Problem 45 - Calculus Volume 1

Question

For each pair of functions, find a. \((f \circ g)(x)\) and b. \((g \circ f)(x)\) Simplify the results. Find the domain of each of the results. \(f(x)=x^{2}+7, g(x)=x^{2}-3\)

Step-by-Step Solution

Answer

(f ∘ g)(x) = x^4 - 6x^2 + 16; domain: all real numbers. (g ∘ f)(x) = x^4 + 14x^2 + 46; domain: all real numbers.

Step 1: Understand the Functions

We have two functions: \(f(x) = x^2 + 7\) and \(g(x) = x^2 - 3\). We will find the compositions \((f \circ g)(x)\) and \((g \circ f)(x)\), which involve substituting one function into another.

Step 2: Compute (f ∘ g)(x)

The composition \((f \circ g)(x)\) means substituting \(g(x)\) into \(f(x)\). So, we substitute \(x^2 - 3\) into \(x^2 + 7\):\[(f \circ g)(x) = f(g(x)) = f(x^2 - 3) = ((x^2 - 3)^2) + 7.\]

Step 3: Simplify (f ∘ g)(x)

Simplify the expression:\[(x^2 - 3)^2 = x^4 - 6x^2 + 9.\]So,\[(f \circ g)(x) = x^4 - 6x^2 + 9 + 7 = x^4 - 6x^2 + 16.\]

Step 4: Domain of (f ∘ g)(x)

The domain of \((f \circ g)(x)\), \(x^4 - 6x^2 + 16\), is all real numbers \(\mathbb{R}\), since any real number can be input into the polynomial.

Step 5: Compute (g ∘ f)(x)

The composition \((g \circ f)(x)\) means substituting \(f(x)\) into \(g(x)\). So, we substitute \(x^2 + 7\) into \(x^2 - 3\):\[(g \circ f)(x) = g(f(x)) = g(x^2 + 7) = ((x^2 + 7)^2) - 3.\]

Step 6: Simplify (g ∘ f)(x)

Simplify the expression:\[(x^2 + 7)^2 = x^4 + 14x^2 + 49.\]So,\[(g \circ f)(x) = x^4 + 14x^2 + 49 - 3 = x^4 + 14x^2 + 46.\]

Step 7: Domain of (g ∘ f)(x)

The domain of \((g \circ f)(x)\), \(x^4 + 14x^2 + 46\), is also all real numbers \(\mathbb{R}\) since it is a polynomial function.

Key Concepts

Domain of a Function

Understanding the domain of a function is vital when dealing with function compositions and expressions. The domain refers to all the possible input values (usually denoted as \(x\)) for which the function is defined. In simpler terms, it's the set of all values you can plug into a function without causing mathematical mishaps, like division by zero or taking the square root of a negative number.
When we talk about polynomial functions like \(f(x) = x^2 + 7\) and \(g(x) = x^2 - 3\), their domains are typically all real numbers, denoted by \(\mathbb{R}\). This is because you can substitute any real number into a polynomial without encountering undefined mathematical operations.
For the function compositions \((f \circ g)(x)\) and \((g \circ f)(x)\) given in the problem, we do not have restrictions either. Both are polynomial functions too, so their domains are also all real numbers \(\mathbb{R}\).

• Polynomials: Domain is all real numbers.
• Watch for undefined operations like division by zero in other types of functions.

Simplifying Expressions

Simplifying expressions is an essential skill in algebra. It involves reducing expressions to a simpler form that is easier to work with or understand. When you simplify an expression, you remove unnecessary parts or combine like terms to make it tidy.
In our function composition example:

• For \((f \circ g)(x)\), we have the expression \((x^2 - 3)^2 + 7\). To simplify, expand the squared term first: \[(x^2 - 3)^2 = x^4 - 6x^2 + 9\]. Then, combine it with the constant 7 to get \(x^4 - 6x^2 + 16\).
• For \((g \circ f)(x)\), the expression \((x^2 + 7)^2 - 3\) is simplified by expanding to \[x^4 + 14x^2 + 49\] and combining with \(-3\) for a result of \(x^4 + 14x^2 + 46\).

Both cases involve expanding squares and adjusting constants, a common task in algebraic simplification.

Polynomial Functions

Polynomial functions form the backbone of many algebraic operations, and understanding them is key to mastering function compositions.
A polynomial function is any mathematical expression comprising variables with whole number exponents and coefficients. They can be as simple as \(f(x) = x + 1\) or more complex like the ones in our exercise: \(x^4 - 6x^2 + 16\) and \(x^4 + 14x^2 + 46\).
Polynomials feature several interesting properties:

• They are continuous and smooth, meaning they're defined for every real number
• No gaps, holes, or vertical asymptotes.

These characteristics make dealing with polynomial domains straightforward. Plus, they have simple derivative and integration properties, which is handy for solving more advanced calculus problems later on.
Understanding the structure and behavior of polynomial functions enhances your ability to simplify expressions and accurately determine domains in compositions.