Problem 82 - College Algebra with Corequisite Support

Question

For the following exercises, use \(f(x)=x^{3}+1\) and \(g(x)=\sqrt[3]{x-1}\). What is the domain of $$ (g \circ f)(x) ? $$

Step-by-Step Solution

Answer

The domain of \\( (g \circ f)(x) \\) is all real numbers.

Step 1: Understand the composition of functions

The composition of two functions \(g\) and \(f\) is written as \((g \circ f)(x) = g(f(x))\). This means we take the output of \(f(x)\) and use it as the input for \(g(x)\). In this exercise, \((g \circ f)(x) = g(f(x)) = g(x^3 + 1)\)."

Step 2: Determine the domain of f(x)

The function \(f(x) = x^3 + 1\) is a polynomial, which means its domain is all real numbers, \(-\infty < x < \infty\)."

Step 3: Determine the domain of g(x)

The function \(g(x) = \sqrt[3]{x - 1}\) is a cube root function. A cube root function is defined for all real numbers because any real number can be a cube. Therefore, the domain of \(g\) is also all real numbers, \(-\infty < x < \infty\)."

Step 4: Find domain of (g ∘ f)(x) using domain constraints

Since \(g(f(x)) = \sqrt[3]{f(x) - 1}= \sqrt[3]{x^3+1 - 1} = \sqrt[3]{x^3}\), we need to make sure that the input to the cube root \(x^3\) is within \(g\)'s domain. Since \(g\) can accept any real number, \(f(x)\) can be any number, which corresponds to the domain of \(f\): all real numbers \(-\infty < x < \infty\)."

Step 5: Conclude the domain of (g ∘ f)(x)

Since both \(f(x)\) and \(g(f(x))\) are defined for all real numbers, the domain of \((g \circ f)(x)\) is all real numbers, \(-\infty < x < \infty\)."

Key Concepts

Understanding Polynomial Functions

Polynomial functions are a fundamental part of algebra. These are mathematical expressions that involve sums of powers of variables, each multiplied by a coefficient. A typical polynomial has the form \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\), where the coefficients \(a_i\) are real numbers, and \(n\) is a non-negative integer indicating the highest power of \(x\).
For example, the function \(f(x) = x^3 + 1\) is a polynomial. It consists of one term with \(x\) raised to the power of 3, plus a constant term of 1.

• Polynomials are continuous and smooth, meaning they don’t have breaks or sharp corners.
• The domain of any polynomial function is always all real numbers, \(-\infty < x < \infty\).

Because they are defined for all real numbers, polynomials can serve as ideal input functions when performing function compositions.

Cube Root Functions in Focus

Cube root functions involve finding a number that, when multiplied by itself three times, gives the original number. The notation for a cube root function is \(g(x) = \sqrt[3]{x}\).
In the exercise problem, \(g(x) = \sqrt[3]{x - 1}\), which means we are taking the cube root of \(x-1\).

• Unlike square root functions, cube roots are defined for all real numbers (positive, negative, or zero) because every real number has a corresponding cube root.
• The domain of a cube root function like \(g(x) = \sqrt[3]{x - 1}\) is \(-\infty < x < \infty\).

This wide applicability makes cube root functions versatile in mathematical operations, especially in compositions with polynomials.

Exploring the Domain of Function Compositions

When composing two functions, like \(g(x)\) and \(f(x)\), it's critical to understand how their domains interact. The domain of the composite function \((g \circ f)(x)\) is shaped by the domains of \(f(x)\) and \(g(x)\).
Here's a step-by-step guide:

• First, identify the domain of the "inside" function, \(f(x)\). Here, since \(f(x) = x^3 + 1\) is a polynomial, its domain is all real numbers.
• Next, note the domain of the "outside" function \(g(x)\). As discussed, cube root functions like \(g(x) = \sqrt[3]{x-1}\) are also defined for all real numbers.
• Finally, ensure the output of \(f(x)\) fits within the input range of \(g(x)\). In this problem, \(g(f(x)) = \sqrt[3]{x^3}\), which is defined for any real number.

Since both functions \(f\) and \(g\) accommodate all real numbers, the composite function \((g \circ f)(x)\) is defined for all real numbers, giving it a domain of \(-\infty < x < \infty\).