Welcome everybody to AP Calculus AB. Today we've got the video review guide for the Chapter 0 review. Now I do have a key posted that's just a PDF of all the solutions from this video.
So please realize that this is if you're going through and checking your work and you have a question on how the problem was carried out. This is probably the video that you're going to want to take a look at. But if you're just here to check your work, you may want to consult that PDF key.
Instead, it'll probably be much faster. This video will go a little bit quicker since we've already been through all of the content. This is just going to kind of be to fine tune any details. as we look to prepare for this test.
All right, so here we go with the key to the chapter zero review. Problem number one, write an equation for a line that passes through the point 4, 2 and has a slope of m equals, and I apologize for that question mark form added in there. That just should not be there.
That was a mistake in equation editor. Write your answer in slope-intercept form. All right, we have a point.
We have a slope, so we will set up slope-intercept form, y minus y1 equals m times x minus x1, and because of the prompt for slope-intercept, We're going to going to distribute our slope. So distributing that 3 4ths gives me 3 4ths times x and 3 4ths times negative 4. And then we're going to finish solving for y. We're going to add the 2 over to both sides. y equals 3 4ths x minus 1. Now silly things like that, you'll notice I did not show like the plus 2 on each side.
You may want to do that if you find yourself making little negative mistakes. It's surprising how many times students will see the minus 2 here and then they'll drop this down to a negative 5, but realize to get rid of this minus 2 We need to add two to both sides. It's just like an easy little negative sign mistake. So we want to be aware of stuff like that.
Problem two, find an equation of the line in slope intercept form passing through these two points. Now remember we've got a lot of linear stuff happening here, but your test will only be about half of the content on this review, but all the concepts on here If you study all of these and feel comfortable with those, you should be in very good shape for the test. So y minus y2 minus y1, 2 minus 4, and then 1 minus a negative 3. That's like 1 plus 3. We get negative 1 half for our slope. All right.
And then we just need to grab one of those points. I'm going to choose the point 1, 2. And because we're told slope intercept form, let's go ahead and distribute. We did talk about how the AP exam itself will not require you to go to slope intercept form frequently. However, it's a really good review of the algebra for us, so that's why the requirement is here in the prerequisites type of work. Kind of the same steps.
Once we have the slope, it's very similar to the one up above. Equations for the vertical and horizontal lines. Remember, vertical lines are x equals some constant. Horizontals are y equals some constant. And so we're just going to grab the x and the y values.
And here there were nice integers. If it were something like an e or a pi or something like that, that would still hold true. The x value, x equals that number, is going to be our vertical line through that point. And then y equals the line that's keeping that height.
That will be our horizontal line. Number four, here we are parallel to the given line. Okay, so I need to know this given line. I need to know its slope.
I don't really care about its y-intercept, but let's set it up like slope-intercept form. I'm going to subtract 2x from both sides, divide both sides by negative 3. Notice the negative divided by negative here. And again, I don't really care about that four-thirds. Since I want to be parallel, I know that I want my slope to perfectly match this given line's slope. And then here's more of the same.
y minus y1 equals m times x minus x1. We'll distribute that 2 thirds, and we'll add 2 to both sides, so y equals 2 thirds x is our answer. And again, this first page probably seems pretty competitive or pretty repetitive.
That's intentional. Okay. Sorry, the equation did not copy here. We had some issues with equation editor, it looks like, when it made this PDF.
So y equals, this should be negative three. 3 halves x minus 3, and you should be seeing that printed on your review guide. It was just when I made this PDF for the video. All right, so now if we want a line that is perpendicular, we want our slope to not perfectly match like the previous problem, but we want the opposite.
opposite reciprocal. So we want positive 2 over 3. We don't need the x there. Okay, y minus y1 equals m times x minus x1.
That's minus a negative. Again, those little things we want to keep track of. We'll distribute the two-thirds.
And then we will add over 2, and I'd like to show that I'm adding 2. And I'm going to just so when I'm dealing with fractions, I can see the common denominator. Yes, I'm adding something that's equivalent to 2 over here. So y equals 2 thirds x plus 8 thirds.
is my answer. Okay, and again there is some reasoning behind this. Yes, it's just lines.
Yes, you probably did work like this in your Algebra 1 class, and you probably saw it in Algebra 2, and again in PreCalculus, but just realize that answers like this with fractions everywhere, they're totally common. It's absolutely fair game, so we do want to be prepared for stuff like that. All right, problem number six. Again, more issues with the equation editor. I apologize for that.
This is just supposed to be an absolute value. That's here. It should be printed correctly in your review guide, but just making this PDF apparently had some issues. Sketch a complete graph, give the domain and range, blah, blah, blah.
Write it as a piecewise function since we can't use calculus with absolute values. So here we go. So with this, realize you'll be allowed your calculator, obviously support our answer there. So we could use the Y equals menu to help us draw this graph.
I'm probably going to guess that we don't need that. if I realize that the hinge point of our single absolute value here is where x equals 2. That would zero everything out. So I want to think what's happening when x is on the left of that value of 2 and when x is on the right of that value of 2. Notice this is a positive absolute value, so it's a v-shape opening up, meaning that when x is less than 2, that absolute value is looking like negative x plus 2. It's just the negated version of everything inside. Okay, and then we still have the plus 3 at the end. It's just the absolute value that gets its contents made next.
negative. And on the right hand side, everything is positive. So the absolute value wouldn't have changed anything.
It's just two times the quantity x minus two and then plus three. So we get negative two x plus four and then three is plus seven. And here we get two x minus four plus three is minus one.
So that is my piecewise function y equals one piece is negative two x plus 7 and we want that to be when x is less than or equal to 2 and then 2x minus 1, we want that to be when x is greater than 2. It does not matter where you tuck the little or equals sign here. Absolute value functions are continuous, so they join right up. There's no jumps around or anything like that.
As long as we only have one of them with the or equals, we're good. We need to have it somewhere. We need to have our function defined when x equals 2, but we get to choose whether it's top or bottom.
Alright, so now for me, I like to draw the graph and then grab the values for the table of values, but you could be plugging in some x values right now that are around 2. So I just think the numbers 1, 2, 3, and 4 are good choices. We've got some that are 2, you know, this way I've got my first two values going into the top and my second two values going into the bottom piece of that piecewise. It's really up to you. So let's get the graph drawn.
So I start off at positive 7, and then my slope is down 2 and to the right 1. All right, and we're going to keep that as long as x is less than or equal to 2. So there. And then we've got 2x minus 1. So we're down here, and then 2. 2, there's where it hooks up, and then it keeps going after. And so this one's going to have an open circle, even though we've already got the point defined there.
And it'll go up that way. I would want to tuck some arrows on this one. I mean, a complete graph just means we go to the edges.
Those arrows aren't necessary. But we would want to know that it does continue going to the left and to the right. Now that we see the graph we can tell that the x-intercept there is not going to be one and My y-intercept well I can see that that's happening at the point when x equals 0 that's the point 0 7 Really nice pulling that from the graph my domain is going to be the set of all numbers x that are elements of the real numbers Okay, or if we used interval notation, that's negative infinity to infinity. We're going to want to know how to use both. The range, this is all y values, such that y is greater than or equal to 3. Or we could say the set of y values including that height of 3 going up to infinity.
And again, that's just pulling the information from the graph, seeing that it continues to rise upwards, and there is a point included at a height of 3. All right, so at 1, 2, 3, and 4, I'm just going to plug those x values into their appropriate places. I get 5, 3, 5, and 7. We could also just look at the graph now that we have that drawn carefully. All right, so there we have problem number 6. Number seven. Yoyoyoy!
x minus 4 under a square root is what should complete that. The negative 2 is there out front, but for whatever reason, this did not get formatted correctly. Again, not an issue. We can work with it that way. So sketching our complete graph here, we can use...
use a calculator's table of values, but we're actually supposed to come up with a table of values ourselves. Just realize that to make this square root work, x has to be a number that's greater than or equal to 4. So I'm going to start with 4, and then I'm going to go up 1, and then I'm going to go up 3, and then I'm going to go up 5. Reason being, take a look at all of these numbers when I subtract 4. Like what's happening underneath the square root, all of these are going to make perfect square numbers underneath. So if I plug in a 4, I will have the square root having a 0 underneath it.
Negative 2 plus that square root of 0 is just negative 2. If I plug in that 5, then I'll have the square root producing a 1, negative 2 plus 1. If I plug in an 8, this will be the square root of 4, negative 2 plus 2. And if I plug in this 13, I'll have the square root of 9, so negative 2 plus 3. like that. And so we'll plot those points. We got 4, negative 2, 5, negative 1, 8, 0, and then 13 way over here-ish would be that next point.
The graph bends to fit those those points. And as long as we're taking that graph to the edge of the grid we're provided, that's a complete graph. All right, so our x and y intercepts, we do have an x intercept, but we do not have a y intercept.
You can see where x equals equals 0 is not on the domain of our function so we're not going to interact with the y-axis. To find the x-intercept we just find out where where y equals 0. So where does 0 equal negative 2 plus the square root of x minus 4? Sorry we would add 2 to both sides, square both sides, and add 4 to the other side. And we did have that point showing up in our table, so that's really nice. But we could also do it algebraically as well, which is very, very good.
The domain for this is all x's such that x is greater than or equal to 4, as we said at the start of the problem. And we would write this as from 4 to positive infinity if we were using interval notation. The range is going to be all y values such that y is greater than or equal to negative 2. And again, we can kind of see that from our graph that once we have that initial starting point, the graph is only increasing from there. It's not going to go down. Number eight, hey this one formatted correctly for us.
Sketch a complete graph here for this one. Here we've got an exponential function. And so let's try and piece some things together.
There's also a lot of prompts here. There's like asymptotes and things to fill in. So let's start with the defining quality of exponential functions, which is they have a horizontal asymptote. If that exponential. part of the graph has a constant added or subtracted.
In this case we are subtracting 1, so y equals negative 1. So we would write down this as a horizontal asymptote of y equals negative 1. I'm going to add that in right now. on the graph like so. And then you'll notice we've got a little bit of transformation work happening here to this exponential function as well.
So instead of a typical like three to the x, which would be rising to the right, this one has a negative x, meaning meaning it's going to be showing us exponential decay. And so we can totally use a calculator table to help with this if needed. And some of these may have some decimal values to them.
We'll just have to see. But just realize we want to keep our powers here pretty close to zero. And they do say at least one point must have a rational output in fraction form.
So realizing that negative powers are perfectly fine. Alright, so I'm going to choose the x values negative 1, 0, 1, and 2. Just because if I have x equaling 1, that's going to be where the exponent here is 0. out, so I want that to be happening in the middle of my table. Alright, so if x equals a negative 1, I would have 3 to the 1 plus 1, that's 9. 3 to the second minus 1 is 8. If I plug in a 0, I would have 3 to the first minus 1, which is 2. Alright, if I were to plug in a 1, that 0s out my exponent.
3 to the 0th is 1 minus 1, 0. And if I plug in a 2, I've got 3 to the negative 1st, that's 1 3rd. Do that off to the side here. 3 to the negative 1st is 1 3rd, and then I take away 1. Well, that would give me negative 2 3rds.
Alright, so let's plot this graph and then we can get the stuff analyzed. So we got the point 1, 0. We got negative 1 up at 8. We got 0, 2. Good, good. And then we have at 2, we're down at negative 2 thirds, so down there. And that's enough to show me that the graph is indeed racing up in the leftward direction.
And as I move to the right, that's where it's being pulled toward that horizontal asymptote. Just extend it to the edges of the graph. That's a complete graph. Now my x-intercept, I can tell that's the point .
Just by looking at the graph, we had a nice clean point in our table. Same thing with our y-intercept being . But just realize we could do this algebraically as well. The x-intercept will happen when y equals zero. So if I add one to both sides, Alright, and then I would take this 1 and I'd call it 3 to the 0th, equaling 3 to the 1 minus x, just to establish a common base.
Or you could do log base 3 on both sides. At any rate, we end up with the equation 0 equals 1 minus x. Add the x, we find that x equals 1. There's our x-intercept. To find my y-intercept, that's when I plug in 0 for x. 3 to the first is 3 minus 1. Oops, sorry, I'm getting ahead of myself.
And then y would equal 2. So there's our y-intercept. The domain for any exponential function is going to be any x's that are elements of the real numbers. And for our range, here we've got all y values such that y is greater than negative 1. The graph will never reach the height of the asymptote, but it will get unimaginably close to it.
So here we're going from negative 1 up to infinity. That one's done, and you'll notice no calculator was needed to do any of these, and that's kind of one of the things I'm hoping that we're getting as practice moving through these, is that the calculator is a great tool to help and check, but they're all doable without the calculator. The more we can pull back on that previously learned algebra, the stronger calculus student we're going to be.
All right, so number nine, here we've got a logarithmic function. It's a log base two. All right, so I'm going to actually, one of the prompts here, and this will be in bold on your test, I believe, is to rewrite this in exponential form.
That's just to help us find. an easy x and y table of values for the original function. Alright so here I want to isolate the logarithm so that means I'm going to add one to the other side so I've got my output plus one I'm just going to call f now just to make it easier to write. Alright, now to get rid of the log base 2, I would call both sides of that equation a power on 2. So 2 to the y plus 1 equals, and then we get our inverse to happen on the right, it's just x minus 2 now. And then we would add 2 to both sides to equal x.
So here is our function in exponential form. To go ahead and make this graph, logarithms, because they're inverses of exponentials, these have vertical asymptotes. And here the vertical asymptote will happen with whatever x value zeros out the argument.
So what makes that x minus 2 inside our logarithm equal 0? Well, it's when x equals 2. So kind of like the last one. Having the asymptote done early really helps give a backbone to the points that we're plotting.
We can see if the graph is fitting appropriately. All right, so now let's talk about our table of values. And remember, these are y values.
that are going in. So I want to keep these y values pretty close to zero. So I'm going to choose y values that are like negative two, negative one, zero, and one, because when those go into this exponent, they're going to...
going to keep that power pretty close to zero. And here's why we convert to exponential form. If I plug in a negative two for my y value, I'd have two to the negative first.
Okay, and again, we have that rational output in fraction form for one of our outputs. We'll make that happen. So if I plug in a negative two, I'd have two to the negative first plus two. Oh, that's easy.
That's just five halves. So I've got 2.5 and then negative 2. If I plug in a negative 1, I'd have 2 to the 0th, which is 1 plus 2, so that's 3. So I've got the point 3, negative 1. And if I plug in a 0 for my y value, that's 2 plus 2, 4. So I have the point 4, 0. And if I plug in a 1, I would have 2 to the second power 4 plus 2, 6. So the point 6, 1. is my next point. You can see here we have a graph that is indeed racing off to the right and approaching an asymptote as we move to the left. Alright, and again the arrow is not necessary there as long as we extend our graph to the edges. The x-intercept, well again from our graph and our table we found that to be the point 4, 0. But if we did want to find that algebraically, the x-intercept happens when y equals 0. So 0 equals log base 2 of x minus 2 minus 1. Add 1 to both sides.
Raise both sides as a power on a base of 2, so 2 to the first equals, cancel out the logarithm, we just have x minus 2, and we'd add 2 over to the other side, we find that 4 equals x. To find our y-intercept, that's what happens when x equals 0. So here, this one's actually easier to do from our original one. Oh, actually, you know what? Look at our asymptote. There is no y-intercept because we do not come close to the y-axis.
And let me show you why. If we were to plug in and say, okay, my y is going to happen when x is equal to 0, look at what's inside the logarithm. We cannot have a negative argument, negative number going into a logarithm.
It doesn't work. It's not defined. So we have something that's off the domain again. The domain of this logarithm that we've just graphed is all x values that are greater than 2. So all x's such that x is greater than 2. Or all the points from 2 to infinity.
Buzz Lightyear's favorite interval. And the range will be all y values that are elements of the real numbers. Logarithms have infinite ranges.
All right, now we're getting to the end of this beefy sketch, sketch a complete graph. Again, please realize this is going through all of the functions that we've covered in chapter zero. And your test is going to be essentially half of this review.
So if you're like, oh, man, this is taking forever. There's a lot that we covered here, pretty much all of the work you've done the past two years in one chapter. So we're condensing it down quite a bit. All right, here you've got the sum of two absolute values. So that means that I've got two absolute values here that have different hinge points.
The first one is where x equals 3 and the second one where x equals negative 1. So I've got... I've got three different places. I've got to the far left, in between these two, and then to the far right. So let's start with x being less than negative 1. When x is less than negative 1, that means I'm on the negative side. sloping sides of both of these absolute values.
So I have a negative x plus 3 and then a negative x minus 1. This is negative 2x plus 2. When I'm in between sorry 3 is the higher value there so when x is between that negative 1 and that 3 now I'm on the positive side of the x plus 1 right I'm past that hinge points. It's now sloping up, but this one is sloping down. So I've got the negative x minus 1. I've made this one negative, but this one remains positive. Oh, yeah, yeah. And the x's cancel, and you find that we just have a negative 4. And then lastly, we're off to the right, we are beyond x equals 3. Now both of these are positive, so we would combine them together.
So we get x minus 3 plus x plus 1, and we find we have 2x and minus 2. I'm sorry, I just caught something here. I think I went backwards in the middle part. When I said that we are in between, we are on the positive side of this x plus 1, so that one is the same, but we're still on the negative side of this one, so it should be minus x plus 3. I think I had those switched. Yeah, yeah, you can see it's, the x is still canceled, but it's a positive.
for instead and I think when we go to plot this we'll probably see that in order for it to be continuous as it should be they should connect up and that wouldn't have happened all right so what we've just done is we've established our piecewise function sorry we're still going to call it f of X though Okay, so we've got a negative 2x plus 2, and that's happening when x is less than negative 1. We've got 4 when x is in between negative 1 and 3. And then we've got 2x minus 2 when x is greater than 3. And so again, we just need to have the or equals. So you'll notice I've got two places where we join up with negative 1 for an x value. I want one of those to be or equals. Not both, not neither, just one of them.
And you could put it up here and omit it here. It doesn't matter. Okay. So let's go ahead and plot this out.
So negative 2x plus 2. I want to keep that one as long as x is less than negative 1, so we're going to keep that one over here. Next, we have a whole bunch of places where y equals 4, and that's all the way up until x equals positive 3. Right, and then from there we have 2x minus 2. Okay, good, yeah, here it connects up like it should. And we only want to keep that graph when we are past x equaling 3. So we get this little trapezoid look up there. Alright, so what can we deduce from that as I go to answer the x-intercept question?
Well, we never interact with the x-axis, so there are none. No places where that graph has an output of 0. The y-intercept is pretty easy. If I plug in x equals 0, 0 it would just give me an output of 4 so we've got the point 0 4 the domain for this one as we look at the graph it's all connected them any X value has a place to go in here so all X's that are elements of the real number is here Or interval notation from negative infinity to infinity. And my range, I've got all y values such that y is greater than or equal to that height of 4, where our trapezoid is kind of leveled off at the bottom.
Okay, and if we were to make some points here to plug in for our table of values, it would just be nice to include at least one point from each piece. So I'm going to choose negative 2, 0, 2, and 4. And my output's there. I get a 6, a 4, a 4, and a 6. Again, just making sure that each of those x's, when I test them, go into the correct place in the piecewise, or you could throw them back into the original absolute value.
That'd probably be the wiser thing to do to make sure everything is plotted and matching up accordingly. Number 11, we're just graphing the function here. So you could make a table of values if you want, but I think we, these are all linear, so I think we should be just fine. So take a look at our domains for each piece, though.
Negative 2. to negative 1 and we are including those in this top one. So negative x minus 2. Here's minus 2. Negative x would be showing us a slope like this and we only want to keep that one when x is between negative 2 and negative 1. So here's negative 2 and negative 1 just right there. Alright, then the middle one.
Okay, here we have just y equals x. So at y intercept of zero, it's going to keep charting out that way. And we want to keep that one between negative one and positive one for our x values.
just goes there and then at negative x plus 2 here's plus 2 and it's going to go down 1 into the right one for its slope and we're only going to keep that one until we reach an x value of 2 there you go so we do have like included end points here the domain works notice we go less than or equal to negative 1 and then not including that negative 1 again so it does satisfy the Vertical line test, we've got a function here. It's this nice little Harry Potter scar of a graph. All right, next, number 12. If Jonita invests $1,500 in a retirement count that earns 8% compounded annually, how long will it take this single payment to grow to $5,000?
All right, well... Here we have this compound. Investment, so we want to find the amount after time has gone by.
She has put in $1,500. 1 plus our rate is 8%. That's 0.08.
And since it's compounded annually, that means only one time per year. So over 1 to the 1 times t, we got this. Okay, so this ends up being pretty simple. If I clean that up a little bit, it's really just 1,500 times 1.08.
We can actually clean this one up. It gets a little bit messier when we have, like, compounded monthly or something in there. But this is a pretty easy one to solve.
Okay. How long will it take this payment to grow to $5,000? So we actually know the final value.
Let's isolate the exponential. We'll divide both sides by 1500. And so that's the same thing as 50 divided by 15, which is 10 thirds. OK, and that's 1.08 to the T power. And if I'm trying to get to an exponent. I got to cancel out the base.
I'm going to use a log base 1.08 on both sides. 1.08. And so this would be what I type into the calculator and I come up with 15.644.
Okay. Let's just say years until $5,000. Again, inputs and outputs, dollars and years. Okay. Next, this section, if you want to, this is something that always seems to catch students.
And not so much in like the find the domain sense, but it also has some little issues where we want to really be careful that we're following the input. So this is one that is there are silly points to lose. But let's take a look. OK, so for the first one. my function f, I'm trying to find the domain of f, that's what that Capital D, F means.
And it's a rational function, but it also has a square root. So the square root is telling me that X has to be at least a positive 1. But because that's also happening in the denominator, I also know that X can't equal 1. Because if X equaled 1 exactly, it would be a zero denominator. And we can't have that either.
So the domain for F is going to be all X's such that X is greater than 1. or from one. 1 to infinity if we use an interval. The domain of g, well thankfully this one's a nice x squared, so it's going to be all x's that are elements of the real numbers, or anything from negative infinity to infinity. Now we need to find some compositions. So before I find the domain of f of g and the domain of g of f, I want to actually find those two functions.
So off to the side here, I'm just going to say f of g of x, which again I like to think of it like this so that I can see what's on the actual end. inside, I'm going to take g and plug it in for f. So I get 1 over the square root of x squared minus 1. Right, I just replaced the input in f with the function g.
All right, and so this one is actually pretty nice because my domain of f of g, it looks like, hey, as long as I've got something where the x squared, if I'm just looking at this function, I would say, hey, this is good. Really, x cannot equal anything that goes from negative 1 to 1. If it's anything in between there, any of those small little values in there, then I'd have a negative under a square root or a 0 under the denominator. So I can't include any of the numbers. numbers on that interval, otherwise I'm fine, which is even better than it was for f, it looks like.
However, this is where you get hung up, because whatever went into g, we have to look at that first. So x is going into g, and then that is going into f. So what's allowed into g? Anything.
Great. So then what actually is allowed into F afterwards is... anything that we can think of here.
So for f of g we've got the set of all x values such that x is less than negative 1 or x is greater than positive 1. In intervals that would be from negative infinity up to negative 1 not including it unioned with 1 to positive infinity. But the same thing does not happen when I do g of f of x. Oops, let me get that a little bit easier to read here. So here I'm taking function f and then putting that into g. In other words, I'm taking 1 over root x minus 1 and then squaring it, which if I square the top, square the bottom, I get 1 over x minus 1. Again, this looks like a really, really nice thing for us because it looks like as long as x doesn't equal 1, we're good.
But that's not the case. So here, sorry, f of g of x, that was 1 over square root of x squared minus 1. And then g of f of x is 1 over x minus 1. The domain of g of f, let's think about it. What was allowed into f, right? That's where x goes first.
My input goes into f and then into g. So what was allowed into f was not all real numbers, it was just any x's that are greater than one. So that's what my domain is here.
we can only allow the numbers from 1 to infinity to go into f, and then when it goes into g, we get all those outputs to happen, right? Then we're okay, but only what was allowed into f can continue on through function g as well. All right, we're getting into the tail end of this review. Here we've got some parametric problems, so we're going to graph these. I've got my sines and cosines again going from 0 to 2 pi for my t interval.
So let's stick with those pi over 2 increments that give us such nice clean values coming out of those sines and cosines. So for x, it's 5 times the cosine. And for y, it's 2 times the sine. Alright, now I ripped through those very, very quickly.
All we're doing is one at a time plugging in t equals 0, t equals pi over 2. If you need to set up a reference triangle or something like that, by all means do so. But my initial point, we have a closed interval, so I do have an initial point. It's the point 5, 0, right there.
Then as we move through the table, we went to 0, 2, and then negative 5, 0, and then 0, negative 2, and then back to 5, 0. All right, good. And as we drew those points, we kind of rotated through in a counterclockwise fashion. So we'll connect those points with some arrows to indicate the direction. Right. All this is in the instructions.
If we kind of get, oh, do I need to put arrows or not? That's all included in the instructions. Next, find a Cartesian equation.
Well, realize that to pull the t values, and that's what we're trying to get rid of, eliminate the parameter, we need to go by the trigonometric identity that will get rid of sines and cosines. And that's Pythagorean identity, sine squared plus cosine squared would equal 1. And so here, my sine is really equal to y over 2. My cosine is really equal to x over 5. So it is a substitution method, but we're substituting into an identity, which makes this a little bit more nitpicky. y squared over 4 plus x squared over 25 equals 1. So there's our Cartesian equation.
Again, it doesn't have to be a function, so it's fine that this fails the vertical line test. This is the equation of an ellipse. one revolution is traced with this interval of t. And it might be something like the top half or something like that, but here it's one full revolution.
Right, number 15, here we've got, well, this is pretty nice, another parametric, so x and y. I know that my t values have to be greater or less than or equal to 2. Okay, so this is interesting. It's like I've got something and then t and then it goes up to 2 and then stops.
So when t gets to 2, that's going to be my terminal point. It looks like there is no initial point. It comes out of nowhere.
But then it will have a terminal point. So the x value is going to be 1 plus t. In this case, that t equals 2, our terminal point, that will be 3. And our y value, the square root of 4 minus 2 times 2, that's 0. So this graph terminates at the point 3, 0. All right, and now let's think of what would be good t values to use that would kind of lead up to this in nice. I'm looking at the square root more than anything.
So if I use t equaling 0 and t equaling 3 halves, I should be good. Okay, so for x, 1 plus 0, 1. 1 plus 3 halves, 5 halves. For y, the square root of 4 minus 0 is just the square root of 4. That's nice.
And then for 3 halves, this would be 4 minus 3. Okay, square root of 1. 1. Okay, so I've got x equaling 5 halves. and we have a height of 1. X equaling 0, we've got a height of 2 right there. We could do some negative values if you wanted to take this one further, but what we're seeing happening here is a graph that's coming in in this direction.
stopping at that point . So if we have to substitute to kind of solve this thing, if x is equal to 1 plus t it means t is equal to x minus 1 and then we can substitute into y, 4 minus 2 times x minus 1 and we get y equals the square root of 4 and then there's a negative 2x and a positive 2 which means plus 6. Alright, and so this would be a Cartesian equation. We've solved for y, which is the beneficial way to do this. That's how we're used to seeing functions, and you could describe it as the top half of a parabola. Okay, and again, we're looking for a very, very casual description there.
Are you analyzing the shape of the graph? Are you looking for something like that? Alright.
Next section is kind of a breeze, change from degrees to radians, 30 degrees, man I hope you know that that's pi over 6 without having any work to do. Next for letter B, here we're leaving our answers in terms of pi, so here negative 140 degrees. Well that's degrees and I want to cancel out degrees so I'm going to multiply by pi over 180 degrees because these are equivalent angles we can kind of use almost like dimensional analysis right this will cancel out the degrees and give us some fraction of pi so negative 14 over 18 pi negative 7 ninths pi or negative 7 pi over 9 Number 17, change from radians to degrees. Well, here, negative 3 pi over 4. Man, I hope we know that that's 135 degrees.
So here's just a negative 135 degrees. If you need to draw a reference or something like that, or use the conversion formula, that's fine. But the part a's in both of these problems are ones that I would hope you would just know. Alright, now 3.7 is kind of an interesting one.
Remember, this is radians. So 3.7 in radians, think of that as like being over 1. I want to cancel out the pi. So I want to multiply by 180 over pi.
And here, we're told to round this to three decimal places. So we would just type this into our calculator and use the pi key. And we get 211.994. And we have turned that into a degree measure. Okay, number 18. Here we have graphing a trig function.
We're really on the tail end. Here's our final page. Alright, so I know these aren't the same order of blanks that we've used.
in class, but hopefully we can still make some sense of it. The amplitude, notice what's multiplied onto cosine here is a negative 2, so my amplitude is the absolute value of that, it's 2. I'm going to try and color code this thing as well. Okay, the center line, this 3 that's just kind of floating around by itself, that's like a plus 3 that was at the end and it's just been written out front.
So the center line is actually y equals 3. So I actually have everything I need for my y-axis with that center line and the amplitude. Next, let's look at the period of the graph. So here we've got cosine of 4 times.
All this stuff is inside that cosine. Man, that's kind of a finicky one to work with, isn't it? So this is actually going four times as fast.
So the period is 2 pi divided by that 4 that's multiplied onto our x. And so 2 pi over 4 is a period of pi over 2. then our increments are going to take that pi over 2 and divide it by 4. So we're actually going to measure out pi over 8. Then lastly my phase shift. Well what would make this the contents of this cosine 0 out?
It's if x were to equal pi over 4. So they kind of did us a favor by factoring out the 4 to begin with. They made it much easier for us to see what zeroes out that. So here I want to count with increments of pi over 8. Okay, and then I'm actually going to start at pi over 4, which is 1, 2 pi over 8. And then for increments, we're just going to go 1, 2. Sorry, that's pi over 2, 3, 4. And that should give us one full period of the graph.
Now this is a negative cosine function at its heart, so usually cosine starts off at a maximum, but since this has a negative it starts at a minimum, then goes to the center line, then a maximum, then a center line. line, then back to a minimum. Here's one period of the graph.
And if need be, every little tick mark along the way, we would just keep that pattern going of center line, maximum, center line, maximum. And we could continue to graph that just, well, as much as we needed it. Alright, but hopefully that was a huge point of emphasis for you last year and not too much of a stretch to work with anymore. Alright, next one.
Find the inverse and then show that you actually have found the inverse or not. So what we have is actually y equals this rational function. Let's flip the roles of x and y.
We'll multiply both sides by y plus 1 to get rid of the denominator. We will factor out. No, we will not factor out. Not yet.
We want to group all of our y terms together, so I'm going to subtract y and subtract x. Then I'm going to factor out a y. And then finally...
My f inverse of x is going to equal that negative x minus 2. We're going to divide out the x minus 1 because it's multiplied on. We'll divide it out to finish solving for y. All right, so now the proving part. First way, f of f inverse of x.
So here's f. I'm going to take my inverse and substitute that in. Negative x minus 2 over x minus 1 minus 2 over negative x minus 2 over x minus 1 plus 1. I'm going to multiply top and bottom by a factor of x minus 1. That's totally allowed. It will cancel out the fractions. Okay, so I get negative x minus 2 minus 2x plus 2. right this negative 2 will distribute into that set of parentheses.
Likewise underneath, negative x minus 2 plus x minus 1. Just multiplying that in. It cleans things up in a hurry, doesn't it? If we combine like terms, constants cancel in the top, x's cancel in the bottom. That's what should be happening.
So we're happy about that. I get negative 3x over negative 3, which equals x. And just to confirm that this works the other way as well, now we've got negative x minus 2 over x plus 1 minus 2 over x minus 2 over x plus 1 minus 1. Just taking my inverse right here and plugging f into each of those x places. Alright, so we'll do that same little nifty trick of cleaning out our denominators. And so I would get negative x plus 2. The denominator cancels, but this negative is still going to distribute in.
That's kind of why I wrote it so obnoxiously far out front there. And negative 2x. minus 2. In the bottom we'll have x minus 2 minus x minus 1. Again distributing negatives are a huge pain in the butt so please take care when you distribute things. We had like a backwards and forwards distribution in here.
Constants cancel, x's cancel, happy to see that. We again get negative 3x for our numerator and negative 3 in our denominator which equals x. Cha-ching! All right, now realize the only answer that you actually had was way back here.
All this extra work was just kind of backing up. Yep, it's an inverse. I can prove it.
You know, kind of work. All right. Number 20, our penultimate problem here.
We are drawing from an inverse. So for letter A, we're drawing the angle whose cosine, well, that means we're stuck right here, is a negative root 3 over 2. If cosine is negative, negative cosine is X left and right that has to be over here and we've got an adjacent over hypotenuse that is a negative root 3 over 2 that's an X on R over here now hopefully we can tell that the angle inside if we're looking at a root 3 right here and a 2 we should know that that missing leg there is a 1 and that this is a 30 degree angle okay so here we're looking for what angle Leads us up to that. We're always starting on the initial side to measure an angle.
So here we've got 150 degrees. Or we could say that that's 5 pi over 6 radians. All right, for letter B, we'll go to green for this one.
Angle whose tangent, so we're down here. Okay, and the tangent is positive. And so that's my y on x. And here we've got to use Pythagorean theorem. It's not a special right triangle.
That's okay. I mean, I guess it's not really imperative that we do this. This would be the square root of 17. But this angle inside is our theta. So when I find that in degrees, I get 14.036 degrees. And then if I convert that to radians, I get 0.245.
Alright, and because our 1 over 4 was falling right here where we wanted it into quadrant 1, you could actually just type this into the calculator to get those decimal answers. So that was a direct calculation problem because we didn't have special right triangles to give us a perfect clean answer. And the final problem on the entire review, solve for y and simplify completely.
So here we're going to condense some logarithms together on the left. So I would have the natural log. Notice subtraction between the logarithms. That means y minus 2 over 3. And then on the right we don't have that luxury.
But that doesn't mean we're out of luck. To undo this natural log, we're solving for y. I'm going to raise that onto a base of e. and I'll do that with everything over here on the right hand. hand side as well.
Get a lovely little cancellation on the left. So that has now become y minus two over three. And on the right, notice I've got e to the and then a sum. So I can call that e to the x squared times e to the natural log of x plus two.
Because when I add an exponent means I had terms with the same base multiplied together. It's using that property very cleverly in reverse. and this comes up again in chapter 6 for us quite a bit bit. So definitely a skill that is nice to be introduced to so early. All right, so where are we at now?
We were able to sneak another cancellation out because we had times e to the natural log of, remember those are inverses, so that has really all just turned into x plus 2. Then we're going to multiply both sides by three And then lastly, add two to both sides. All right, and believe it or not, this is the simplified answer. This is the simplified answer. Some of you may be thinking, well, can't I distribute this in there?
Let me show you what that answer would look like. 3e to the x squared times x, so we'd probably call that 3x. e to the x squared plus 6 e to the x squared plus 2. Notice how with our answer that we came up with leaving this factored out notice how many fewer places we notice an x in our answer.
Here we only have two x's here we have one. two, three, that leaves this being a much easier answer to work with than this one down below. Alright, so in terms of calculus this is what we're going to refer to when we say a simplified answer.
Alright, that brings us to the end of this chapter zero review. Hopefully this helps us kind of button down some of those kind of loose end issues and we're coming into this first test of the year feeling pretty confident. Thank you for watching.
Thank you.