Problem 43 - College Algebra Essentials

Question

Find \(f+g, f-g,\) fg, and \(\frac{f}{g} .\) Determine the domain for each function. $$f(x)=\frac{5 x+1}{x^{2}-9}, g(x)=\frac{4 x-2}{x^{2}-9}$$

Step-by-Step Solution

Answer

\((f+g)(x) = \frac{9x-1}{x^{2}-9}\), \((f-g)(x) = \frac{x+3}{x^{2}-9}\), \((fg)(x) = \frac{20x^2 - 6x - 2}{x^{4}-18x^{2}+81}\), \(\frac{f}{g}(x) = \frac{5x+1}{4x-2}\). The first three functions are defined on all real numbers except ±3, and the fourth function is defined for all real numbers except \(x = \frac{1}{2}\).

Step 1: Define the functions

We have two functions to work with: \(f(x) = \frac{5x+1}{x^{2}-9}\) and \(g(x) = \frac{4x-2}{x^{2}-9}\). Notice that both functions share the same denominator, which will simplify the calculations.

Step 2: Find \(f+g\)

Adding functions simply involves adding the corresponding numerators: \[(f + g)(x) = \frac{5x + 1 + 4x - 2}{x^{2} - 9} = \frac{9x - 1}{x^{2} - 9}\]. The domain of \(f+g\) is all real numbers except for \(+3\) and \(-3\) where the denominator becomes zero.

Step 3: Find \(f - g\)

Following the same logic as in the previous step, we subtract the numerators: \[(f - g)(x) = \frac{5x + 1 - (4x - 2)}{x^{2} - 9} = \frac{x + 3}{x^{2} - 9}\]. The domain of \(f - g\) is all real numbers except for \(+3\) and \(-3\) where the denominator becomes zero.

Step 4: Find \(fg\)

Multiplying functions involves multiplying the numerators together and the denominators together: \[(fg)(x) = \frac{(5x + 1)(4x - 2)}{(x^{2} - 9)^2} = \frac{20x^2 - 6x - 2}{x^{4} -18x^2 + 81}\]. The domain of \(fg\) is all real numbers except ±3.

Step 5: Find \(\frac{f}{g}\)

Dividing one function by another involves multiplying the numerator of the first function by the denominator of the second function and the denominator of the first function by the numerator of the second function: \[\frac{f}{g}(x) = \frac{(5x + 1)/(x^2 -9)}{(4x - 2)/(x^2 - 9)} = \frac{5x + 1}{4x - 2}\]. Notice that the common denominator cancels out. The domain of \(\frac{f}{g}\) is all real numbers except \(\frac{1}{2}\) where the denominator becomes zero.

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values (usually "x" values) that the function can accept without leading to undefined or impossible outputs. When dealing with rational functions, we must be cautious about the values that make the denominator zero since division by zero is undefined.
In the exercise, both functions, \( f(x) = \frac{5x+1}{x^2-9} \) and \( g(x) = \frac{4x-2}{x^2-9} \), have a common denominator of \( x^2-9 \). This simplifies our job significantly because we only need to consider when this denominator equals zero.
The expression \( x^2-9 \) is zero when \( x = 3 \) or \( x = -3 \). Therefore, the domain for any combination or operation involving \( f \) and \( g \), such as \( f+g \), \( f-g \), or \( fg \), will be all real numbers except \( x = 3 \) and \( x = -3 \).
When it comes to \( \frac{f}{g} \), there's an additional constraint due to the new denominator, \( 4x-2 \), used in the division process. Thus, \( x = \frac{1}{2} \) is also excluded because it makes the denominator zero, alongside the previously noted constraints.

Rational Functions

Rational functions are a class of functions defined by the division of two polynomials. The general form of a rational function is \( \frac{p(x)}{q(x)} \), where both \( p(x) \) and \( q(x) \) are polynomials. These functions can have complex domains due to the restrictions created by the denominator.
In our example, the rational functions \( f(x) = \frac{5x+1}{x^2 - 9} \) and \( g(x) = \frac{4x-2}{x^2 - 9} \) are particularly notable because they share the same denominator. This is critical as it guarantees that

• Simplification is more straightforward when performing operations.
• We can easily determine their common domain restrictions.

Rational functions can also have vertical asymptotes, which occur at the values that make the denominator zero. In our case, \( x = 3 \) and \( x = -3 \) are examples of where vertical asymptotes would occur, hence they are not included in the domain.

Function Operations

Function operations involve performing standard mathematical operations on functions, such as addition, subtraction, multiplication, and division, just as you might with regular numbers. Each operation carries its own rules, especially when dealing with rational functions. Here's a quick guide:

• Addition \((f+g)\): You add by combining like terms from the numerators only, as the denominator remains the same (assuming it's common as in this case).
  Example: \( \frac{5x+1}{x^2-9} + \frac{4x-2}{x^2-9} = \frac{9x-1}{x^2-9} \)
• Subtraction \((f-g)\): Similarly, you subtract the numerators. Make sure to distribute negative signs correctly.
  Example: \( \frac{5x+1}{x^2-9} - \frac{4x-2}{x^2-9} = \frac{x+3}{x^2-9} \)
• Multiplication \((fg)\): Multiply the numerators together and denominators together. This often leads to a more complex denominator.
  Example: \( \frac{(5x+1)(4x-2)}{(x^2-9)^2} = \frac{20x^2-6x-2}{x^4-18x^2+81} \)
• Division \((\frac{f}{g})\): Multiply by the reciprocal of the divisor, which affects the domain restrictions as the new denominator could introduce new problematic values.
  Example: \( \frac{\frac{5x+1}{x^2-9}}{\frac{4x-2}{x^2-9}} = \frac{5x+1}{4x-2} \)

Understandably, each operation could affect the function's domain, particularly when divisions result in zero denominators. Always reassess the domain after performing each operation.