Problem 44 - 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. \(f(x)=2 x+4, g(x)=x^{2}-2\)

Step-by-Step Solution

Answer

(f \circ g)(x) = 2x^2; (g \circ f)(x) = 4x^2 + 16x + 14; both have domain \( (-\infty, +\infty) \).

Step 1: Define the Function Compositions

For the functions \(f(x) = 2x + 4\) and \(g(x) = x^2 - 2\), we'll first define the compositions:a. \((f \circ g)(x) = f(g(x))\)b. \((g \circ f)(x) = g(f(x))\)

Step 2: Calculate \((f \circ g)(x)\)

Substitute \(g(x) = x^2 - 2\) into \(f(x)\):\[(f \circ g)(x) = f(g(x)) = f(x^2 - 2) = 2(x^2 - 2) + 4\]Simplify the expression:\[2(x^2 - 2) + 4 = 2x^2 - 4 + 4 = 2x^2\]Thus, \((f \circ g)(x) = 2x^2\).

Step 3: Calculate \((g \circ f)(x)\)

Substitute \(f(x) = 2x + 4\) into \(g(x)\):\[(g \circ f)(x) = g(f(x)) = g(2x + 4) = (2x + 4)^2 - 2\]Expand \((2x + 4)^2\):\[(2x + 4)^2 = (2x)^2 + 2(2x)(4) + (4)^2 = 4x^2 + 16x + 16\]Simplify the expression:\[4x^2 + 16x + 16 - 2 = 4x^2 + 16x + 14\]Thus, \((g \circ f)(x) = 4x^2 + 16x + 14\).

Step 4: Determine the Domain for \((f \circ g)(x)\)

Since \(f(g(x)) = 2x^2\) involves a polynomial, which is defined for all real numbers, the domain of \((f \circ g)(x) = 2x^2\) is all real numbers, \((-\infty, +\infty)\).

Step 5: Determine the Domain for \((g \circ f)(x)\)

Similarly, since \((g \circ f)(x) = 4x^2 + 16x + 14\) is also a polynomial, it is defined for all real numbers. Therefore, the domain of \((g \circ f)(x) = 4x^2 + 16x + 14\) is all real numbers, \((-\infty, +\infty)\).

Key Concepts

Domain of a Function

In mathematics, the domain of a function is the set of all possible input values (usually denoted as "x") that the function can accept without causing any kind of anomaly, like division by zero. For example, with the polynomial function \(f(x) = 2x + 4\), the domain includes all real numbers, symbolized by \((-fty, +fty)\).
This means you can substitute any real number into the function, and it will provide a valid output.
The same applies to the function \(g(x) = x^2 - 2\). Since it's a polynomial, its domain is all real numbers as well.

• If a function has no specific restrictions (like a square root or a denominator that could become zero), then its domain is typically all real numbers.
• This makes polynomials very straightforward when discussing domains, as they allow any real number as input.

Understanding the domain is essential to perform operations like composition accurately, ensuring you are using valid inputs throughout.

Polynomials

Polynomials are a type of mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. The simplest polynomial is a constant, while more complex polynomials include terms like \(2x^2\), \(3x\), and so forth.
The functions discussed here, \(f(x) = 2x + 4\) and \(g(x) = x^2 - 2\), are both examples of polynomials.

• Polynomials are straightforward to handle since they are defined for all real numbers.
• The highest power of the variable (in these cases, \(x^2\) for \(g(x)\) and \(x\) for \(f(x)\)) determines the degree of the polynomial.

Computing compositions such as \((f \circ g)(x)\) or \((g \circ f)(x)\) involves substituting one polynomial into another.
Since both original function and result are polynomials, they remain defined for all real inputs.

Mathematical Functions

Mathematical functions are relations that uniquely associate members of one set (the domain) with another set. Functions like \(f(x) = 2x + 4\) and \(g(x) = x^2 - 2\) exhibit simple relations that map any input \(x\) to exactly one output.
In the context of functions:

• Each element of the domain maps to a unique element of the output, called the range.
• The types of functions can vary widely, from simple linear ones to more complex quadratic or exponential functions.

One can also perform function composition, combining functions to form a new function.
This exercise shows compositing \(f(x)\) and \(g(x)\) so that the output of one function becomes the input of another.
Understanding how to accurately combine functions is essential to solving complex mathematical problems.