Problem 41 - Calculus Volume 1

Question

For the following exercises, 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

Find f+g, f-g, f⋅g, and f/g, and note domain exclusions (x=0).

Step 1: Understanding the Functions

The functions given are \( f(x) = 6 + \frac{1}{x} \) and \( g(x) = \frac{1}{x} \). We'll find \( f+g \), \( f-g \), \( f \cdot g \), and \( \frac{f}{g} \), as well as determine the domains for each.

Step 1: Addition of Functions \( f+g \)

To find \( f(x) + g(x) \), add the functions together: \[ f(x) + g(x) = \left( 6 + \frac{1}{x} \right) + \frac{1}{x} = 6 + \frac{1}{x} + \frac{1}{x} = 6 + \frac{2}{x}. \]The domain of \( f(x) + g(x) \) is all real numbers except \( x = 0 \), as division by zero is undefined.

Step 2: Subtraction of Functions \( f-g \)

To find \( f(x) - g(x) \), subtract the functions:\[ f(x) - g(x) = \left( 6 + \frac{1}{x} \right) - \frac{1}{x} = 6. \]The domain of \( f(x) - g(x) \) is all real numbers except \( x = 0 \).

Step 3: Multiplication of Functions \( f \cdot g \)

To find \( f(x) \cdot g(x) \), multiply the functions:\[ f(x) \cdot g(x) = \left( 6 + \frac{1}{x} \right) \cdot \frac{1}{x} = \frac{6}{x} + \frac{1}{x^2}. \]The domain of \( f(x) \cdot g(x) \) is all real numbers except \( x = 0 \).

Step 4: Division of Functions \( \frac{f}{g} \)

To find \( \frac{f(x)}{g(x)} \), divide the functions:\[ \frac{f(x)}{g(x)} = \frac{6 + \frac{1}{x}}{\frac{1}{x}} = 6x + 1. \]The domain for \( \frac{f(x)}{g(x)} \) is all real numbers except \( x = 0 \), where \( g(x) = 0 \) since it causes division by zero.

Key Concepts

Addition of Functions

Adding functions involves combining two functions so that their output values are also added together for each input. For two given functions, like in our exercise where we have \( f(x) = 6 + \frac{1}{x} \) and \( g(x) = \frac{1}{x} \), the addition \( f + g \) is performed by simply adding corresponding outputs from each function. This results in:

• \( f(x) + g(x) = \left( 6 + \frac{1}{x} \right) + \frac{1}{x} = 6 + \frac{2}{x} \).

The domain for the sum of these functions remains all real numbers except \( x = 0 \), because \( x = 0 \) would require us to divide by zero, which is undefined.
When adding functions, always ensure that the operations within the functions can be performed without causing any mathematical inconsistencies, such as division by zero.

Subtraction of Functions

Subtraction of functions involves taking one function's output and subtracting the output of another function for any given input value. With our functions \( f(x) = 6 + \frac{1}{x} \) and \( g(x) = \frac{1}{x} \), subtraction is straightforward:

• \( f(x) - g(x) = \left( 6 + \frac{1}{x} \right) - \frac{1}{x} = 6 \).

Here, the result is a constant function \( 6 \), as the \( \frac{1}{x} \) terms cancel out. Even though \( 6 \) is simple, the domain needs consideration. The domain still excludes \( x = 0 \), since both original functions are undefined at that point.
Always take note that while subtraction can often simplify functions, one must track how operation affects the domain.

Multiplication of Functions

When multiplying functions, each output value of one function is multiplied by the output value of the other. Considering our example, we input:

• \( f(x) \cdot g(x) = \left( 6 + \frac{1}{x} \right) \cdot \frac{1}{x} = \frac{6}{x} + \frac{1}{x^2} \).

The resulting function combines rational components, and so extra care must be taken to understand the domain. Here, the domain remains as all real numbers, excluding \( x = 0 \), due to division by zero being undefined.
Multiplying functions can lead to new rational expressions, and it's important to simplify these to identify their domains properly. Always check for restrictions from both functions involved.

Division of Functions

Division of functions can be tricky. It involves dividing the output of one function by the output of another. Here, for our functions \( f(x) = 6 + \frac{1}{x} \) and \( g(x) = \frac{1}{x} \):

• \( \frac{f(x)}{g(x)} = \frac{6 + \frac{1}{x}}{\frac{1}{x}} = 6x + 1 \).

During division, simplifying complexities often reveals a function in a different form, as seen here. The domain is consistent with previous results, still lacking \( x = 0 \), as this causes the functions involved to be undefined.
In division, look carefully at where the denominator is zero, as this defines key restrictions in the function's domain.

Domain of a Function

The domain is a set of all possible inputs that allow a function to be calculated without error. When finding the domain for functions like those in our exercise, it’s crucial to look for points where operations like division by zero happen. For example, in \( f(x) = 6 + \frac{1}{x} \), \( x = 0 \) is undefined because division by zero is not possible.
For each operation \((f+g), (f-g), (f \cdot g), \text{and} \frac{f}{g})\), \( x = 0 \) remains an excluded value across all functions derived.
When determining the domain:

• Identify any values that make a denominator zero.
• Review any other operations that are similarly undefined, like taking square roots of negative numbers in real numbers.

Understanding and correctly defining the domain is essential so the function can be applied accurately in any problem-solving context.