g \circ f\) and give its domain. Do not use a calculator. $$f(x)=\sqrt[3]{6-3 x}, g(x)=2 x^{3}+1$$

Step-by-Step Solution

Answer

The functions \(f+g(x), f-g(x), \) and \(fg(x)\) have domains of all real numbers. \(\frac{f}{8}\) has a domain of all real numbers. \(f\circ g\) and \(g\circ f\) have domains of all real numbers.

Step 1: Determine (f+g)(x)

To find \((f+g)(x)\), simply add the two functions together: \(f(x) + g(x) = \sqrt[3]{6-3x} + (2x^3 + 1)\). So, \((f+g)(x) = \sqrt[3]{6-3x} + 2x^3 + 1\).

Step 2: Determine (f-g)(x)

To find \((f-g)(x)\), subtract \(g(x)\) from \(f(x)\): \(f(x) - g(x) = \sqrt[3]{6-3x} - (2x^3 + 1)\). So, \((f-g)(x) = \sqrt[3]{6-3x} - 2x^3 - 1\).

Step 3: Determine (fg)(x)

To find \((fg)(x)\), multiply \(f(x)\) and \(g(x)\): \(\sqrt[3]{6-3x} \cdot (2x^3 + 1)\). This equals \((fg)(x) = (2x^3 + 1)\sqrt[3]{6-3x}\).

Step 4: Domain of (f+g)(x), (f-g)(x), and (fg)(x)

For \((f+g)(x)\), \((f-g)(x)\), and \((fg)(x)\), the domain includes all real numbers since \(\sqrt[3]{6-3x}\) is defined for all real \(x\). Thus, the domain is \(( -\infty, +\infty )\).

Step 5: Find \(\frac{f}{8}\) and its domain

\(\frac{f}{8} = \frac{\sqrt[3]{6-3x}}{8}\). The domain of \(\frac{f}{8}\) is the same as \(f(x)\) itself, which is all real numbers, \(( -\infty, +\infty )\).

Step 6: Calculate f∘g(x) and its domain without a calculator

\(f\circ g = f(g(x)) = \sqrt[3]{6 - 3(2x^3 + 1)}\). Simplifying, \(f\circ g(x) = \sqrt[3]{6 - 6x^3 - 3}\) which becomes \(\sqrt[3]{3 - 6x^3}\). Since cube roots accept any real numbers as inputs, the domain is all real numbers \(( -\infty, +\infty )\).

Step 7: Calculate g∘f(x) and its domain without a calculator

\(g\circ f = g(f(x)) = 2(\sqrt[3]{6-3x})^3 + 1\). The cube and cube root cancel, leading to \(g\circ f(x) = 2(6-3x) + 1\), which simplifies to \(12 - 6x + 1 = 13 - 6x\). This is defined for all real numbers. Thus, the domain is \(( -\infty, +\infty )\).

Key Concepts

Domain of Functions

The domain of a function is the set of all possible input values (often represented by \(x\)) that can safely be used without causing any mathematical problems, such as division by zero or taking an even root of a negative number.
For example, when we talk about cube roots, like in the function \(f(x) = \sqrt[3]{6-3x}\), cube roots are allowed for any real number.
This is because cubes and cube roots can work with any real number, whether positive, negative, or zero. So, the domain of \(f(x)\) is all real numbers.

When finding the domain of a composed function, like \(f+g, f-g,\) or \(fg\), we consider the domains of the individual functions involved. For instance:

• The functions \(f(x)\) and \(g(x) = 2x^3 + 1\) both have domains that include all real numbers.
• Since both functions are well-defined for all real \(x\), the domain for \((f+g)(x), (f-g)(x),\) and \((fg)(x)\) is also all real numbers \((-\infty, +\infty)\).

This means you can use any real number for \(x\) in these expressions without running into any issues.

Composition of Functions

Composition of functions is like putting one function inside another to create a new function.
This is denoted by \(f \circ g\), which is read as \(f\) of \(g\).
Essentially, you take the result of \(g(x)\) and use it as the input for \(f(x)\).

For example, in the exercise, we have:

• \(f(g(x)) = f\) of \(g(x) = \sqrt[3]{6 - 3(2x^3 + 1)} = \sqrt[3]{3 - 6x^3}\)
• This newly composed function is also defined for all real numbers because it only involves cube roots, which can handle any real input.
• Similarly, \(g(f(x)) = 2(\sqrt[3]{6-3x})^3 + 1\) simplifies to \(13 - 6x,\) which is defined for all \(x\).

By finding the compositions and their domains, you can see how functions interact with each other and verify that the domain remains all-encompassing, \((-\infty, +\infty)\), in this case.

Cube Root Function

The cube root function, represented as \(\sqrt[3]{x}\), is unique among root functions because it works smoothly with every real number.
Unlike the square root function, which is undefined for negative numbers, the cube root function simply finds a number that, when cubed, will give you the original number back.

When solving problems involving cube roots, like \(f(x) = \sqrt[3]{6-3x}\), here’s what you should keep in mind:

• The domain is all real numbers since there are no restrictions on \(x\).
• The cube root of a negative number is also negative (e.g., \(\sqrt[3]{-8} = -2\)).

Understanding the behavior of cube roots helps when composing functions, especially when they are paired with polynomial functions, as seen in our exercise. They give results that remain smoothly defined across the entire number line. Therefore, for the cube root function, you can confidently use any real number, making problems involving cube roots highly approachable.