Solving factored form rational inequalities by testing intervals
When testing f(-1), the narrator says the denominator factor is '-2 minus 5' / '-7'; it should be -1 - 5 = -6. The sign conclusion (denominator positive, f(-1) negative) is still correct.
The check-ins use the correct version. The video is Khan Academy's, unchanged.About this video
To solve f(x) = (x - 3)(x + 2) / ((2x + 1)(x - 5)) >= 0, the video finds the zeros (x = 3 and x = -2) and the vertical asymptotes (x = -1/2 and x = 5), which are all places the sign can change. It tests a value in each interval and concludes the solution is (-infinity, -2] together with (-1/2, 3] and (5, infinity), including zeros but never the asymptotes.
You'll be able to
- find where a rational function equals zero and where it has vertical asymptotes.
- test intervals between these points to find where a rational function is positive or negative.
- decide which boundary points to include, using closed circles for zeros and open circles for asymptotes.
- write the solution set in interval notation.
Key words
- rational function
- A function written as one polynomial divided by another.
- vertical asymptote
- An x-value where the denominator is zero and the function is undefined, so the graph shoots up or down.
- sign transition point
- A zero or asymptote where the function might switch between positive and negative.
- interval notation
- A way to write a set of x-values using brackets [ ] for included ends and parentheses ( ) for excluded ends.
Transcript
- [Instructor] We are told that f of x is equal to all of this. Identify when f of x is greater than or equal to 0. So pause this video and have a go at that before we do this together. All right, so what we're going to be doing is thinking about all of the places where f of x could be equal to 0, and then also the places where f of x has a vertical asymptote, and those are all possible sign transition places, and some of them might even be included 'cause this is not just greater than 0, this is greater than or equal to 0, and then we can test the intervals in between to see what's actually going on.
So when does f of x equal 0? Well, that's if either of these expressions that we see up here, either of these expressions are equal to 0. So that first one, if that's equal to 0, that means x equals 3. This second one, if x plus 2 is equal to 0, that means that x is equal to -2.
So these are both interesting values. They're definitely places that we would include in the solution because the function is equal to 0 at both of these points. But also, let's think about where we might hit vertical asymptotes. So that's when the denominator is equal to 0.
So for example, let's look at when 2x plus 1 is equal to 0. 2x plus 1 is equal to 0, that means 2x is equal to -1, that means x is equal to negative one half. And then last, but not least, we could look at when x minus 5 is equal to 0. X minus 5 is equal to 0 or it's at x equals 5.
All right, so let's think about... I will just draw these out on a number line and we can maybe think about what we want to include or not include. So we go from -2, all the way up to 5. So maybe if this is 0, let's see, we go 1, 2, 3, 4, 5, and -1, 2, 3, 4, 5.
And so if we are thinking about, let's see,
- 2 right over here, so that is -2, that's that first point or one of the first points, it's the lowest of the points. And then we go to x equals negative one half, which is right over here, negative one half. And actually, I'm gonna make sure that we don't include that point.
So actually, I'm not even gonna fill it in, I'm gonna put an open circle there, that'll help us. So negative one half, but that is a place where we might see some type of a sign switch. We will include x equals -2 'cause that's a place where the function is equal to 0, so it's nice that I filled that point in.
And then next we go to x equal 3. That's also another point where the function is equal 0, so I'll include that 1, 2, 3. So x equals 3. And then last, but not least, have x equals 5, but we can't include it 'cause that's actually a restriction to the domain. It makes the denominator equal 0, so that's 5.
So let's think about when we're greater than 0. So let's try out -3 right over here. Let me do these in neutral colors actually. So let's try this point, -3. So if I were to try that out, f of -3 is equal to -3 minus 3, times -3 plus 2, all of that over 2 times -3, which is -6 plus 1, times -3 minus 5.
And really, all we care is the sign whether we are greater than or less than 0. So this is going to be -6 times -1. So our numerator's going to be positive in this situation, and then this is going to be -5, and then this is going to be -8. So our numerator's going to be positive, it's going...
Negative times a negative, and our denominator's going to be positive. So this is going be a positive value and we are assuming we're not gonna have anymore sign switches if we go to the left of -2. So I'm going to include all of that when we are greater... Saying that we are going to be greater than 0 here.
So I'm gonna include all of this in our solution set, and let's see what's happening in between here. So let's try at -1. So f of -1 is equal to -1 minus 3, times -1 plus 2, all of that over 2 times -1, that's -2 plus 1, times -2 minus 5. Our denominator, let's see, this is a negative, it's a -1.
This is a -7. Our denominator is still positive, but our numerator over here is now... This is going to be -4 while this is going to be positive 1. So our numerator is now negative, a negative or a positive... This is a negative, so we're gonna be less than 0 here.
So let's exclude this interval. This interval right over here, we're gonna exclude it. Now, let's go to this section, and a number that's always fun to test out is 0. The math tends to be pretty straightforward. So let's see, f of 0 is going to be -3 times 2 over 1 times -5.
And so that is equal to -6 over -5, which is equal to positive 6 over 5. So I'm feeling pretty good. I'm gonna shade that in the green again. So this interval right over here 'cause I don't expect any sign changes. And we know that 3 is also included because that's where we equal 0.
Now, let me try out 4, see what's going on there. I'm guessing that we might have a sign switch and this might not work, but let's try f of 0. F of... Not F of 0, F of 4, I should say. F of 4 is going to be 4 minus 3, which is 1, times 4 plus 2, which is 6, over 2 times 4, which is 8 plus 1, which is 9, and then 4 minus 5 is -1, So this is gonna be a positive over negative, so this is less than 0.
Yep, this interval doesn't count. But now, let's go to values larger than 5, and let's try 6. So let's try... Let me do this in a color I have not used. Well, I'll stay neutral. F of 6 is going to be 6 minus 3, which is 3, times 6 plus 2, which is 8, over 12 plus 1, which is 13, and 6 minus 5, which is 1.
So everything here is positive, so this checks out. So I think we have our interval now. Our interval we can go from, let me just write it, I'll write it in a neutral color, we can go from negative infinity and we're going to include up to and including -2. And then to that, we are going to add, we're not going to include negative one half 'cause our domain is restricted there, that's a vertical asymptote, all the way to and including 3, and then to that, we're going to also add, if we're thinking about all of the intervals that work, we're gonna go from 5 to positive infinity, and we are done.
Video © Khan Academy, used under CC BY-NC-SA. Watch on Khan Academy. Puzzicle is not affiliated with Khan Academy.