How do I create a rational inequality with solutions (-2,-1) and (1, ∞)?

Answer 1

First thing is to recognize that these are not points, they are intervals. So we only want to accept values that are inside those boundaries. That tells us that there are vertical asymptotes at -2 and -1 and 1.

So the denominator of our fraction will have (x + 2), (x + 1) and (x - 1) in it.

Something had to go in the numerator. So I put x² in the numerator and used DESMOS to graph it. (A hand held graphing calculator will also work.).

This gave me an interesting graph where the asymptotes were in the right places and my desired values for y were all positive. So set the fraction >0 and you will only consider those values above the x-axis.

Answer: x² > 0

(x + 2)(x + 1) (x - 1)

Answer 2

We can use the endpoints of the given solution intervals as our starting point to generate a rational inequality with the required solutions:

What we want is a rational expression the numerator or denominator of which has zeros at the given endpoints, in this case x = -2, -1, and 1. So, the factors (x + 2) , (x + 1), and (x - 1) will appear in our inequality.

Because the solution set is comprised of open intervals (that don't include the endpoints), it might be easiest to place all of these factors in the denominator of the rational expression (zeros of the denominator will always be excluded from the solution set, since they make the rational expression undefined). We can use either < 0 or > 0 in our inequality, and if we needed, we could just multiply the expression by -1 to make it work for the given intervals.

We're ready to create the rational expression, which will be 1 / [(x + 2)(x + 1)(x - 1)]. Because this has 3 linear factors and a positive lead coefficient, its end behavior will be the same as a cubic (odd degree) with a positive lead coefficient, whose graph starts in QIII and ends in QI. So it starts with negative y-coordinates and finishes with positive ones.

That means 1 / [(x + 2)(x + 1)(x - 1)] > 0 will have the given solution set.

We can double-check this by graphing the rational function on the left-hand side of the inequality and noting that it has vertical asymptotes at x = -2, -1, and 1, and that it changes signs at each of these. Its graph is above the x-axis (positive) on the 2 given intervals, (-2 , -1) and (1 , ∞).