uc3rmw_0205e04

Hi there. In this example, we will solve the rational inequality 2 over x minus 1 is greater than or equal to 1 over x minus 2. The solution to a polynomial inequality involves a five-step process. In step 1, we want to arrange the inequality so that the 0 is on the right-hand side. So let's subtract 1 over x minus 2 from the right-hand side. We have that the left-hand side is equal to 2 over x minus 1 minus 1 over x minus 2. And that's all now greater than or equal to 0. We can simplify the expression on the left-hand side of the inequality symbol by finding the common denominator. We multiply the first fraction, 2 over x minus 1, by x minus 2 over x minus 2. And then the second fraction, 1 over x minus 2, we'll multiply by x minus 1 over x minus 1. And that's all greater than or equal to 0. Now we have the common denominator x minus 1 times x minus 2 for both fractions. And we'll go ahead and combine the numerators. We have 2 times x minus 2 minus x minus 1, in parentheses, all over that common denominator, x minus 1 times x minus 2. This is all still greater than or equal to 0. We can simplify the numerator now. We have to expand everything out and combine like terms. We have 2x minus 4 minus x plus 1, all over our common denominator, x minus 1 times x minus 2, greater than or equal to 0. 2x minus x is equal to x, and minus 4 plus 1 is equal to minus 3. And that's all over, again, x minus 1 times x minus 2. That's all greater than or equal to 0. So now we have a simplified single fraction expression on the left-hand side of the inequality symbol. We're going to go ahead and assign that expression the functional assignment R of x. In step 2, we're going to factor R of x and find the real zeros of the numerator and the denominator. These numbers will form the boundary points. So the real zeros of the numerator are found by setting the numerator equal to 0. The numerator is x minus 3. So we're going to set that equal to 0 and solve for x. We have that x equals 3. We'll also find the zeros of the denominator. So we're going to set the denominator equal to 0. The denominator is our common denominator x minus 1 times x minus 2. So we're going to set that equal to 0 and solve for x. We set each linear factor equal 0. We have x minus 1 equals 0 or x minus 2 equals 0. The solutions are x equal 1 and 2 from both of those linear equations. And so the boundary points are the points 1, 2, and 3. So now we move on to step 3, and we want to locate the boundary points on the number line. These boundary points are going to divide the number line into separate intervals. Again, our boundary points are 1, 2, and 3. So we plot that on the real number line. Here's the point 1. Here's the point 2. And here's the point 3. That separates the real number line into four separate intervals. We have the first interval from minus infinity to 1, the second one from 1 to 2, the third from 2 to 3, and a fourth from 3 to infinity. We see that the boundary points at 1 and 2 are marked as open circles to indicate that they do not belong to the domain of the rational function R of x. The boundary point at 3 is a solid circle. Now, in step 4a, we're going to use the graph method. We want to determine the sign of the rational function R of x equal to 1 N of x over D of x and form the corresponding polynomial P of x equal N of x times D of x. We're going to graph P of x using the end behavior and multiplicity where it crosses or touches the x-axis of the zeros of P of x and determine the sign of P of x on the intervals determined by the boundary points. Now, the sign of P of x is the same as the sign of R of x on the intervals found in step three. So we want to calculate the leading term for P of x. Again, P of x is equal to the product of the numerator and denominator of our rational function, R of x, so that is it's x minus 3 times x minus 2 times x minus 1. The leading term is calculated by just taking the product of the leading terms of each of these linear factors. So we have x times x times x equals x cubed. And therefore, P of x is equal to x cubed plus, then lower degree powers of x. And that is that P of x is a polynomial of degree 3, and the leading coefficient is a sub 3 equal to 1, where 1 is greater than 0. That is, as x becomes large positive, P of x becomes large positive. And as x becomes large negative, P of x becomes large negative, as well. Now, because 1, 2, and 3 are zeros of odd multiplicity the graph of P of x crosses the x-axis at these points. We're summarizing all that information that we gathered in the previous slide here. We see the four different intervals that the real number line, the x-axis, was divided into based on those boundary points, 1, 2, and 3. So we have the interval minus infinity to 1, 1 to 2, 2 to 3, and 3 to infinity. The end behavior is now used. We see that as x tends towards large negative numbers, we have that P of x becomes large negative in y. And as x becomes large positive, P of x becomes large positive, as well. The odd multiplicity means that the graph crosses the x-axis at those boundary points, 1, 2, and 3, as shown. Therefore, we generate a sign table at the bottom here. We indicate that the graph is negative from the interval from minus infinity to 1, positive from 1 to 2, negative again from 2 to 3, and positive from 3 to infinity. In step 4b, we're going to use the Test Points Method to select points on each interval determined by the boundary points and evaluate R of x at each of these test points to find the sign of R of x on that interval. We're going to choose for the first interval from minus infinity to 1, a test 0, which is an easy one to evaluate. We're going to calculate the value of R of x at that test point. And the value of R of x at x equals 0 is given as follows. We get negative 3/2, which is negative, and we go ahead and summarize that in the table here. For the next interval from 1 to 2, we're going to choose a test point 1.5 on that interval. We evaluate R of x at x equal 1.5. We get a value of 6, which is positive, and we go ahead and summarize that in the table. Next, for the interval from 2 to 3, we choose a point 2.5 on that interval. R of x at x equal 2.5 is evaluated to be minus 2/3, a negative value. And we go ahead and summarize that there. And finally, for the interval from 3 to infinity, we choose a test point 4. We have R of x at x equal 4 evaluated to be 1/6, which is positive. And so that's summarized here, as well. Now, the value of R of x at each of these test points dictates the sign of R of x on those intervals containing those test points. So we found that at the test point 0 on the interval from minus infinity to 1 that that value is negative. Therefore, all the values of R of x for points on the interval from minus infinity to 1 are negative. For the second interval from 1 to 2, our test point gave us a positive value for R of x. And therefore, all the values of R of x on the interval from 1 to 2 are positive. For our test point 2 and 1/2 on the interval from 2 to 3, we got a negative value for R of x at that point. And therefore, all the points for R of x on that interval from 2 to 3 are negative. Finally, for 3 to infinity, R of x at x equal 4 gave us a positive value, and so R of x is positive everywhere on that interval from 3 to infinity. Now, in step 5, we want to write the solution set by selecting the intervals in step 4 that satisfy the given inequality. If the inequality is R of x greater than or equal to 0 or R of x less than or equal to 0, we want to include only the boundary points corresponding to the zeros of the numerator of the rational function R of x. Otherwise, we're going to exclude them if it's just greater than 0 or just less than 0. Then we're going to graph the solution set. So in our case, we're solving the inequality R of x greater than or equal to 0, and so we're going to find the intervals on which the function is positive. Now, since the inequality involves greater than or equal to, we're including the zeros of the numerator of R of x, that is the boundary points, in the solution set for the numerator. The only boundary point that is a 0 of the numerator is 3. And so the solution set is the interval from 1 to 2 union the interval from 3 to infinity, where, again, we're including the boundary point 3. That's the 0 of the numerator. We show that graphically. Again, we have parentheses here to indicate that 1 and 2 are not included in our solution set. We've got a bracket here to indicate that 3 is included. And the intervals include all the values from 1 to 2 and all the values from 3 to infinity.