Problem 43 - Calculus Volume 1

Question

For the following exercises, 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. $$ f(x)=x+4, g(x)=4 x-1 $$

Step-by-Step Solution

Answer

Both compositions simplify to linear functions with domains of all real numbers.

Step 1: Define Composition of Functions

The composition of two functions is denoted by \((f \circ g)(x)\), which means \(f(g(x))\), and \((g \circ f)(x)\), which means \(g(f(x))\). This involves substituting one function into another.

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

Substitute \(g(x) = 4x - 1\) into \(f(x)\):\[ f(g(x)) = f(4x - 1) = (4x - 1) + 4 \]Simplify the expression:\[ f(g(x)) = 4x + 3 \]

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

Substitute \(f(x) = x + 4\) into \(g(x)\):\[ g(f(x)) = g(x + 4) = 4(x + 4) - 1 \]Simplify the expression:\[ g(f(x)) = 4x + 16 - 1 = 4x + 15 \]

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

The domain of \((f \circ g)(x) = 4x + 3\) is the set of all real numbers since there are no restrictions in the expression such as division by zero or square roots of negative numbers.

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

The domain of \((g \circ f)(x) = 4x + 15\) is also the set of all real numbers for the same reason: the expression has no restrictions.

Key Concepts

Domain of a Function

The domain of a function refers to the set of all possible input values (often real numbers) that you can safely use within that function. When determining the domain, it's crucial to ensure that the function's expression remains valid for those inputs.
For example:

• Functions with denominators should not have zero in their domain since division by zero is undefined.
• Functions involving square roots should avoid negative numbers, as their real values are not defined in this context.

In the context of our exercise, for both \((f \circ g)(x) = 4x + 3\) and \((g \circ f)(x) = 4x + 15\), these expressions are linear functions, which have no restrictions on their domain. Consequently, the domain for both is all real numbers, as there are no divisions or even roots that could impose restrictions.

Simplifying Expressions

Simplifying expressions is a fundamental skill in mathematics that involves reducing expressions to their simplest form. This step is crucial in solving problems efficiently, revealing clearer insights into the behavior of a function.
In the exercise, we began with expressions like \(f(g(x)) = (4x - 1) + 4\) and simplified it to \(4x + 3\). This simplification removes redundancy and helps us present the function more concisely.
To simplify an expression:

• Combine like terms, such as numbers or variables that have the same exponent.
• Follow the order of operations (PEMDAS/BODMAS).
• Keep an eye out for opportunities to factor or cancel terms, although not applicable here, it is good practice.

Simplifying helps in recognizing the nature of the function, such as identifying it as linear, which informs us about its behavior and domain.

Substitution in Functions

Substitution in functions involves replacing a variable in one function with another expression from a second function. It is the key process in composing functions, allowing you to build new functions.
In the exercise, we demonstrate this by calculating \((f \circ g)(x)\) and \((g \circ f)(x)\). With \(f(x) = x+4\) and \(g(x) = 4x-1\), the substitutions are:

• For \((f \circ g)(x)\), replace \(x\) in \(f(x)\) with \(g(x) = 4x-1\).
• For \((g \circ f)(x)\), replace \(x\) in \(g(x)\) with \(f(x) = x+4\).

This method allows us to explore how changing input variables can transform outputs, offering a useful way to manipulate and analyze complex functions in both mathematical studies and practical applications.