Problem 30 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\frac{1}{x-1}, \quad g(x)=x-1 $$

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = \frac{1}{x-2}\), domain: \(\{ x \in \mathbb{R} \,|\, x \neq 2 \}\); (b) \((g \circ f)(x) = \frac{1}{x-1} - 1\), domain: \(\{ x \in \mathbb{R} \,|\, x \neq 1 \}\).

Step 1: Finding \((f \, \circ \, g)(x)\)

To find \((f \, \circ \, g)(x)\), we substitute \(g(x) = x-1\) into \(f(x)\). Thus, \((f \, \circ \, g)(x) = f(g(x)) = f(x-1) = \frac{1}{(x-1) - 1} = \frac{1}{x-2}\).

Step 2: Determining the Domain of \((f \, \circ \, g)\)

The function \(f(x)\) has a domain restriction where the denominator cannot be zero. Therefore, \((f \, \circ \, g)(x) = \frac{1}{x-2}\) is undefined when \(x-2=0\), i.e., \(x eq 2\). Hence, the domain of \((f \, \circ \, g)(x)\) is \(\{ x \in \mathbb{R} \, | \, x eq 2 \}\).

Step 3: Finding \((g \, \circ \, f)(x)\)

To find \((g \, \circ \, f)(x)\), we substitute \(f(x) = \frac{1}{x-1}\) into \(g(x)\). Thus, \((g \, \circ \, f)(x) = g(f(x)) = g \left( \frac{1}{x-1} \right) = \frac{1}{x-1} - 1\).

Step 4: Determining the Domain of \((g \, \circ \, f)\)

The function \(f(x) = \frac{1}{x-1}\) is undefined when \(x-1=0\), i.e., \(x eq 1\). Additionally, the expression \(g(f(x))\) is valid for all except where \(f(x)\) is initially undefined. Therefore, the domain of \((g \, \circ \, f)(x)\) is \(\{ x \in \mathbb{R} \, | \, x eq 1 \}\).

Key Concepts

Domain of a Function

The domain of a function is the complete set of possible input values (usually real numbers) that a function can accept without causing mathematical issues like division by zero or negative square roots. Understanding the domain is crucial because it tells us where the function is well-defined and can operate properly.

• Consider a basic function like a square root function, the domain consists of all values that make the expression inside the root non-negative.
• For a rational function, which is a fraction of two polynomials, the domain excludes values that make the denominator zero, as division by zero is undefined.

In our problem, we first identify the domain restrictions for

• \(f(x) = \frac{1}{x-1}\): Here, the denominator becomes zero when \(x = 1\), so \(x = 1\) is not in the domain.
• \(g(x) = x-1\): This function is defined for all real numbers since it is a simple linear equation.

By identifying the restrictions for each composed function, we determine their domains, ensuring that neither function encounters issues like dividing by zero.

Function Composition

Function composition is a method used to combine two functions, say \(f\) and \(g\), into a single new function. This process involves applying one function to the results of another, written as \((f \circ g)(x) = f(g(x))\). It is a powerful tool in mathematics because it allows the integration of multiple processes into one.

• The notation \((f \circ g)(x)\) implies that you take the output of \(g(x)\) and use it as the input for \(f(x)\).
• It's important to note that function composition is not necessarily commutative, meaning \((f \circ g)(x)\) does not always equal \((g \circ f)(x)\).

In the exercise provided:

• For \((f \circ g)(x)\), we substitute \(g(x) = x-1\) into \(f(x)\), resulting in \((f \circ g)(x) = \frac{1}{x-2}\).
• For \((g \circ f)(x)\), we substitute \(f(x) = \frac{1}{x-1}\) into \(g(x)\), giving \((g \circ f)(x) = \frac{1}{x-1} - 1\).

Function composition requires care in determining the domain constraints, as restrictions from both functions apply.

Rational Functions

Rational functions are expressions that are ratios of two polynomials, represented in the form \(\frac{P(x)}{Q(x)}\). They possess several unique properties and considerations, particularly concerning their domain and behavior around undefined points.

• The domain of a rational function is all real numbers except where the polynomial denominator is zero.
• These functions can have vertical asymptotes, which occur at points where the denominator equals zero.
• Rational functions may also exhibit horizontal or oblique asymptotes based on the degrees of their numerator and denominator polynomials.

In the exercise at hand, both \(f(x)\) and \((f \circ g)(x)\) are rational functions due to their form:

• For \(f(x) = \frac{1}{x-1}\), the denominator \(x-1\) defines its domain, excluding \(x = 1\).
• Similarly, \((f \circ g)(x) = \frac{1}{x-2}\) is undefined at \(x = 2\), as the composition introduces a new restriction.

Understanding rational functions and their traits helps in analyzing their graphs and determining the behavior as they approach undefined values.