Problem 9 - The Calculus with Analytic Geometry

Question

In Exercises 7 through 12, the functions \(f\) and \(g\) are defined. In each problem define the following functions and determine the domain of the resulting function: (a) \(f+g ;\) (b) \(f-g ;\) (c) \(f \cdot g ;\) (d) \(f / g ;\) (e) \(g / f\); (f) \(f \circ g ;(\mathrm{g}) g \circ f\). $$ f(x)=\frac{x+1}{x-1} ; g(x)=\frac{1}{x} $$

Step-by-Step Solution

Answer

All new functions have the domain \( x \in \mathbb{R} \setminus \{0,1\} \).

Step 1: Define the Functions

Given: \( f(x) = \frac{x+1}{x-1} \) and \( g(x) = \frac{1}{x} \)

Step 2: Find the Domain of f(x)

The function \( f \) is defined as \( \frac{x+1}{x-1} \). For this function to be defined, the denominator must not be zero. Therefore, \( x-1 eq 0 \) which gives us \( x eq 1 \).

Step 3: Find the Domain of g(x)

The function \( g \) is \( \frac{1}{x} \). For this function to be defined, the denominator must not be zero. Therefore, \( x eq 0 \).

Step 4: Define and Find the Domain of f + g

The function \( f+g \) is \( \frac{x+1}{x-1} + \frac{1}{x} \). The domain of \( f+g \) is the intersection of the domains of \( f \) and \( g \): \( x eq 1 \) and \( x eq 0 \). Hence, the domain of \( f+g \) is \( x \in \mathbb{R} \setminus \{0,1\} \).

Step 5: Define and Find the Domain of f - g

The function \( f-g \) is \( \frac{x+1}{x-1} - \frac{1}{x} \). The domain of \( f-g \) is the same as that of \( f+g \): \( x \in \mathbb{R} \setminus \{0,1\} \).

Step 6: Define and Find the Domain of f \cdot g

The function \( f \cdot g \) is \( \frac{x+1}{x-1} \cdot \frac{1}{x} = \frac{x+1}{x(x-1)} \). The domain of \( f \cdot g \) is \( x \in \mathbb{R} \setminus \{0,1\} \).

Step 7: Define and Find the Domain of f / g

The function \( \frac{f}{g} \) is \( \frac{\frac{x+1}{x-1}}{\frac{1}{x}} = \frac{(x+1)x}{x-1} \). The domain of \( \frac{f}{g} \) is \( x \in \mathbb{R} \setminus \{0,1\} \).

Step 8: Define and Find the Domain of g / f

The function \( \frac{g}{f} \) is \( \frac{\frac{1}{x}}{\frac{x+1}{x-1}} = \frac{x-1}{x(x+1)} \). The domain of \( \frac{g}{f} \) is the same as \( \frac{f}{g} \); \( x \in \mathbb{R} \setminus \{0,1\} \).

Step 9: Define and Find the Domain of f \circ g

The function \( f \circ g \) is \( f(g(x)) \). Substitute \( g(x) \) into \( f \): \( f \left( \frac{1}{x} \right) = \frac{\frac{1}{x} + 1}{\frac{1}{x} - 1} \). The domain follows from ensuring \( g(x) \) is defined and the output of \( g \) does not make \( f \) undefined: \( x \in \mathbb{R} \setminus \{0, \pm 1\} \).

Step 10: Define and Find the Domain of g \circ f

The function \( g \circ f \) is \( g(f(x)) \). Substitute \( f(x) \) into \( g \): \( g \left( \frac{x+1}{x-1} \right) = \frac{1}{\frac{x+1}{x-1}} = \frac{x-1}{x+1} \). The domain follows from ensuring \( f(x) \) is defined and the output of \( f \) does not make \( g \) undefined: \( x \in \mathbb{R} \setminus \{0, 1\} \).

Key Concepts

Domain of Functions

The domain of a function is the set of all possible input values that make the function work correctly. Let’s take a closer look at how to find the domain of two functions:

Consider the given functions: \(f(x) = \frac{x + 1}{x - 1}\) and \(g(x) = \frac{1}{x}\). To determine the domain of \(f(x)\), we need to ensure the denominator is not zero, as zero makes the function undefined. Therefore, \(x - 1 eq 0\), which simplifies to \(x eq 1\). This means the domain of \(f\) includes all real numbers except 1.

For \(g(x)\), we also need the denominator to not be zero: \(x eq 0\). So, the domain of \(g\) includes all real numbers except 0.

By understanding the domain of each function separately, we can combine them to see where both functions are defined. This will be the basis for analyzing their combinations, like addition, subtraction, multiplication, and division.

Function Operations

Function operations include addition, subtraction, multiplication, and division. Let’s see how these operations affect the domains:

When adding or subtracting two functions, the resulting domain is the intersection of the individual domains. For instance: \(f + g = \frac{x + 1}{x - 1} + \frac{1}{x}\), the domain can’t include 1 or 0. Thus, it is \(x \in \mathbb{R} \setminus \{0,1\}\).

Similarly, for subtraction, \(f - g = \frac{x + 1}{x - 1} - \frac{1}{x}\), the domain will be the same as for addition: \(x \in \mathbb{R} \setminus \{0,1\}\).

For multiplication, \(f \cdot g = \frac{x + 1}{x - 1} \cdot \frac{1}{x} = \frac{x + 1}{x(x - 1)}\), again, we exclude the values 0 and 1. Therefore, the domain remains \(x \in \mathbb{R} \setminus \{0,1\}\).

Division is a bit trickier, as we need to ensure the denominator is not zero. For \(f/g = \frac{\frac{x + 1}{x - 1}}{\frac{1}{x}} = \frac{(x + 1)x}{x - 1}\), and \(g/f = \frac{\frac{1}{x}}{\frac{x + 1}{x - 1}} = \frac{x - 1}{x(x + 1)}\), both critical values (0 and 1) are excluded from the domain. Hence, the domain is \(x \in \mathbb{R} \setminus \{0,1\}\).

Function Composition

Function composition involves plugging one function into another. We need to consider the domains of both functions to avoid undefined regions.

For \(f \circ g = f(g(x))\), we substitute \(g(x)\) into \(f\): \(f \left( \frac{1}{x} \right) = \frac{\frac{1}{x} + 1}{\frac{1}{x} - 1}\). The domain requires that \(x eq 0\) from \(g(x)\) and implies that \(\frac{1}{x} eq 1\) from \(f\). Therefore, \(x eq \pm 1\). Thus, the domain of \(f \circ g\) is \(x \in \mathbb{R} \setminus \{0, -1, 1\}\).

For \(g \circ f = g(f(x))\), substitute \(f(x)\) into \(g\): \(g \left( \frac{x + 1}{x - 1} \right) = \frac{1}{\frac{x + 1}{x - 1}} = \frac{x - 1}{x + 1}\). The domain must exclude \(x = 1\) and consider the output to ensure \(x + 1 eq 0\) from \(g\). Thus, we avoid \(x = -1\). The domain is \(x \in \mathbb{R} \setminus \{1, -1\}\).

Rational Functions

Rational functions are fractions where the numerator and the denominator are polynomials. An essential part of understanding rational functions is identifying their domains and potential restrictions.

For \(f(x) = \frac{x + 1}{x - 1}\), it is crucial to note the denominator can’t be zero, i.e., \(x eq 1\). This restriction is necessary to prevent undefined behavior.

Similarly, for \(g(x) = \frac{1}{x}\), we must ensure \(x eq 0\) so that the denominator remains non-zero.

When combining rational functions through operations like addition or composition, we face combined domain restrictions. For example in \(f \circ g\), we must consider constraints from both \(f\) and \(g\) to ensure no undefined outputs.

Analyzing these constraints in steps ensures we systematically avoid undefined areas, providing clear, well-defined function outcomes.