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 - 4} ≤ 1 /{ 2 - x}

Accepted Answer

Hey, everyone, this problem asks us to provide the solution set for the given rational inequality and interval notation. And the inequality we're given is one divided by X squared minus is less than or equal to two divided by four minus X. We're given four answer choices A through D and each of the solutions contains the union of two different intervals. And we are gonna come back to those intervals as we work through this problem. Now, let's start by writing what we were given one divided by X squared minus 16 is less than or equal to two divided by four minus X. Now, when we're given an inequality like this, what we wanna do is move everything to one side and just have zero on the other side. So in this case, let's start by moving our two divided by four minus X to the left hand side. And if we can get that right hand side to be zero, it's gonna be a lot easier to compare our function to zero. So doing that and we're gonna also factor at the same time, one divided by X squared minus 16. OK X squared minus 16 is a difference of squares. So we can factor that denominator. We have one divided by X minus four, multiplied by X plus four. OK? And we're gonna subtract two, divided by four minus X. And this is all gonna be less than or equal to zero. OK. So we have this everything on one side, zero on the other side that we wanted. Now, I want to simplify the left hand side to write it as one rational function. And we want all of these factors either multiplied or divided. We don't want to be subtracting or adding full functions. Now, in order to do that, we need a common denominator. Let's rewrite this first. We can see that we have an X minus four term in the denominator of our first function or our first term and we have four minus X in this second function. OK? Well, we can rewrite four minus X as and let me write it on the side here as negative X minus four. Doing that allows us to have at least one similar term in the denominator. OK? So if we write it as negative X minus four, then the entire thing can be written as one divided by X minus four, multiplied by X plus four plus two, divided by X minus four A because we have that negative and then another negative that's gonna become a positive and this is less than or equal to zero. Now, we want to find our common denominator, we can see that our common denominator will be X minus four, multiplied by X plus four. Now, the term on the left hand side already has that denominator. So we don't need to do anything special there. On the right hand side, we have X minus four, but we need to multiply this denominator by X plus four in order to get that same denominator. If we do it to the denominator, we need to do it to the numerator. OK? And then we can combine these functions. So we get one plus two, multiplied by X plus four, all divided by X minus four, multiplied by X plus four is less than or equal to zero. And so we've combined those functions with their common denominator. And now we are going to simplify that numerator and we can expand these terms and simplify. So we're gonna have one plus in expanding the brackets, you get two X plus eight all divided by that common denominator. X minus four multiplied by X plus four. A less than or equal to zero. Simplifying one more time, we have one plus eight. And so we're gonna end up with two X plus nine all divided by X minus four, multiplied by X plus four is the last center equal to zero. And now we have a single rational function on the left hand side, we have zero on the right hand side. So we can figure out how to solve this. No, let's take a look and find our points of interest and our points of interest are gonna be any points where one of our factors of a rational function is equal to zero. OK? So in the numerator we have two X plus nine, if we set that equal to zero, we get that X is equal to negative nine halves in the denominator. OK? We have X plus four and X minus four. And so that's gonna give us X is equal to negative four and X is equal to positive four. So if X is equal to negative nine halves, negative four or positive four, we have a point of interest. And why we want to look at these points of interest is because that's where our function is gonna change sides and where these factors equal zero. OK? On either side of that, our function may change signs from being positive to negative. And that's gonna give us where our solutions are because we're comparing this to zero. So let's go ahead and draw a number line and we want to include all of these points of interest. OK? So we have negative nine halves, we have negative four, working from left to right and then we have positive four. And this might not be quite to scale, but it shows us all of those intervals. All right. And I think that drawing the number line and working on it, this way is just a really nice clean way to look at all of these intervals and keep your work organized. We're gonna start on the left hand side and this first interval is gonna be everything to the left of negative nine halves. So our X values are gonna be less than negative nine halves. What that means is that two X plus nine, the numerator? Well, that's gonna be less than zero. It's gonna be negative X minus four. Well, that's also gonna be less than zero. If we substitute in an X value, that's negative, negative minus a negative is gonna give us a negative and X plus four is also gonna be negative. Hm All right. So what that means, OK, the numerator function which we're gonna call N is less than zero. The denominator of function is a negative value multiplied by a negative value which gives us a positive value. So the left hand side which we call LHs of our inequality is going to be negative and negative divided by a positive, gives us a negative. And this is what we want. OK? We want the left hand side to be less than or equal to zero. So this interval actually satisfies our inequality. We're gonna give it a green check. Now, before we move on to the next interval, we wanna look at our endpoint and figure out if it should be included if X is equal to negative nine half and the numerator is gonna be zero. And our inequality says that zero is less than, or equal to zero. Now, because we have that less than equal to zero. OK. Having the left hand side equal to zero does satisfy. And so we're gonna have a closed circle on negative nine halves to indicate that we are including this point in our interval. Hey, and that's just gonna be a reminder when we go to write this an interval notation that we need to include that point moving to the next interval. OK? And this is gonna be X values between negative nine halves and negative four A. And we're gonna do the same thing two X plus nine. Well, that's gonna be positive. Now X minus four is still gonna be negative X plus four is still going to be negative. So our numerator is now positive, our denominator is still positive because we have a negative multiplied by a negative, which means that the left hand side of our rational function is going to be positive, it's bigger than zero. So that does not satisfy our inequality. So this interval, we're gonna give a red X and we move to the next. Now this interval is from negative four all the way up to positive four. OK? Two X plus nine. Well, that's going to be positive X minus four. Well, this is still going to be negative but X plus four is going to be positive now. And you can think of choosing any point in this interval. OK. So think about 00 is right in the middle of this interval. If you were to substitute zero into each of these factors to X plus nine, you get nine, that's positive bigger than zero, zero minus four, negative four. That's less than 00 plus 44. That's bigger than zero. OK? So that's what we're doing in each of these intervals. Now we have that the numerator is greater than zero. The denominator is gonna be negative multiplied by a positive. So it's gonna be less than zero, a positive divided by a negative gives us that the left hand side of our inequality is negative less than zero. That's exactly what we want. And so this is gonna be included in our interval. So we give a green check mark to this interval. OK. This interval is part of our solution. Now, we need to look at the end points again, in this case, if we have X equals negative four or X equals positive four and we substitute that into the left-hand side of our inequality. OK? The denominator is gonna be zero. We do not want to divide by zero. If the denominator is zero, the left hand side is undefined. And so these are points are not included in our solution. OK? So we want to have an open circle on those points to indicate that we are not including these end points because the function is not defined there. And one last interval to look at now and that is the interval for X greater than four, everything to the right of four, all of these are going to be positive two, X plus nine, positive greater than zero, X minus four, positive greater than zero. And X plus four is also gonna be positive greater than zero. We're multiplying dividing values that are positive. So the left hand side is going to be greater than zero, it's gonna be positive, which does not satisfy our inequality. So this interval is not included in our solution. So looking at what we've done on our number line, we have two intervals that we're including, OK, we've already indicated what needs to happen with these end points. Now, it's just a matter of converting this into interval notation like the question asked. So the first interval we have is X values to the left of negative nine halves. So what that means is that from negative infinity all the way up to negative nine halves is our first interval, negative nine halves we've drawn as a solid circle. And so we include that value. So we use a square bracket to indicate that. Then we're gonna take the union with the second interval we found and we're gonna use round brackets here because we've indicated that we are not including either endpoint. And this interval goes from negative four up to positive four OK. And so this is this set, the union of these two intervals is going to be the solution set for this problem. If we compare this to our answer choices, we can see right off the bat that option B and option C are not going to be the correct option because they are not including the end point negative nine halves may option A and D have that correct first interval. And then for the second interval, option A is including one of the endpoints A four which we know we can't include because the function is not defined there. And so option A is not the correct answer either. And we have that the correct answer is option D, the solution set is the union of the two intervals. The first one is negative infinity up to and including negative nine halves with the interval, negative 4 to 4 where we don't include the end points. Thanks everyone for watching. I hope this video helped see you in the next one.

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

Key Concepts

Rational Inequalities

Interval Notation

Critical Points

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