Construct a rational function that has a vertical asymptote at x = 3 and a removable discontinuity at . Explain how you determined your answer.
One possible rational function that satisfies the given conditions is:
f(x) = (x - 3)/(x - 1)
To determine this function, we start by analyzing the conditions given:
1) A vertical asymptote at x = 3: A vertical asymptote occurs when the denominator of a rational function equals zero and the numerator does not. In this case, we want a vertical asymptote at x = 3, so the denominator should contain a factor of (x - 3) and the numerator should not. Therefore, (x - 3) should be the denominator.
2) A removable discontinuity at x = 1: A removable discontinuity occurs when both the numerator and the denominator of a rational function equal zero at some point, but can be simplified or factored so that the common factor cancels out. In this case, we want a removable discontinuity at x = 1, so both the numerator and the denominator should have a factor of (x - 1). However, since we only want a removable discontinuity, we need to ensure that the same factor cancels out in both the numerator and the denominator. Therefore, we include (x - 1) in both the numerator and the denominator.
By incorporating these conditions, we obtain the rational function f(x) = (x - 3)/(x - 1), which has a vertical asymptote at x = 3 (since the denominator is zero at x = 3) and a removable discontinuity at x = 1 (since both the numerator and the denominator are zero at x = 1).
Please note that there could be other rational functions that satisfy the given conditions, and the function provided above is just one example.