Problem 41 - Calculus Volume 1

Question

For each pair of functions, find a. \(f+g\) b. \(f-g\) c. \(f \cdot g\) d. \(f / g\). Determine the domain of each of these new functions. \(f(x)=6+\frac{1}{x}, g(x)=\frac{1}{x}\)

Step-by-Step Solution

Answer

The domains for all resulting functions are \((-\infty, 0) \cup (0, \infty)\).

Step 1: Understand the Functions and Perform Preliminary Analysis

We are given the functions \(f(x) = 6 + \frac{1}{x}\) and \(g(x) = \frac{1}{x}\). Notice that both functions have a variable \(x\) in the denominator, which means \(x eq 0\) for both functions as division by zero is undefined. Hence, the domain for both individually is \((-\infty, 0) \cup (0, \infty)\).

Step 2: Calculate f + g

To find \(f+g\), we add the two functions: \[(f+g)(x) = f(x) + g(x) = \left(6 + \frac{1}{x}\right) + \frac{1}{x} = 6 + \frac{2}{x}.\] The domain of \(f+g\) is the same as the domain of \(f\) and \(g\), which is \((-\infty, 0) \cup (0, \infty)\).

Step 3: Calculate f - g

To find \(f-g\), subtract \(g(x)\) from \(f(x)\): \[(f-g)(x) = f(x) - g(x) = \left(6 + \frac{1}{x}\right) - \frac{1}{x} = 6.\] Interestingly, this simplifies to 6, a constant function with a domain of \((-\infty, 0) \cup (0, \infty)\).

Step 4: Calculate f * g

To find \(f \cdot g\), multiply the two functions: \[(f \cdot g)(x) = f(x) \cdot g(x) = \left(6 + \frac{1}{x}\right) \cdot \frac{1}{x} = \frac{6}{x} + \frac{1}{x^2}.\] The domain is \((-\infty, 0) \cup (0, \infty)\) because \(x\) cannot be zero.

Step 5: Calculate f / g

To find \(f/g\), divide \(f(x)\) by \(g(x)\): \[(f/g)(x) = \frac{f(x)}{g(x)} = \frac{6 + \frac{1}{x}}{\frac{1}{x}} = x(6 \cdot x + 1) = 6x + 1.\] The original division posed a constraint \(x eq 0\); hence the domain remains \((-\infty,0) \cup (0,\infty)\).

Key Concepts

Domain of a Function

When exploring functions in calculus, one of the first things to look at is the domain. The domain of a function is the complete set of values that the input, often represented as "x," can take to produce a valid output. In simple terms, it is the set of all possible x-values that won’t lead to any undefined results.

For instance, if a function has a denominator, like in the functions given such as \(f(x)=6+\frac{1}{x}\) and \(g(x)=\frac{1}{x}\), we must ensure that "x" does not make the denominator zero, as division by zero is undefined. This means "x" cannot be zero. Consequently, the domain for these functions, and indeed their resulting operations, is anything except zero: \( (-\infty, 0) \cup (0, \infty) \).

Understanding the domain is crucial because it specifies where the functions are applicable and prevents errors in calculations, especially when involving operations like addition, multiplication, or division.

Addition of Functions

When adding two functions together, like \(f(x)=6+\frac{1}{x}\) and \(g(x)=\frac{1}{x}\), you sum their outputs for each value of x in their domain. This creates a new function: \((f+g)(x) = f(x) + g(x) = 6 + \frac{2}{x}\).

The domain of the newly created function remains \( (-\infty, 0) \cup (0, \infty) \). This is because the restrictions on the x-values are inherited from both original functions, ensuring that no division by zero occurs.

A crucial point in the addition of functions is to ensure that terms are correctly combined and simplified where possible, as seen with the terms involving fractions here.

Multiplication of Functions

Multiplying functions involves creating a new function by taking the product of their outputs for the same x-value. For example, with \((f \cdot g)(x) = (6 + \frac{1}{x}) \cdot \frac{1}{x}\), you get \((f \cdot g)(x) = \frac{6}{x} + \frac{1}{x^2}\).

The domain of the multiplied function is still \( (-\infty, 0) \cup (0, \infty) \), as both functions have x in the denominator, demanding that x must not be zero.

When dealing with multiplication, watch for results that could potentially simplify further or result in complex fractions. It's helpful to break the expression down and ensure you're keeping track of any restrictions initially posed by each function.

Division of Functions

In dividing functions, the goal is to create a function where the value of one function at "x" is divided by the value of another. This can complicate things somewhat, as seen in \((f/g)(x) = \frac{6 + \frac{1}{x}}{\frac{1}{x}}\), which simplifies to \((f/g)(x) = 6x + 1\).

Even after simplification, the domain restriction remains the same: \( (-\infty, 0) \cup (0, \infty) \). This stems from the fact that initially x was a zero in both functions' denominators, regardless of any simplifications made afterward.

Whenever performing division, double-check both for mathematical accuracy and to adhere to the initial domain restrictions. Simplifying accurately without losing sight of what makes fractions undefined is vital in these problems.