Introduces the fundamental definition of a rational function, how to determine its domain, and identifies the concept of a point of discontinuity where the function is undefined.
A rational function, \(f(x)\), is any function that can be written as the ratio of two polynomial functions, \(P(x)\) and \(Q(x)\). The general form is \(f(x) = \frac{P(x)}{Q(x)}\), where \(Q(x) \neq 0\).
How do you determine the domain of a rational function?
The domain consists of all real numbers except for the x-values that make the denominator equal to zero. To find these exclusions, set the denominator polynomial \(Q(x) = 0\) and solve for \(x\).
What is a 'point of discontinuity' in a rational function?
A point of discontinuity is an x-value for which the function is undefined. These are the values of \(x\) that make the denominator zero.
What are the two types of discontinuities found in rational functions?
The two types are:
1. Removable Discontinuities (Holes)
2. Non-removable Discontinuities (Vertical Asymptotes)
Find the domain of the function $$f(x) = \frac{x+5}{x-3}$$
Set the denominator to zero: \(x-3=0\), which gives \(x=3\). The domain is all real numbers except \(3\), written as \((-\infty, 3) \cup (3, \infty)\) or \(\{x \in \mathbb{R} | x \neq 3\}\).
Find the points of discontinuity for \(g(x) = \frac{x}{x^2 - 4}\).
Set the denominator to zero: \(x^2 - 4 = 0\). Factoring gives \((x-2)(x+2) = 0\). The points of discontinuity occur at \(x=2\) and \(x=-2\).
What is the domain of \(h(x) = \frac{3x^2}{x^2+1}\)?
Set the denominator to zero: \(x^2+1=0\). This equation has no real solutions (\(x^2 = -1\)). Therefore, there are no real numbers that make the denominator zero. The domain is all real numbers, \((-\infty, \infty)\).
What is the fundamental difference between the cause of a hole versus a vertical asymptote?
A hole occurs from a factor that is common to both the numerator and denominator and can be canceled. A vertical asymptote occurs from a factor that remains in the denominator after all simplification is done.
