Problem 42 - Precalculus Essentials

Question

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

Step-by-Step Solution

Answer

\(f+g = 6, f-g = 6 - 2\frac{1}{x}, fg = 6\frac{1}{x} - 1, \frac{f}{g} = 6x - 1\). The domain for each of these functions is all real numbers except 0.

Step 1: Calculation of \(f+g\)

Add the corresponding expressions of f(x) and g(x).\n So, \(f(x) + g(x) = (6-\frac{1}{x}) + \frac{1}{x}\).\n This simplifies to f(x) + g(x) = 6. The domain here is all real numbers except 0.

Step 2: Calculation of \(f-g\)

Subtract g(x) from f(x).\n So, \(f(x) - g(x) = (6-\frac{1}{x}) - \frac{1}{x}\).\n This simplifies to f(x) - g(x) = 6 - 2\frac{1}{x}. The domain here is all real numbers except 0.

Step 3: Calculation of fg

Multiply the expressions of f(x) and g(x).\n So, f(x) * g(x) = (6-\frac{1}{x}) * \frac{1}{x}\). \nThis simplifies to f(x) * g(x) = 6\frac{1}{x} - 1. The domain here is all real numbers except 0.

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

Divide f(x) by g(x).\n So, \(\frac{f(x)}{g(x)} = \frac{(6-\frac{1}{x})}{\frac{1}{x}}\).\n This simplifies to \(\frac{f(x)}{g(x)} = 6x - 1\). The domain here is all real numbers except 0.

Key Concepts

Domain of a Function

When discussing functions, the domain is the set of all possible input values (usually "x values") that a function can accept without causing any division by zero, or other undefined mathematical operations. For example, in the function

• \(f(x) = 6 - \frac{1}{x}\)

The term \(\frac{1}{x}\) means that x cannot be zero, as division by zero is undefined.
This is crucial when you determine the domain of a function involving fractions or roots.
For the given functions, both \(f(x)\) and \(g(x) = \frac{1}{x}\) fail at \(x = 0\), thus the domain avoids this value, using the notation: all real numbers except zero.

Addition of Functions

The process of adding functions involves combining two expressions to form a new function.
When you add two functions, you simply add their respective formulas together.
The addition of our example functions is:

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

Notice how the \( \frac{1}{x}\) and \(- \frac{1}{x}\) cancel each other out, leaving us with a simple constant, 6.
The domain in this situation remains all real numbers except zero, since the original components both contained \( \frac{1}{x}\), which is undefined at zero.

Subtraction of Functions

Subtraction of functions works similarly to addition. You subtract the formula of one function from another:

• \(f(x) - g(x) = (6 - \frac{1}{x}) - \frac{1}{x} = 6 - 2\frac{1}{x}\)

The subtraction results in \(6 - 2\frac{1}{x}\), which simplifies based on the algebraic operations between the two functions.
Just like with addition, the domain must exclude \(x = 0\) to avoid division by zero, thus remains all real numbers except zero.

Multiplication of Functions

To find the product of two functions, you multiply their expressions together.
The resulting function from our example is:

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

In this step, remember: you must distribute multiplication across each term in the bracket. This yields a new expression \( 6\frac{1}{x} - 1\).
The domain, as in previous examples, excludes zero to prevent undefined values from \( \frac{1}{x} \).

Division of Functions

Dividing functions involves placing one expression over the other:

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

This transformation simplifies because the division by a fraction is equivalent to multiplication by its reciprocal.
Remember: never divide by zero, as it remains undefined. Hence, the domain excludes both instances where the denominator could be zero. For both \(f(x)\) and \(g(x)\) at zero, the domain is all real numbers except zero.