Problem 55 - Algebra 2

Question

Let \(f(x)=3 x^{2}+2 x-8\) and \(g(x)=x+2 .\) Perform each function operation and then find the domain. $$ \frac{f(x)}{g(x)} $$

Step-by-Step Solution

Answer

The result of the division of \(f(x)\) by \(g(x)\) is \(h(x) = \frac{3x^2+2x-8}{x+2}\). The domain of \(h(x)\) is all real numbers except \(x=-2\). \(-2\) is excluded from the domain because it makes the denominator equal to zero, and division by zero is undefined.

Step 1: Perform the operation

Perform the division of the function \(f(x)\) by the function \(g(x)\). This results in the function \(h(x)\) such that: \[h(x) = \frac{f(x)}{g(x)} = \frac{3x^2+2x-8}{x+2}\]

Step 2: Simplify the function

Check if the function \(h(x)\) can be further simplified. Here, since the denominator 'x+2' does not factorize the numerator '3x^2+2x-8', the function \(h(x)\) itself is the simplest form.

Step 3: Determine the domain

The domain of a function is all the values of 'x' for which the function is defined. Here, since this is a division of functions, the only restriction on the domain is that the denominator should not be zero. Therefore, solve the equation \(x+2 = 0\) to find the value that x cannot take. Solving that gives \(x=-2\).

Key Concepts

Domain of a Function

The domain of a function represents the set of all possible input values (typically represented by \(x\)) for which the function is defined. For most polynomial functions, the domain is all real numbers because they are defined everywhere on the real number line. However, when we divide functions, the domain can be restricted due to the possibility of division by zero.
To determine the domain of a function created by dividing two functions, like our function \(h(x) = \frac{3x^2+2x-8}{x+2}\), we need to make sure that the denominator is never equal to zero. Since division by zero is undefined in mathematics, identifying and excluding these points from the domain is crucial.
In this exercise, solve for when the denominator, \(x+2\), equals zero:

• Set \(x+2=0\).
• Solve to find \(x = -2\).

Thus, the domain of \(h(x)\) is all real numbers except \(x = -2\). You can also express this in interval notation as \((-\infty, -2) \cup (-2, \infty)\).

Division of Functions

Function operations allow us to create new functions from existing ones. Among these operations, division involves creating a quotient function. Given two functions \(f(x)\) and \(g(x)\), their division is represented by \(\frac{f(x)}{g(x)}\).
To divide two functions as in our problem, you simply place one function in the numerator and the other in the denominator:

• Numerator: \(f(x) = 3x^2 + 2x - 8\)
• Denominator: \(g(x) = x + 2\)

This results in the new function \(h(x) = \frac{3x^2+2x-8}{x+2}\).
This kind of division can introduce restrictions into the domain because the denominator (\(g(x)\)) must not be zero. So it's not just about performing the division; it's also critical to ensure the resulting function remains mathematically valid by checking the denominator.

Simplifying Expressions

Simplifying expressions is an important skill in algebra, especially when dealing with functions. The purpose of simplifying is to make the expression as easy to work with as possible while not altering its value.
In the context of dividing functions, simplification often involves reducing the numerator and the denominator to create the simplest form of the quotient. However, it’s essential to recognize when a function is already in its simplest form, as with \(h(x) = \frac{3x^2+2x-8}{x+2}\).
To check simplification:

• Examine if the numerator can be factored in such a way that it has a common factor with the denominator.
• Try to identify any common factors between the top and bottom.

If no common factors exist, you have reached the simplest form. For our example, the polynomial in the numerator doesn't factor neatly with the \(x+2\) in the denominator, thus \(h(x)\) remains unchanged.