Problem 80 - Precalculus : Building Concepts and Connections

Question

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=|x| ; g(x)=\frac{x}{x-3}$$

Step-by-Step Solution

Answer

\( (f \circ g)(x) = | \frac{x}{x-3} | \) with a domain of \( (-\infty, 3) \cup (3, +\infty) \) and \( (g \circ f)(x) = \frac{|x|} {|x|-3} \) with a domain of \( (-\infty, 3) \cup (3, +\infty) \)

Step 1: Find \( (f \circ g)(x) \)

When we do \( (f \circ g)(x) \), we are actually applying the function g into the function f. To do this, replace any instance of x in the function f with g(x). In mathematical terms, \( (f \circ g)(x) = f(g(x)) = | \frac{x}{x-3} | \). This is the expression for \( (f \circ g)(x). \)

Step 2: Figure out the domain of \( (f \circ g)(x) \)

Remember that the domain is the set of x-values that we can put into a function. For this exercise, we've to infer the domain from g(x). The only value x is not allowed to take is 3 because it would make g(x) undefined as a result of division by zero. Hence, the domain of \( (f \circ g)(x) = ) (-\infty, 3) \cup (3, +\infty) \)

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

In this step, we apply the function f into the function g. This is equivalent to replacing any instance of x in the function g with f(x). In mathematical terms, \( (g \circ f)(x) = g(f(x)) = \frac{|x|} {|x|-3} \) . This is the equation of \( (g \circ f) \).

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

Just as before, let's consider the domain from g(x). However this time, we've to ensure the denominator in g(f(x)) is not zero. So we need the absolute value of x not to be 3. The absolute value of a number is always positive, and the absolute value of a number can never be negative. Therefore, the domain of \( (g \circ f)(x) \) is \( (-\infty, 3) \cup (3, +\infty). \)

Key Concepts

Function Composition

When studying functions in mathematics, one often encounters the concept of function composition, which involves combining two or more functions to create a new function. In simpler terms, this is the process of feeding the output of one function into another.

Considering our example with functions f and g, the composite function (f \( \( \circ \) \) g)(x) is found by applying g(x) as the input for the function f. This is denoted mathematically as f(g(x)), indicating that g(x) is substituted into f wherever there is an \(x\). Likewise, (g \(\( \circ \)\) f)(x) means that f(x) is placed into g, giving g(f(x)).

Understanding how to construct a composite function is a fundamental skill in calculus and higher mathematics, where the properties of the resulting function may be used for various applications, from solving equations to modeling real-world situations.

Absolute Value Function

An absolute value function is a mathematical expression that describes how far a number is from zero on a number line, without considering which direction from zero the number lies. The absolute value of a number is always a non-negative value.

The function f(x) = |x|, used in our example, is a basic form of the absolute value function. It outputs the absolute value of the input \(x\). When this function is part of a composite function, such as (g \( \) f)(x) or (f \( \) g)(x), its role is to ensure that the input provided to the subsequent function is non-negative, which can significantly affect the resulting function’s domain and properties.

Domain Determination

The concept of domain determination is critical to understanding the functionality and limitations of a composed function.

The domain of a function refers to the set of all possible input values (\(x\)-values) for which the function is defined. In the composite function (f \( \) g)(x), we determine the domain by looking at the restrictions on \(g(x\)) because its values are what we're putting into \(f\). If \(g(x\)) could result in division by zero or the square root of a negative number, those \(x\)-values must be excluded from the domain to avoid undefined or imaginary results.

In our exercise, for the composite (f \( \) g)(x), we exclude \(x=3\) due to the division by zero in \(g(x\)). For (g \( \) f)(x), we must also consider values that make the denominator zero after applying the absolute value function, \(f(x\)), which again excludes \(x=3\) and negatives of three even though they do not appear due to the nature of absolute value. The domain, in this case, is the same for both composites: \( (-\infty, 3) \cup (3, +\infty) \).

Proper domain determination ensures that all operations within a function yield real and defined results, maintaining the function's integrity.