In solving a rational inequality there are times when we do not include an endpoint although the inequality is?-Collection of common programming errors

  • Can’t be certain of your question, but you might be refering to those problems involving a nonstrict inequality (including the “or equal to” option) in which a particular value would correspond to a zero in the denominator. For example, consider the inequality (x+1)/(x-2) ≥ 0 To solve this, we would look for those values of x for which the numerator and denominator are zero. These are -1 and 2. We can split the real numbers into three intervals (-∞, -1), (-1,2), and (2, ∞) For -∞ < x < -1, the expression on the left is positive. For -1 < x < 2, it is negative. And for 2 < x < ∞, it is again positive. So our solution set should include (- ∞, -1) and (2,∞). The inequality is not strict, so we need to include those x-values for which it is zero. It is zero if x = -1. Our solution set contains -1. The solution set is (-∞, -1] U (2,∞)

    Note, that the solution set does not contain 2. Why? Because the expression is not zero at 2, it is undefined. For a nonstrict inequality, the solution set will contain those values that make the numerator zero. It can not contain those that make the denominator zero as these are not in the domain of the express. Most importantly, the substitution x=2 doesn’t make the expression greater than or equal to zero—it doesn’t make it any number at all.

  • Originally posted 2013-11-09 21:43:03.