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Rational functions are mathematical expressions representing a ratio of two polynomials. They are written in the form \(f(x) = \frac{p(x)}{q(x)}\), where \(p(x)\) and \(q(x)\) are polynomials and \(q(x) \eq 0\). A key characteristic of rational functions is their potential to exhibit points of discontinuity, which occur where the denominator \(q(x)\) equals zero. Understanding rational functions is essential for analyzing their behavior, including identifying their domain (all possible \(x\)-values where they are defined), range, asymptotes, and intervals of continuity.

In the provided exercise, \(f(x)=\frac{x+1}{x-1}\), \(p(x)\) is \(x+1\) and \(q(x)\) is \(x-1\). The domain of a rational function is all real numbers except the zeros of the denominator, which in this case is \(x \eq 1\) as solving \(x-1=0\) yields the excluded value.

Points of discontinuity in a function are values of \(x\) at which the function fails to be continuous. For rational functions, discontinuities typically occur where the denominator is zero, since division by zero is undefined. These points are also known as singularities. Depending on the nature of the polynomials, discontinuities can be removable, where there's a hole in the graph, or non-removable, associated with vertical asymptotes or infinite discontinuities.

To identify points of discontinuity, one must solve for \(x\) in the denominator's equation when set to zero. If the numerator does not also become zero at these points, the function has non-removable discontinuities. In the exercise, \(x-1=0\) yields \(x=1\) as the point of discontinuity for \(f(x)\), and since \(x+1\) does not equal zero when \(x=1\), we confirm that \(x=1\) is a non-removable point of discontinuity.

Real values refer to the set of all possible numbers on the continuum of the real number line, including all rational and irrational numbers. In the context of function continuity, we're interested in which real values a function can take without interruption or gaps. A function is continuous at a point if the function value is defined at that point, the limit exists, and both the function value and the limit are equal.

For rational functions, while they can take many real values, they are not continuous across all real numbers due to points of discontinuity. In our example, \(f(x) = \frac{x+1}{x-1}\) is defined for all real values of \(x\) except \(x=1\), making it continuous everywhere else on the real number line. This exclusion creates a domain for our function which is every real number except \(x=1\), thus ensuring the function's continuity wherever it is defined.