Solve each rational inequality. Give the solution set in interval...

Question

Solve each rational inequality. Give the solution set in interval notation. See Examples 4 and 5. 1 /{x^2 - 4x + 3} ≤ 1 /{ 3 - x}

Accepted Answer

Welcome back. I am so glad you're here. We are asked to provide the solution set for the given rational inequality. In interval notation. Our given rational inequality is on the left. We have the fraction one divided by X squared minus 13, X plus 12. That fraction is less than, or equal to the fraction on the right, which is one divided by 12 minus X. Our answer choices are answer choice. A, the solution set open parentheses negative infinity to zero. Close bracket in union with open parentheses, 1 to 12 close bracket. Answer choice B the solution set open parentheses negative infinity to zero. Close parentheses in union with open bracket, 1 to 12, close parentheses. Answer choice C the solution set open parentheses, negative infinity to zero. Close parentheses in union width, open bracket, 1 to 12, close bracket and answer choice D the solution set open parentheses, negative infinity to zero. Close bracket in union width, open parentheses, 1 to 12 closed parentheses. All right. So in order to even begin to solve for this rational inequality, we need to first make sure we have zero on one side of the inequality. And then we need to ensure that we have one single fraction on the other side. So let's begin by subtracting one divided by 12 minus X from both sides of the inequality. And that will give us on the right hand side, zero E A on the left, we have one divided by X squared minus 13 X plus 12 minus one divided by 12 minus X all less than are equal to zero. And now we have to get that single fraction on the left. And yay, I mean, we need to have a common denominator in order to combine these two. And let's factor the first denominator see if we have any common factors with the second fraction. And so we have X squared minus 13 X plus 12. That factors to be the quantity of X minus 12 multiplied by the quantity of X minus one and X minus 12 is so similar to 12 minus X, but they're not the same. So what we do need to do is to get a common denominator which will be the quantity of X minus 12 multiplied by the quantity of X minus one multiplied by the quantity of 12 minus X. So the factors from all three, in order to get that common denominator, we need to multiply the first fraction by minus X divided by 12 minus X. And we need to multiply the second fraction by the denominator from the first, which is X minus 12 multiplied by the quantity of X minus one all divided by the quantity of X minus 12, multiplied by the quantity of X minus one. And so, in doing that, we get our common denominator yay. And then from the first fraction what's left in the numerator is 12 minus X. And then we need to subtract that numerator from the second fraction. We know that X minus 12 multiplied by the quantity of X minus one is the same as X squared minus 13, X plus 12. And it's a little bit easier to distribute the subtraction when it's already been multiplied out. So we'll put for the fraction, the second fractions numerator that will be the quantity of X squared minus 13 X plus 12. And now we can distribute this negative to the X squared minus 13 X plus 12. Remembering that this is all still less than or equal to zero. So in our numerator, we've got 12 minus X minus X squared N negative multiplied by negative 13, X is plus 13 X and the negative multiplied by the positive 12 is now minus 12 all over our same three factors in the denominator X minus 12 multiplied by X minus one multiplied by 12 minus X all still less than or equal to zero. Now, we can combine like terms and put these in order in our numerator, we've got that negative X squared which will be first and then we have a negative X plus 13 X, which will be plus 12 X and then we have 12 minus 12, that's zero. And this is all still over the quantity of X minus 12, multiplied by the quantity of X minus one multiplied by the quantity of 12 minus X. And that's all still less than or equal to zero. One more thing we can do here, we can factor in the numerator. We can factor out a negative AX A negative X squared divided by negative X leaves us with X and a positive 12, X divided by negative X will be a negative 12. All still over those same three factors X minus multiplied by X minus one multiplied by 12 minus X all still less than or equal to zero. Now, before we cancel out the factors that occur in both the numerator and the denominator, we wanna make sure that we note for our domain that this X minus 12. If we set that equal to zero, add 12 to both sides, we get X equals a positive 12, X cannot equal a positive 12 because that would make it undefined. So before we cancel those out, we just want to note that X cannot equal 12 or we'd have zero in the denominator and that would be undefined. So our simplified inequality is a negative in the numerator divided by the two factors that are left in the denominator X minus one multiplied by the quantity of 12 minus X all less than, or equal to zero. Hey, we finally met those two conditions and we can now start to solve for this inequality. Yay. All right, here's how we do that. We're going to set both our numerator and our denominator equal to zero. And those are going to be our end points for our intervals. So let's do that. We can set the numerator negative X equal to zero, divide both sides by negative one and we get X equals zero. That's one of our end points. Now are two factors from the denominator X minus one equals zero and 12 minus X equals zero for X minus one equals zero. We'll add one to both sides and we get X equals one for minus X equals zero. We could add X to both sides and we get equals X noting that that's that same that we had before. So we do have the end point of 12 that we already discussed for the domain issue. So we've got these three end points for our intervals. We can draw a number line and it's not gonna be to scale. We go from negative infinity to positive infinity. Now we can note our first endpoint which is X equals zero. The next end point not to scale here is going to be an X equals one and the next end point is going to be an X equals 12. I made that. So not to scale make it slightly more to scale. OK? And then what we have are four intervals, the first one from negative infinity to zero, the next interval from 0 to 1, the next interval from 1 to 12 and the final interval from 12 to infinity. What we do is choose a test point from each of those four intervals. And then we will plug that value of the test point in for X into our inequality. If the inequality is true, then that interval is true for the inequality. If the test point is false, then that in interval is not included for our inequality. So for the interval interval from negative infinity to zero, I'm going to choose the test point of negative one from 0 to 1. I'm choosing one half or 10.5 from 1 to 12. I'm choosing two and from 12 to positive infinity. I'm choosing 13. So plugging the test points in for X start off with negative one. So in the numerator, we've got a negative negative one all divided by the quantity of not X but negative one minus one multiplied by the quantity of 12 minus not X but negative one all supposedly less than are equal to zero. Let's see if it is in the numerator negative negative one is a positive one. For the first quantity negative one minus one that's negative two and for 12 minus negative one that's positive multiplying in the denominator negative two multiplied by 13 is negative 26. So we've got one divided by a negative 26 which is the same as negative one 26th. Is that less than or equal to zero? Yes, it is. So it's our inequality is true for that interval. Now let's test the next one. So we've got in the numerator, a negative 0. all divided by 0.5 minus one, multiplied by the quantity of minus 120.5, supposedly less than are equal to zero. Simplifying in the numerator, we just have negative 0.5 in the denominator 0.5 minus one is a negative 10.5. And that's multiplied by 12 minus 120.5 which is 11. in the denominator negative 0.5, multiplied by 11.5 is a negative 5.75. So we have a negative 0.5 divided by a negative 5.75, negative divided by a negative is a positive. This simplifies to two 23rd. Is that less than or equal to zero? No, it is not. So the inequality is not true for that interval. Now, let's test from 1 to 12 with the test 120.2. So in the numerator, we've got a negative two, that's all divided by the quantity of two minus one, multiplied by the quantity of 12 minus two, all supposedly less than are equal to zero. Keeping negative two in the numerator two minus one is one in the denominator. And 12 minus two is 10, one multiplied by 10 is 10. So we have negative 2/10 which is negative 1/5. Is that less than or equal to zero? Yes, it is. So that interval is true for our inequality. Now, let's test the last one from 12 to infinity with in the numerator, we've got negative 13 divided by the quantity of minus one multiplied by the quantity of minus 13. All supposedly less than are equal to zero. Simplifying will keep our numerator negative 13 divided by minus one is 12 in the denominator and 12 minus 13 is a negative one in the denominator 12 multiplied by negative one is negative 12. So we have negative 13 divided by negative 12, which is a positive 13 12th. Is that less than or equal to zero? No, it is not. So the inequality is not true for that interval either. Now, all we need to do is write this in interval notation. OK? So our first interval is from negative infinity to zero. We say the negative infinity part with an open parentheses, open parentheses means it's not included. We'll never quite get to negative infinity. And then we put comma and then zero is zero included. Well, this whole inequality is less than or equal to and we found zero when we set the numerator equal to zero. And if the numerator is equal to zero, the whole fraction is zero. And that's true. Zero would be less than, or equal to zero. So zero is included. We say that zero is included with a bracket. And then that's in union with, because it's not directly connected to our next interval, which is from 1 to 12 or one in included. Well, we found one and 12 when we set the denominator equal to zero. And if our denominator is zero, then this whole thing is undefined. We said X cannot equal 12. So 12 and one are not included. And again, we say those are not included with parentheses. We say open parentheses, one comma 12, close parentheses. And that's how we express our answer in interval notation. Let's check out our answer choices. All right, we look at our answer choices and this matches with answer choice. D well done. We'll catch you on the next one.

Solve each rational inequality. Give the solution set in interval notation. See Examples 4 and 5. 1 /{x^2 - 4x + 3} ≤ 1 /{ 3 - x}

Key Concepts

Rational Inequalities

Interval Notation

Finding Critical Points

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