Problem 42 - 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. \(\quad f(x)=3 x, g(x)=x+5\)

Step-by-Step Solution

Answer

(f∘g)(x) = 3x + 15 with domain ℝ; (g∘f)(x) = 3x + 5 with domain ℝ.

Step 1: Compose the Functions: (f ∘ g)(x)

To find \(f(g(x))\), substitute the function \(g(x) = x + 5\) into \(f(x) = 3x\). This gives us \((f \circ g)(x) = 3(x + 5)\). Simplify the expression to get \((f \circ g)(x) = 3x + 15\).

Step 2: Define the Domain of (f ∘ g)(x)

The function \(f(g(x)) = 3x + 15\) is a linear function, and linear functions typically have a domain of all real numbers. So, the domain of \((f \circ g)(x)\) is \(\mathbb{R}\).

Step 3: Compose the Functions: (g ∘ f)(x)

To find \(g(f(x))\), substitute the function \(f(x) = 3x\) into \(g(x) = x + 5\). This yields \((g \circ f)(x) = 3x + 5\).

Step 4: Define the Domain of (g ∘ f)(x)

The function \(g(f(x)) = 3x + 5\) is also a linear function, which means its domain is all real numbers. Hence, the domain of \((g \circ f)(x)\) is \(\mathbb{R}\).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values (usually 'x' values) for which the function is defined. Understanding domains is crucial because it tells us the range of values that can be plugged into a function without causing any mathematical errors, such as division by zero or taking the square root of negative numbers.

For most simple functions, like polynomials, the domain is all real numbers. This is often denoted by the symbol \(\mathbb{R}\). When dealing with function compositions, such as \((f \circ g)(x)\), you first apply one function and then the other. It's important to ensure that each step remains defined.

Linear functions, which we'll discuss in the next section, typically have domains of all real numbers unless specified otherwise. For our exercise examples, both composed functions \((f \circ g)(x)\) and \((g \circ f)(x)\) were linear, so their domains were both \(\mathbb{R}\). It's good practice to always check your function type to determine its domain.

Linear Functions

Linear functions are among the most straightforward types of functions in mathematics. They can be written in the form \(f(x) = mx + b\), where \(m\) and \(b\) are constants. Here, \(m\) is the slope of the line, determining its steepness, and \(b\) is the y-intercept, showing where the line crosses the y-axis.

These functions graph as straight lines, which is why they're called linear. Because they have constant rates of change, they don't curve. This simplicity makes them easy to work with and their domains are typically all real numbers, as there are no restrictions like division by zero or roots.

Understanding linear functions is essential because they appear in many scenarios beyond just algebra, like in economics for cost functions or in physics for velocity. In our exercise, both composed functions resulted in linear equations, which made them straightforward to analyze and determine the domain.

Substitution Method

The substitution method is a crucial tool for simplifying or solving mathematical expressions, particularly in function compositions. It involves replacing one part of an equation, like a variable or function, with another expression. This is done to transform the equation into a simpler form, often making it easier to solve.

In function compositions, substitution allows us to combine two functions into one. For instance, with \((f \circ g)(x)\), the function \(g(x)\) is substituted for the input of \(f(x)\). This results in a single function as the output. It's a method that requires precision to ensure each step logically follows the mathematical rules.

For our exercise, we used substitution to find \((f \circ g)(x) = 3(x+5)\), which simplifies to \(3x + 15\). Similarly, by substituting \(f(x)\) into \(g\), we found \((g \circ f)(x) = 3x + 5\). Mastering substitution gives you a powerful way to handle complex compositions quickly.