Problem 14 - College Algebra

Question

Find \(f+g, f-g, f g,\) and \(\frac{f}{g}\). Determine the domain for each function. $$f(x)=6-\frac{1}{x}, g(x)=\frac{1}{x}$$

Step-by-Step Solution

Answer

The sum of the functions is 6, the difference is \(6 - \frac{2}{x}\), the product is \(6 \frac{1}{x} -1\), the quotient is \(6x -1\). The domain for each function is all real numbers except 0.

Step 1: Find the Sum of f and g

Sum of the functions \(f\) and \(g\) is given by \(f(x) + g(x)\). \nSo, \(f(x) + g(x) = 6-\frac{1}{x} + \frac{1}{x} = 6\)

Step 2: Find the Difference of f and g

Difference of the functions \(f\) and \(g\) is given by \(f(x) - g(x)\). \nSo, \(f(x) - g(x) = 6 -\frac{1}{x} - \frac{1}{x} = 6 - \frac{2}{x}\)

Step 3: Find the Product of f and g

The product of the functions \(f\) and \(g\) is given by \(f(x)g(x)\). \nSo, \(f(x)g(x) = (6-\frac{1}{x}) \times \frac{1}{x} = 6 \times \frac{1}{x} -1\)

Step 4: Find the Quotient of f and g

The quotient of the functions \(f\) and \(g\) is given by \(\frac{f(x)}{g(x)}\). \nSo, \(\frac{f(x)}{g(x)} = \frac{6-\frac{1}{x}}{\frac{1}{x}} = 6x -1\)

Step 5: Determine the Domain for Each Function

For the sum, difference, product, and difference quotient functions, x can't be 0, because you can't divide by zero. Consequently, the domain will be all real numbers except 0.

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values (or "x" values) that allow the function to produce a real number output.In mathematical terms, it's the set of all "x" for which the function is defined.
For example, if you have a function with a denominator, and that denominator can become zero for a certain "x" value, then that "x" value is excluded from the domain.

In our example with functions \(f(x) = 6 - \frac{1}{x}\), and \(g(x) = \frac{1}{x}\), both expressions have \(x\) in the denominator. This means that they are undefined for \(x=0\) since division by zero is not defined in mathematics. Hence, the domain for these functions is all real numbers except zero: \(x eq 0\).

When determining the domain of new functions created by combining other functions, such as their sum, difference, or product, the domain must respect the limitations present in the original functions. If either of the original functions is not defined at a certain point, the new function cannot be either.

Sum of Functions

The sum of two functions \(f\) and \(g\) refers to adding their outputs for any given input \(x\).Mathematically, it is represented as \((f + g)(x) = f(x) + g(x)\).Calculating the sum involves performing basic arithmetic on the right-hand side of the functions.

In our original exercise, this means:

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

To find \(f(x) + g(x)\), add the two functions:\[f(x) + g(x) = 6 - \frac{1}{x} + \frac{1}{x} = 6\]
The plus and minus \(\frac{1}{x}\) terms cancel each other out, leaving the constant \(6\).

The domain of the sum is determined just like the initial functions: \(x eq 0\). It's important to consider where the original functions were defined.

Difference of Functions

The difference of two functions \(f\) and \(g\) means subtracting \(g(x)\) from \(f(x)\), shown as \((f - g)(x) = f(x) - g(x)\).Utilize arithmetic to simplify the expression:

Given:

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

Subtract the functions:\[f(x) - g(x) = 6 - \frac{1}{x} - \frac{1}{x} = 6 - \frac{2}{x}\]
Here, the \(\frac{1}{x}\) terms combine to \(\frac{2}{x}\), resulting in a cleaner expression.

As with the sum, the domain remains \(x eq 0\).This restriction is due to the division by \(x\) initially present in the component functions.

Product of Functions

When multiplying two functions together, you find the product \(f \cdot g\), which is calculated by multiplying their outputs for a given \(x\): \((f \cdot g)(x) = f(x) \cdot g(x)\).This operation uses simple multiplication, respecting ordinary algebraic rules.

For the functions:

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

The product is:\[f(x) \cdot g(x) = \left(6 - \frac{1}{x}\right) \cdot \frac{1}{x} = \frac{6}{x} - 1\]Here, distribute \(\frac{1}{x}\) across the terms within the brackets.

For the domain of the product, consistency underlines the rule: \(x eq 0\).This ensures the function remains valid and defined similarly to its original components.