Rational Expression - Coursework Example

Week One Discussion: Domains of Rational Expressions In simple words, the domain is the values (input) that go into a function (or a relation). In Real Number System, division by a zero is not defined that is it cannot be done. A rational number is expressed as a fraction that is a numerator divided by a denominator (p/q, q≠0). Many functions are defined using both a numerator and a denominator. Therefore, if we take the denominator value as a zero, the function (or relation) will be undefined according to the Real Number System.

This is the reason why a denominator cannot be zero.My first rational expression is (g^2 – 6g – 55)/g. The denominator value is g. To find the excluded value(s) for the rational expression, I need to set the denominator equal to zero.g = 0Thus, domain (D) for my first expression is the set of all real numbers excluding 0. In set notation, this can be written as D = {g | g ∈ ℜ, g ≠ 0}My second expression is (k^3 + k) /(k^2 + k – 42). The denominator value is (k^2 + k – 42). To find the excluded value(s) for the rational expression, I need to set the denominator equal to zero.

k^2 + k – 42 = 0 –42 is the product of 7 and –6, and k is the sum of 7k and –6k. Therefore, k^2 + (7 – 6)k + (7)(–6) = 0 k^2 + 7k – 6k + (7)(–6) = 0 Taking k common from the first two terms and –6 coming from last two terms k(k + 7) – 6(k + 7) = 0 (k + 7)(k – 6) = 0 Setting each factor equal to zero. (k + 7) = 0 or (k – 6) = 0From both sides, subtracting 7 from first expression and adding 6 to second expression. k = –7 or k = 6Thus, domain (D) for my second expression is the set of all real numbers excluding –7 and 6.

In set notation, this can be written as D = {k | k∈ ℜ, g ≠ –7, 6}Now, both of my rational expressions have excluded values in their domains. In the first expression, the excluded value is 0. The value of zero is excluded from the domain of the first expression because if inserted will make denominator value equal to zero, and thus, whole expression would become undefined. In the second expression, the excluded values are –7 and 6. The value of –7 and 6 are excluded from the domain of the second expression because if inserted will make denominator value equal to 0, thus, whole expression would become undefined.