Problem 12 - College Algebra

Question

Find \(f+g, f-g, f g,\) and \(\frac{f}{g}\). Determine the domain for each function. $$f(x)=\frac{1}{x}, g(x)=x-5$$

Step-by-Step Solution

Answer

The functions \(f+g\), \(f-g\), \(f*g\) and \(f/g\) are \(x + \frac{1}{x} - 5\), \(-x + \frac{1}{x} + 5\), \(1 - \frac{5}{x}\) and \(\frac{1}{x(x-5)}\) respectively. The domain for \(f+g\), \(f-g\) and \(f*g\) is \(R - \{0\}\) while the domain for \(f/g\) is \(R - \{0,5\}\).

Step 1: Function Addition (f+g)

Add the functions, \(f(x)\) and \(g(x)\), together. \[f(x)+g(x) = \frac{1}{x} + (x-5) = x + \frac{1}{x} - 5.\] The domain for this function will be all real numbers except zero, since dividing by zero is undefined. Therefore, the domain is \(R - \{0\}\).

Step 2: Function Subtraction (f-g)

Subtract the function \(g(x)\) from \(f(x)\). \[f(x)-g(x) = \frac{1}{x} - (x-5) = \frac{1}{x} - x + 5 = -x + \frac{1}{x} + 5.\] The domain for this function is also all real numbers except zero, so the domain is \(R - \{0\}\).

Step 3: Function Multiplication (f*g)

Multiply the functions, \(f(x)\) and \(g(x)\), together. \[f(x)*g(x) = \frac{1}{x} * (x-5) = (x-5)/x = 1 - \frac{5}{x}.\] Similar to the above functions, the domain for this function is all real numbers except 0. Therefore, the domain is \(R - \{0\}\).

Step 4: Function Division (f/g)

Divide the function \(f(x)\) by \(g(x)\). \[f(x)/g(x) = \frac{1/x}{x-5} = \frac{1}{x(x-5)}.\] Here, the function is undefined when x is either 0 or 5. Therefore, the domain of this function is all real numbers except 0 and 5. So, the domain is \(R - \{0,5\}\).

Key Concepts

Function Addition

When we talk about function addition, our goal is to combine two functions into one by simply adding their expressions together. For instance, given the functions \(f(x) = \frac{1}{x}\) and \(g(x) = x - 5\), we can add these functions by performing the operation \(f(x) + g(x)\).

This results in:

• \(f(x) + g(x) = \frac{1}{x} + (x - 5) = x + \frac{1}{x} - 5\).

It's important to determine the domain, which includes all the values of \(x\) for which the expression is defined.

• For this function addition, \(x\) cannot be zero because division by zero is undefined.
  
• Thus, the domain is all real numbers except zero, or \(R - \{0\}\).

Function Subtraction

Function subtraction follows a similar process as addition, but instead, you subtract one function from another. Given \(f(x) = \frac{1}{x}\) and \(g(x) = x - 5\), subtracting these gives us \(f(x) - g(x)\).

The subtraction looks like this:

• \(f(x) - g(x) = \frac{1}{x} - (x - 5) = \frac{1}{x} - x + 5 = -x + \frac{1}{x} + 5\).

As with addition, we must ensure the expression is defined across the domain:

• The function is undefined at \(x = 0\), so the domain is \(R - \{0\}\).

Function Multiplication

Function multiplication is about taking two functions and multiplying them to form a new function. If \(f(x) = \frac{1}{x}\) and \(g(x) = x - 5\), then multiplying them gives us \(f(x)g(x)\).

The result is:

• \(f(x) \cdot g(x) = \frac{1}{x} \cdot (x - 5) = \frac{x - 5}{x} = 1 - \frac{5}{x}\).

Check for domain constraints, as these determine where the function is valid:

• For multiplication, the expression is undefined at \(x = 0\).
• Thus, the domain is all real numbers except zero, \(R - \{0\}\).

Function Division

Function division involves dividing one function by another, forming a quotient function. Given \(f(x) = \frac{1}{x}\) and \(g(x) = x - 5\), dividing them, we find \(\frac{f}{g}\).

The division process is as follows:

• \(\frac{f(x)}{g(x)} = \frac{\frac{1}{x}}{x - 5} = \frac{1}{x(x-5)}\).

The domain requires checking where the expression is undefined:

• Division by 0 occurs when \(x = 0\) or \(x = 5\).
• Therefore, the domain excludes these values, leaving \(R - \{0, 5\}\).

Domain of a Function

The domain of a function identifies the complete set of possible input values (\(x\)) that will still produce valid outputs (\(f(x)\)). It’s crucial to determine these values to understand where a function is defined.

In the operations above:

• For addition, subtraction, and multiplication of our given functions, the domain excludes \(x = 0\).
• For division, additional restrictions apply since \(g(x)\) becomes 0 not only at \(x = 0\) but also at \(x = 5\).

This concept ensures that mathematical operations on functions are clear, and it helps us identify the valid intervals or set of numbers we can safely use in these functions.