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Contents

- 1 Concert Hall Volume
- 2 Concert Hall Volume
- 3 Number of Surfaces
- 4 Number of Surfaces
- 5 Concert Hall Surface Area
- 6 Concert Hall Surface Area
- 7 Reverberation Time
- 8 Reverberation Time
- 9 Cancel Out Common Factors
- 10 Cancel Out Common Factors
- 11 Hall Vertical Asymptotes
- 12 Hall Vertical Asymptotes
- 13 First Term Form
- 14 First Term Form
- 15 Divide the Long Way
- 16 Divide the Long Way
- 17 Approaching Infinity
- 18 Approaching Infinity
- 19 What does the function approach
- 20 What does the function approach
- 21 All Kinds of Asymptotes
- 22 All Kinds of Asymptotes
- 23 Check off Points
- 24 Check off Points
- 25 Slant Asymptote of T
- 26 Slant Asymptote of T
- 27 The Domain of Common Sense
- 28 The Domain of Common Sense
- 29 Domain and Range of T
- 30 Domain and Range of T
- 31 How many real numbers
- 32 How many real numbers
- 33 Forbidden Values of x and y
- 34 Forbidden Values of x and y
- 35 Remember Rational Inequalities
- 36 Remember Rational Inequalities
- 37 All the Asymptotes
- 38 All the Asymptotes
- 39 Add a Line
- 40 Add a Line
- 41 Visualizing the Solution Set
- 42 Visualizing the Solution Set

Athena is ready to keep pushing forward with her understanding of acoustics to help out with Concert Hall design. As before, she's still pretty concerned with how well audience members will be able to hear the music from different places in the concert hall. Now, however, she's worried about Reverberation time. This is basically how long a sound echoes for after it's been produced. Athena has figured out a simplified equation for reverberation time in the hall. The equation is, T equals 0.322 V over A. Here, T stands for Reverberation time, V is the volume of the hall, and A is the surface area of the hall. So, to figure out the reverberation time then, Athena is going to need to find the volume and the surface area of the room. On this diagram of the hall, I've written in the dimensions of each of the sides. It's not super important right now what x stands for but just so you know, it's the depth of the stage back here. This is how long x is. Considering each of these lengths, what is the volume of our concert hall?

We can find the volume of the concert hall by just multiplying these three different lengths together, since this is a rectangular prism. So we'll have 3 x times 2 x times x plus 10. And of course it doesn't matter what order you multiply these together in. We could leave the answer like this, or we could simplify. I like this form a bit better, we have V equals 6 x squared times the quantity x plus 10. Awesome.

Next, we're going to need to find the surface area of the hall. Now, the surface area of an object is the sum of the areas of all of its surfaces. So for this hall, how many sides do we have? Don't forget to count the ceiling and the floor.

The concert hall has 6 surfaces. We have the ceiling and the floor and each of these four sides for the walls.

Now I'd like you to work out what the total surface area of the hall is. To do that, find each of the areas of the four walls and of the ceiling and of the floor. And then add all those areas together. Make sure that you simplify your answer as much as you can.

We actually have a bunch of pairs of faces that have the same area. Each face has the same area as the one opposite it. So we have two faces whose area is 3x times 2x. Two that are 3x times x plus 10. And another two that are 2x times x plus 10x. Let's multiply all of this out. After a few steps of simplification, we end up with a surface area of 22x squared plus 100x.

That means that in terms of x, we remember x is the depth of her stage. Our equation for reverb time is this: T equals 0.322 times 6 x squared, times x plus 10, all over 22 x squared plus 100 x. This is a rational function. Now, as we've done with other rational functions, let's factor the numerator and denominator. Write the fully factored forms of each side of the fraction here and here. I'd also like you to make a note of any values of x for which the function isn't defined.

The fully factored form of this rational function is T equals 1.932 x squared times x plus 10 for the numerator and in the denominator, 2 x times 11 x plus will be undefined are x equals 0, and x equals negative 50 over 11. So x is not allowed to equal either of these values.

To make sure we have our rational function written in simplest form, I'd like you to cancel out any common factors that are shared by the numerator and the denominator, then write the simplified form of the function right here. Note that I've still included our domain restrictions here, which means we won't lose any information, even if we cancel out some factors. I also would like you to make sure that in this final form down here, there's no constant factor left in the denominator. I'd like the only constant factor to be in the numerator.

We can notice that we have an x squared and an x here. So, one of our x's from the top will cancel with the x from the bottom. We can also combine our two coefficients by just dividing 1.932 over 2. That gives us a reduced form of

Now does the reduced form of this function have any vertical asymptotes? If so, please write their equations in this box.

Our vertical asymptote is x equals negative 50 over 11. In this reduced form of the equation, this is the only value of x that will make the denominator equal to 0.

Now that we factored and found our vertical asymptote, let's use long division just like we did in the last lesson, to find the horizontal asymptote of this function. First though, I'm going to need to multiply the numerator out again, so that I have something in a form that I can divide with. Okay, so now I have that T equals 0.966x squared plus 9.66x divided by 11x plus 50. Let's use long division to carry out this division that we see in the fraction. So here I've written out our division in long division form. If we solve this long division problem, what is the first term of our answer going to look like? Is it going to be some coefficient times x. Will it be a constant? Will it be a fraction with a denominator of 11x plus 50? Or, will it not be anything because you refuse to do anymore long division at all.

We're going to end up with sum coefficient times x plus a bunch of other stuff at the end. That's what we were trying to find by doing long division, after all. Well, we're actually not going to have one. I think we're on to something pretty interesting here, so let's pursue this for a bit.

We know that in the last example we were working with, we didn't have a horizontal asymptote. But do we have something else instead? Well, we're going to check out a slightly simpler example right now to get some ideas about that. Let's consider f of x equals 2x squared plus 3x minus 7 Over X minus 2. What do we get if we use long division to divide the numerator by the denominator?

When we complete the long division, we get 2x plus 7 plus 7 over x minus 2. Hm? Well what does this mean? Again, this is not a horizontal Asymptote.

The answer we get when we carry out this long division and rewrite the function in this new way, is definitely different than what we would get if we had a horizontal asymptote. But once again, we have a fraction as one of our terms. What value does this fraction 7 over x minus 2 get very close to as x heads to infinity.

As x approaches infinity, 7 over x minus 2 will approach 0.

We've seen now that as x gets very large, 7 over x minus 2 becomes very small, and it pretty much goes to 0. So, as this fraction gets tiny, what expression does that leave us with for the function?

f of x will approach the function 2x plus 7, since that's what we have if this entire fraction disappears, just 2x plus 7.

So a x gets bigger and bigger, our function is going to approach the line y equals 2x plus 7. This is called a slant asymptote. We'll see in a second how this shows up on our graph, but first, I'd like for us find the vertical asymptote for this function. Please write its equation in this box.

The vertical Asymptote for this ratioinal function will be x equals 2, as we can see from the demoninator right here.

Where on the coordinate plane does this graph lie, then? We know that the Asymptotes, which I've drawn here, divide the plane into sections. And we can find the graph of the rational function in some of those areas, but not all of them. Please check off all the spots that the function will pass through.

The function will pass through these points here, and these points here. It won't be at all in this part of the graph, or in this part of the graph. Interesting. Slant asymptotes really change things up. Now since this is a major tease, let me just show you the full graph of the function. And here it is. We can see our slant asymptote right here. The graph is getting closer and closer to it, as x gets bigger on both sides, positive and negative. And we can also see our vertical asymptote right here. Pretty cool. Asymptotes clearly are what determines the behavior of rational functions.

Now let's get back to Athena's example for a second. Going through this long division would have been really messy, so I've gone ahead and done it for you. What we end up with is still something that's not super pretty, but at least now you have the numbers. Considering this answer, what is the slant asymptote of our function T? I've already written y equals for you, so just fill in an expression on the right-hand side of the equation.

A slant asymptote is going to be these first two terms, the two that are not fractions that would reduce to 0 as x gets really big. So, the slant asymptote is y equals 0.0878182x plus 0.479008.

It'll be really great if we could draw the graph of Athena's function, and we are actually really close to being able to do that. First however, I'd like to know what domain makes sense for the concert hall situation. This is the original equation we had, written in a slightly different form, since I've Factored out a two from both the numerator and the denominator. It's important to remember here what our variables stand for. T of course is the reverberation time, measured in seconds as a matter of fact. And X here more importantly, stands for the depth of the stage in the concert hall measured in feet. So what restrictions based on what x stands for, do we need to place on the values of x, or on the domain? Use your common sense here.

We need to have x greater than 0 here, since x is measuring length in feet. We also don't want a stage that is 0 feet deep, then we wouldn't have a stage at all.

I know that you've been absolutely dying to see the graph of our famous reverberation time function. So here it is, now notice that I've ignored the restrictions brought on by the application. That's going to let us look at the function as a whole. Otherwise we'd only be looking at these positive values of x, and that wouldn't be quite as interesting. What are the domain and the range of the entire function T? I've given you 2 options for each. You can either say that the domain of the range is all the real numbers, or you can say that it's all the real numbers, except for certain values. If you pick the second option, I'd like you to fill in those values that either X or T cannot equal.

Looking back at our original equation, the two values of x that make the denominator of our function equal to 0, are 0 and negative 50 over 11. Just as we found earlier on. since the domain is all real numbers execpt for those values, the range is just all the real numbers. There's nothing that T can't equal.

Remember our old friend, the function f of x equals 4 over x minus 1? Well, what are its domain and range? Since that's what we've been talking about. I've given you three options each for the domain and for the range. Remember that the asymptotes here are x equals 1 and y equals 0.

The domain is all real numbers except for 1, and the range is all real numbers except for 0. In this case, the function does not cross its horizontal asymptote.

Let's find the domain and range of another rational function. This time I want us to look at y equals x times x squared minus 1 over 2x squared times x squared minus 3x plus 2. To figure out the domain You'll need to look for values of x where the function is undefined, focusing of course on the denominator. To figure out the range, first see if this function has a horizontal asymptote, and if it does, see if any values of x that we can plug in, End up giving us that value Y that we saw in the asymptote. For this quiz I would like you to tell me the domain and the range, by simply writing in the values that X and Y cannot equal. If more than one value belongs in a single box, separate your answers by commas. If you want to leave up box blank, then write the word none instead.

After we completely factored, here's what we end up with. Looking at the denominator, x can equal 0, 2, or 1. So I'll write those values in this box to tell us about the domain. X can equal 0, 1, or 2. Great, that takes care of half the problem. Now onto the range. Let's look at our horizontal asymptote. If there is one at all. First though, I need to multiply out what's in the denominator and the numerator. So we have a form that we can use for long division. In the numerator I'll have x cubed minus x. And in the denominator, we have 2x to the 4th minus 6 x cubed plus 4x squared. Great. Now we can divide. Setting up our long division problem. I need to see how many times 2x to the 4th goes into x cubed. Well, that's just 0 times. I could carry on with my long division. But I know already that we're going to have a horizontal asymptote at. That y equals 0. Now the question is is this an asymptote that the function crosses or not. To find out let's plug in 0 for y in our function. I know that this equation will be satisfied if the numerator is just equal to to do this. So let's just set x times x squared minus 1 equal to 0. Two of these factors x and x minus 1. Are also in the denominator, so I'm going to cross them out. But we still one left over, x plus 1. This means that x equals negative 1, the graph of our function crosses through the horizontal asymptote. Since the horizontal asymptote, then, is not restricting our range, the range is just all the real numbers. So we can write a none in this box. Awesome, that was a super complicated quiz, so if you got through it, great job.

The first time you saw rational expressions in this course, was way early on when we looked at rational inequalities. At this point I think we've setup a great framework for us to understand rational inequalities a lot better than we did before. So let's look at one now. How about this one. 2x minus 3 over x minus 5 is greater than 0. Now I'd like us to find the solution set for this inequality. All of the values of x that make this inequality true. To refresh your memory on how to do this, first I'd like you to find the critical values for this inequality. The values for x that make either the denominator or the numerator equal 0. These critical values split the number line into intervals. So check a value within each interval to see if it satisfies the inequality. Based on this, you can write the solution in inequality form. Remember that you can write backslash or and then a space, if you want to join inequalities or chain them together, if there's more than one region in the solution. Write your answer in this green box.

The critical values for this inequality are x equals 3 halves, and x equals 5. So this splits the number line into three different intervals. Lets pick a value from each interval and test it. From over here on the left, I'll pick 0. And if I plug that in, I end up with a negative number over a negative number, or a positive number which is greater than 0. So yes, this interval checks off. If I pick a number in this middle region, like three, then they'll end up with a positive number over a negative number, which makes a negative number, which is not greater than 0. So this interval does not work. Over on the right, plugging in six for x does check out. So this region should be shaded as well. On a number line then, our solution looks like this. With of course, rounded open brackets showing that 5 and 3 halves are not included. Written as an inequality then, our solution should be x is less than 3 halves joined with x is greater than 5.

But this inequality is intimiatly related to a rational function. Logically, that rational function, is just 2 x minus 3, over x minus 5, equals f of x of course. Let's just take a second to look at this rational function on it's own then. See if there are any vertical, horizontal, or slant asimtotes for this function. If there are no asymptotes of one of these types, simply type the word none in its box. Otherwise we're going to need equations of lines here.

Our vertical asymptote is x equals 5. Our horizontal asymptote is y equals 2. And because we have a horizontal asymptote, we don't have any slant asymptotes.

Since we already found it's asymptotes, I've gone ahead and given you the graph of this function. f of x equals 2x minus 3 over x minus 5. We still have, haven't shown how the solution set of the inequality related to this relates to our function. Remember that the inequality we were working with was 2x minus 3 over x minus 5 is greater than 0. What line do you think we need to add to our graph to help us figure out how the solution of this inequality relates to this function over here? Shall we draw the line x equals 0, y equals 0, x equals 5, or y equals 2?

We need to draw the line y equals 0 as well, since our inequality has the function set as greater than 0.

I've gone in and drawn that line that we said we should add y equals 0. So now we're in the perfect place to finally figure out the connection between our rational function here, and the solution set of this inequality. In fact, the connection is that the x coordinates of some points on the plane over here, represent the solution set of this inequality. Which points are those? Is it all points above y equals 0? All points below y equals 0? All points on the function itself, all points on the function above y equals 0, or all points on the function below y equals 0.

We can find these points by looking at the curve of the function above the line y equals 0. And look at this, the boundaries of those intervals are here at x equals 3 halves, where the graph is intersecting y equals 0. And here at the vertical asymptotes of x equals 5, those were our critical values. All the x coordinates of this part of the function and all of the x coordinates of this part of the function will satisfy our inequality. Awesome.