Problem 43 - Precalculus Essentials

Question

Find \(f+g, f-g,\) fg, and \(\frac{f}{x}\). 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

\( (f+g)(x)=\frac{9x-1}{x^{2}-9} \) with domain all real numbers except \(x = \pm 3\), \n (f-g)(x)=\frac{x+3}{x^{2}-9} \) with domain all real numbers except \(x = \pm 3\), \n (fg)(x)=\frac{20x^{2}+2x-2}{x^{4}-18x^{2}+81} \) with domain all real numbers except \(x = \pm 3\), \n \(\frac{f}{g}(x)=\frac{5x+1}{4(x-0.5)} \) with domain all real numbers except \(x = 0.5\)

Step 1: Find \(f + g\)

Add the functions \(f\) and \(g\) together: \n\( (f + g)(x)=f(x)+g(x)=\frac{5x+1}{x^{2}-9}+\frac{4x-2}{x^{2}-9}=\frac{5x+1+4x-2}{x^{2}-9}=\frac{9x-1}{x^{2}-9}\). The domain is all real numbers except \(x = \pm 3\)

Step 2: Find \(f - g\)

Subtract \(g\) from \(f\): \( (f - g)(x)=f(x)-g(x)=\frac{5x+1}{x^{2}-9}-\frac{4x-2}{x^{2}-9}=\frac{5x+1-4x+2}{x^{2}-9}=\frac{x+3}{x^{2}-9}\). The domain is all real numbers except \(x = \pm 3\).

Step 3: Find \(fg\)

Multiply \(f\) and \(g\) together: \( (fg)(x)=f(x)g(x)=\frac{(5x+1)(4x-2)}{(x^{2}-9)^2}=\frac{20x^{2}-2x+4x-2}{x^{4}-18x^{2}+81}=\frac{20x^{2}+2x-2}{x^{4}-18x^{2}+81}\). The domain is all real numbers except \(x = \pm 3\).

Step 4: Find \(\frac{f}{g}\)

Divide \(f\) by \(g\): \( \frac{f}{g}(x)=\frac{f(x)}{g(x)}=\frac{5x+1}{4x-2}=\frac{5x/4+1/4}{x-2/4}=\frac{5x+1}{4(x-0.5)}\). The domain is all real numbers except \(x = 0.5\).

Key Concepts

Domain of a Function

Understanding the domain of a function is essential in precalculus. It represents the set of all input values, or 'x' values, for which the function is defined. In other words, these values can be plugged into the function without causing any mathematical errors, such as division by zero. For instance, the domain of a simple function like \( f(x) = x + 1 \) includes all real numbers, since adding one to any real number is always possible.

In our exercise, the functions \( f(x) \) and \( g(x) \) have denominators that include the term \( x^2 - 9 \) which factors to \( (x-3)(x+3) \). Therefore, the domain for both functions excludes \( x = \)3 and \( x = -3 \) to prevent division by zero. Thus, the domain for \( f + g \) and \( f - g \) is all real numbers except \( x = \pm 3 \). Meanwhile, for \( \frac{f}{g} \) the domain excludes \( x = 0.5 \), where \( g(x) \) would be zero, leading to undefined division.

Combining Functions

Combining functions involves creating a new function by adding, subtracting, multiplying, or dividing other functions. This transformation allows us to explore complex relationships between functions and understand how they interact with each other.

For example, by adding \( f(x) \) and \( g(x) \) from our exercise, we form the combined function \( (f + g)(x) \) by simply adding the numerators over the common denominator. This is possible because the functions share the same denominator of \( x^2 - 9 \). The process is similar for subtraction, where \( (f - g)(x) \) is found by subtracting the numerators. The key is always to combine the functions in a way that respects the arithmetic operation involved and considers potential restrictions on the domain caused by the combination.

Rational Functions

Rational functions are fractions where the numerator and denominator are polynomials. The function \( f(x)=\frac{5x+1}{x^{2}-9} \) is a rational function because both the numerator \( 5x+1 \) and the denominator \( x^2 - 9 \) are polynomials.

For rational functions, we cannot have zero in the denominator since division by zero is undefined. The values that would make the denominator zero are excluded from the function's domain. In our original exercise, since the denominator is the same for \( f(x) \) and \( g(x) \), it becomes straightforward to combine these functions through addition, subtraction, or multiplication without changing the denominator.

Operation on Functions

Operation on functions, such as those shown in the exercise, can extend beyond addition and subtraction to multiplication and division. When we multiply functions \( f(x) \) and \( g(x) \), denoted by \( (fg)(x) \) or \( f(x) \cdot g(x) \), we multiply the numerators together and do the same for the denominators. Care must be taken to expand and simplify the resulting expressions where possible.

Division of functions, written as \( \frac{f}{g}(x) \) or \( \frac{f(x)}{g(x)} \) is performed by multiplying \( f(x) \) by the reciprocal of \( g(x) \). This operation introduces a new consideration for the domain since we must exclude any 'x' values that make \( g(x) \) zero, as this would lead to division by zero, which is undefined.