91Ó°ÊÓ

Rational functions are a fascinating type of function in mathematics. They are expressed as the ratio of two polynomials. A polynomial can be something as simple as a number or a variable, or more complex expansions involving both, like \(2x^2 + 3x - 5\).
A rational function takes the form \(f(x) = \frac{P(x)}{Q(x)}\), where both \(P(x)\) and \(Q(x)\) are polynomial functions and \(Q(x) eq 0\). The non-zero condition for \(Q(x)\) is crucial to ensuring the function is defined.
In our exercise, the rational function \(f(x) = \frac{1}{x+8} + \frac{3}{x-10}\) is made up of two parts. Each fraction has its own denominator, and understanding these denominators is key to finding the domain of the function. This function requires us to exclude any values that would make either denominator zero.

Interval notation is a simplified way of expressing the set of numbers we are interested in, particularly useful in describing the domain or range of functions. It often uses parentheses \(()\) and brackets \([]\) to denote intervals of real numbers.
Here's a quick breakdown of how to use it:

• Parentheses \(()\) mean that a number is not included in the set.
• Brackets \([]\) mean that a number is included.

In our exercise, to express that the domain of the function excludes the points \(x = -8\) and \(x = 10\), we use interval notation.
It is written as \((-∞, -8) \cup (-8, 10) \cup (10, ∞)\). This statement means all real numbers except \(-8\) and \(10\) are included. Each segment of the interval has parentheses, indicating the exclusion of exactly these points.

The exclusion of values from the domain of a rational function is essential to ensure the function remains defined. In a rational function, any value that causes the denominator to become zero must be excluded.
A division by zero would make the function undefined, as dividing by zero is not possible within the set of real numbers.
In the exercise you worked on, the rational function \(f(x) = \frac{1}{x+8} + \frac{3}{x-10}\) tells us that if \(x = -8\) or \(x = 10\), the function cannot be defined. That's because substituting either of these values will cause a zero in the denominator of one of the fractions:

• \(x = -8\) makes the first denominator zero.
• \(x = 10\) does the same for the second denominator.

Thus, we exclude these values from the domain, making sure any x-value that results in a zero denominator is left out.