Problem 7 - Intermediate Algebra

Question

Determine \((f \circ g)(x)\) and \((g \circ f)(x)\) for each pair of functions. Also specify the domain of \((f \circ g)(x)\) and \((g \circ f)(x)\). (Objective 1\()\) \(f(x)=2 x^{2}-x+2\) and \(g(x)=-x+3\)

Step-by-Step Solution

Answer

\((f \circ g)(x) = 2x^2 - 11x + 17\); \((g \circ f)(x) = -2x^2 + x + 1\); domains are \((-\infty, \infty)\).

Step 1: Understanding Function Composition

Function composition involves plugging one function into another. If you want to find \((f \circ g)(x)\), you'll replace every instance of \(x\) in \(f(x)\) with \(g(x)\). Similarly, for \((g \circ f)(x)\), replace every \(x\) in \(g(x)\) with \(f(x)\).

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

Given \(f(x)=2x^2-x+2\) and \(g(x)=-x+3\), calculate \((f \circ g)(x) = f(g(x))\).Substitute \(g(x)\) into \(f(x)\):\[ f(g(x)) = 2(-x+3)^2 - (-x+3) + 2 \]First, calculate \((-x+3)^2 = x^2 - 6x + 9 \).Now substitute:\[ f(g(x)) = 2(x^2 - 6x + 9) + x - 3 + 2 \]Simplify:\[ = 2x^2 - 12x + 18 + x - 1 \]\[ = 2x^2 - 11x + 17 \]

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

Now calculate \((g \circ f)(x) = g(f(x))\).Substitute \(f(x)\) into \(g(x)\):\[ g(f(x)) = - (2x^2 - x + 2) + 3 \]Distribute the negative:\[ = -2x^2 + x - 2 + 3 \]Simplify:\[ = -2x^2 + x + 1 \]

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

Since both given functions \(f(x) = 2x^2 - x + 2\) and \(g(x) = -x + 3\) are polynomials, they are defined for all real numbers. Therefore, the domain of \((f \circ g)(x)\) is all real numbers, represented mathematically as \((-\infty, \infty)\).

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

Similarly, because \(g(f(x))\) does not have any denominators, roots, or logarithms that could restrict its domain, \(g(f(x))\) is also defined for all real numbers. Thus, the domain of \((g \circ f)(x)\) is \((-\infty, \infty)\).

Key Concepts

Domain of a Function

When you hear about the "domain of a function," think of it as the set of all possible input values (or x-values) that you are allowed to use in a function. This concept is crucial because it tells you when a function is properly defined.
For polynomial functions, like those in the exercise where

• \( f(x) = 2x^2 - x + 2 \)
• \( g(x) = -x + 3 \)

the domain is typically all real numbers. This is because polynomials are defined everywhere on the real number line.
More formally, we represent this domain as \(( -\infty, \infty )\), meaning any real number you choose can be put into these functions without running into any trouble.
Function compositions like \((f \circ g)(x)\) or \((g \circ f)(x)\) follow the same principle. Since both \( f \) and \( g \) are polynomials, the domains of their compositions remain \(( -\infty, \infty )\). There's no restriction because we're simply mixing two polynomials together, which stay nice and friendly across the entire real number spectrum.

Polynomials

Polynomials are one of the building blocks in algebra and understanding them is key to tackling many mathematical problems. A polynomial is an expression consisting of variables, coefficients, and exponents. The exponents must be non-negative integers. You'll see them in forms like

• \( ax^n + bx^{n-1} + \, \ldots \, + cx + d \)

where each term is a product of a coefficient and a variable raised to an exponent.
In the exercise, we have:

• \( f(x) = 2x^2 - x + 2 \)
• \( g(x) = -x + 3 \)

Both are polynomials, with \( f(x) \) being a quadratic polynomial (it has an \(x^2\) term) and \( g(x) \) being a linear polynomial (its highest power of \(x\) is 1).
The simplicity of these expressions is a huge benefit when it comes to calculations and function operations, like addition, subtraction, and composition. They're straightforward, so calculating products or compositions (putting one polynomial inside another) doesn't bring any unexpected complexities.

Function Operations

Function operations include activities like addition, subtraction, multiplication, division, and composition of functions. Composition is a way to combine two functions, say \( f \) and \( g \), into a single function. Here, in the exercise, we dealt with function composition which involved two main operations:

• Finding \((f \circ g)(x)\), which means putting \(g(x)\) into \(f(x)\).
• Finding \((g \circ f)(x)\), which means putting \(f(x)\) into \(g(x)\).

When you perform \((f \circ g)(x)\), you substitute \(g(x)\) in every place \(x\) appears in \(f(x)\). Similarly, for \((g \circ f)(x)\), substitute \(f(x)\) into \(g(x)\). These operations are not too tricky when dealing with polynomials, as the calculations mostly involve basic algebraic manipulations.
Understanding how to compose functions is a valuable skill. It lets you build complex functions from simple ones and is used extensively in calculus and real-world applications, like computing the outcome when one process follows another.