Welcome everybody to the Algebra 2 Chapter 5 Review 2017 edition. Let's get started. Okay, so what did Chapter 5 encompass?
We looked at a lot of radical expressions and exponential expressions, and we found overall that roots and powers undo one another. And we also had quite a few properties of radicals and exponents that apply as well. Alright, so our first concept to solve and find real solutions, this just means using roots to undo powers.
So for problem number one, I want to isolate the term that has the exponent on it. In this case, it means multiplying both sides by 9. Then x to the third equals 729. And then to undo that third power, we will take the cubed root of both sides. And x will equal, well, 9. Okay, on the right-hand side, the thing to the fourth power is a quantity. I still want to isolate that thing to its power, so I will add 7 to both sides. So x minus 11 to the fourth equals 256. Then to undo that fourth power, I will take the fourth root, keeping in mind that this is an even root I'm taking.
x minus 11 equals the fourth root of 256 is 4. Comes out clean. We could get that from a calculator. But since it was an even root and we put that root there, we took the fourth root of both sides. We need plus or minus.
All right. Our final step is to add 11 to both sides to finish solving for x. So x equals, I've got two solutions here, negative 4 plus 11 and positive 4 plus 11. And, of course, if we wanted to check our solutions, all we would have to do, plug those x values back in and test them in a calculator. When I plug those in, does the left side make a 249? So we're off to a good start.
This section right here on your test is going to be the most troublesome. Take heart in the fact that each question is only worth two points. However, this was that 5.2 section. We spent three days on this in class. This is where all the properties of exponents and radicals, simplifying them, all that comes together.
So let's take a look at all these. They all kind of employ some different techniques. Problem number three. We notice in the instructions it says we have to show work.
We have to show steps that show that this was done without a calculator. And you can see it's just a number to a fractional exponent. That just means break it up into its root and its power. And remember, we want to do the root first. Taking a root of a number makes it smaller and easier to work with.
So we want to do that first, even though technically we could call this 32 squared and then take the fifth root afterwards. That does the same thing. Fifth root of 32 is a 2, so really, if I simplify what's in parentheses, I'm just taking 2 squared.
My answer is 4. And we could always use a calculator to check, but showing those simplifications along the way, that ensures that we'll get full credit for that problem. Problem number four, best way to approach this, I've got two roots. I have the same index there, both fourth roots.
And since I'm dividing them, I can simplify that division. as taking place under one root. And 32 divided by 2 comes out to be 16. Fourth root of 16 is 2. Now, I did not put fourth root there.
I did not take the fourth root of both sides of an equation, so I do not need plus or minus here. 2 is the only root that I have for this. Number 5. Here we've got a cubed root. And what we're looking for are perfect cubes underneath the cubed root.
So... What I notice is that if I break this apart, the cubed root of 16, well, I can break that down using a factor tree. So I've got 2 times 8, 2 times 4, and 2 times 2. Since it's a third root, I'm looking for groupings of 3. That is a hidden perfect cube underneath that cubed root. So I could call the cubed root of 16 a freed 2, and then 1, 2 left under a cubed root.
For the next term, that A to the sixth. Well, that's also a cubed root, but I can think of that as a to the, sorry, second power cubed. So cubed root cancels out the cube. It just becomes a squared. Another way would be to think of that as having a fractional exponent.
It would be the 6 over 3 power, which also simplifies to 2. For the next one, I've got a perfect b to the third underneath the cubed root. So that cancels perfectly, leaving me with just b. And then lastly, the cubed root of c to the fifth, I'm going to think of that as the cubed root of c to the third times another cubed root of c squared, right? If they're multiplied together, I would add the exponents.
So I end up with this perfect cancellation, but then a cubed root of c squared remaining. Let's hook up all these things that we have, put what's left under radicals together and what's freed, put that together out front. So freed, I have a 2a squared b. C and then left underneath the cubed root I have a 2 and I have a c squared That should be our final answer of all these problems and remember they're all worth two points That one is the most difficult hands down. Okay, the next one we've got two cubed roots being multiplied together Well, I can multiply that together under one cubed root 120 that makes 136 And then we break that apart with a factor tree.
See if we can't find any perfect cubes hidden underneath. That's 2 times 68, 2 times 34, 2 times 17. Ah, sure enough, we can free a 2. So I get 2 times the cubed root of 17 for my final answer. For the next one, I notice I'm adding radicals, and I can only add those together if they are the same radicals, like combining like terms.
So I've got a fifth root of 2. I can't break that down anymore. But this fifth root of 64, I can break that down. That's 2 times 32, 2 times 16, 2 times 8, 2 times 4, 2 times 2. Since it's a fifth root, I'm looking for a grouping of 5, and sure enough, I do have one. There's a hidden 2 to the fifth power underneath that. So notice I already had a 2 out front, and I freed another 2. That means 2 times 2, 4 is out front, and I'm left with the fifth root of 2, because there was one 2 left underneath.
And now I can combine that with that 13 times the fifth root of 2 to make a grand total of 17 times the fifth root of 2. It's just like combining like terms again. All right. For the next one...
Notice that our denominator has a square root in it, which is considered improper, but it's not just a single square root term. It is a sum of two terms, meaning that the only way to cancel that out is multiplying by the conjugate. So across the top, I will distribute, and I'll get 5 times 3, 15, minus 5 times root 2. And across the bottom, first times first, that's 3 times 3, 9. minus second times second, the square root of 2 times itself is just 2. I get that 15 minus 5 root 2 all over 7. And that's my final answer. Notice how the conjugates produced just a nice rational number underneath. For number 9, here for this one, I don't like this cubed root in the denominator, so I need to find a way to cancel that out.
Well, the cubed root of x, it would cancel out perfectly if it were the cubed root of x cubed. So I'm going to multiply in the cubed root of x squared. And if I do that to the denominator, I have to do that to the numerator as well. So the fraction maintains its value.
So across the top, I've got the fourth root of 16. Hey, we just did that in problem four. That's a nice clean two times the cubed root of x squared. All over in the denominator, we will have the cubed root of x cubed. That's just a regular x.
There's our answer. And lastly, for problem 10, here we have two terms with the same base. They're exponential terms, and I'm multiplying them together, which means I add their exponents. So I've just got to take 2 fifths plus 3 tenths. And if I do that, 2 fifths is the same thing as 4 tenths.
I'm just getting a common denominator to add these fractions up. I get x to the 7 tenths power. for my final answer. Since the problem began with fractional exponents, my answer should end that way.
And you'll notice that with each of the problems here, if it started off as a root, my answer ended up as a root. If I change the form to be a fractional exponent along the way, I just want to make sure my answer matches the format of what I started with. So this is the section that takes some time to get through. But again, we kind of knew that coming in because this was the tough section.
of the entire chapter. Next, there's a word problem here. The time it takes a pendulum to make a complete swing is given by this formula.
Okay, they tell us what each variable means. Okay, then they say, how long does it take? So we're solving for t.
So that equals 2 pi times the square root of, okay, g is the acceleration due to gravity. So that's 32 feet per second. and then The huge pendulum measures 73 feet and 9.75 inches.
Okay, so 73 feet plus 9.75 inches. I want to convert that into feet as well. So I have the entire thing as just some decimal measure of feet. So I'm just going to take 73 plus 9.75 divided by 12. Find out what that is.
And it's 73.8125. That is what's going to go here. And it's okay because it's a, you know, these decimals should not frighten us because this is just a calculator problem in the long run anyway. So we're trying to find that time. We're going to want to round this answer to the nearest hundredth.
So here we go. I'm taking 2 times pi times the square root of 73.8125 divided by 32. And I get 9.54. Okay, so it takes that many seconds, right?
They mentioned that time is measured in seconds in the problem, that many seconds for it to complete one swing. That's a long time. That's got to be just an absolutely awesome swing to ride on. All right, numbers 12 and 13. Here we get into graphing our radical functions.
And if you know your parent functions and transformations, this is very easy. But if you don't, do not worry. You can always use a calculator table to graph these as well.
So do not worry if we are unsure of how to graph these. All right, so number 12, we've got a square root function that is shifted to the left three units and down two units. So again, using my standard parabolic growth, I'm able to quickly graph that with transformations. But again, you could get this out of a calculator, don't worry about that.
Once we have the graph, we can see that our domain are all x values that are greater than or equal to negative 3. We could also find that by taking what's underneath the radical and saying, hey, that x plus 3 has to be greater than or equal to 0 to make the square root happy, and just solve for x. You would get the same domain. For my range, those are the y values that I see the graph at, and if I scan the graph from bottom to top, I don't see the graph until I get to a height of negative 2. And then I'm going to see the graph the rest of the way up. So my range is all y values greater than or equal to that starting point at negative 2, that starting height at negative 2, I should say.
For number 13, here I've got a cubed root function that is going three times as fast, it's twice as tall, and it's shifted down one unit. So what we'll see is, and this one I may actually refer to a calculator for to get a better look at. But I should be seeing something that looks along the lines of this.
And again, we can make a better graph using our calculator table. This is just to make the video more efficient. That's the only reason I'm kind of zipping through this.
The nice thing about cube root functions, they accept any x value. So all real numbers are acceptable for my domain. And the range is also going to continue opening from bottom to top. So it will output.
all real numbers as well. It's really not too bad. This should be a pretty quick section of the test for you. Continuing with our graphical understanding, now applying transformations number 14. Here we just have to analyze what are the transformations being done.
A negative sign out front indicates a vertical flip or a reflection across the x-axis. A two-thirds out in front of the radical function, that sixth root function, that's indicating that it's two-thirds as tall, or a vertical shrink by a factor of two-thirds. That minus one grouped in with x means that it is being shifted or translated to the right one unit. And the minus 13 at the end means that it is experiencing a translation or a shift down 13 units. And this should be just...
just a walk in the park for us because of all that we've done with transformations so far in the class. Number 15, here we have to construct our translation. So first of all, we have a horizontal stretch by a factor of 4. So I'm going to make that transformation kind of just piece by piece. So I'm going to call this function h. And to stretch it horizontally by a factor of 4, that means it needs to take a lot more time to cover the x-axis.
It's moving more slowly. So I want f of 1 fourth x. I want it to travel at 1 fourth the speed. That way it will take four times as long to cover the x-axis.
And all that means is I'll have the square root of not x, but 1 fourth x plus 1 to represent f of x with that transformation. Next, what I need to do is translate that three units to the left. Okay, I'll call this function j of x. All right, and what that's going to do is that's just going to translate it three units to the left. All right, that's going to take my function h, and to translate it to the left, I need to add to the x value.
So it's going to take the x, and it's going to add 3 to it. So looking up above, up in my function h, I'm going to take that same function, but instead of the x, I'm going to replace that with an x plus 3. All right, next we need to translate that 4 units down. I'll call this one k of x.
This has a lot of steps to it. That's going to take my function j that I just made, and to translate it down, it means outside I will subtract 4. So just kind of building this function along, I've got my 1 fourth times the quantity x plus 3 plus 1, and then outside the square root, I'm going to subtract 4. And then finally, we have our reflection in the y-axis of the graph. So finally, I can call this g of x. It's going to take that function k that I just had and to reflect it in the y-axis, I'm negating the x. So I'll have the square root of 1 fourth times not x, but negative x plus 3 plus 1. And then I will subtract 4 afterwards.
I can assure you that the test will not have this many followed by prompts in it. But what you can see is we can definitely follow step by step and just keep applying. Am I adjusting x or am I adjusting the function outside?
If you can do this problem and understand it fully, you should have no worries by the time you sit down to take the test. Next, solving radical equations and checking solutions. So for problem 26, the first thing I would do, I see a one-third power.
I would call that a cubed root. That's what a one-third power means. And I want to isolate that cubed root, so first I'm going to add 4 to both sides, and then divide by 2, and then cube both sides to undo that cubed root on the left, and then finish solving for v. I'll add 3 and divide by 6. The instructions say that we want to check our solutions, and that is absolutely true. However, really the only time we need to check our solutions are when we're taking even powers of both sides.
Because of the way domains and ranges work with odd radicals, odd indexed radicals, I should say, I know that this answer is valid. Number 17. Here we've got perfect fourth roots on both sides. So. They're isolated, and I can raise both sides to the fourth power to wipe out the fourth root. That gives me x plus 12 equals 5x minus 4. And if I subtract x from both sides and add 4 to both sides, find 16 equals 4x, that means x has to equal 4. Now, this is the only solution I came up with.
I'm pretty confident it's going to work, but let's plug it in and check. On the left, I would have the fourth root of 16. And on the right, I would have the fourth root of 16. Yes, that checks out, so I can be confident in that answer. Up until now, we've had a lot of problems, and they've gone pretty quickly. Once we get to number 18, this is the next one after that simplifying section that takes some time.
I'm going to rewrite this 1 half power as a square root. And notice that that square root's already isolated. So I'm able to square both sides, and that does not mean squaring each term on the right. It means squaring the quantity that's on the right. On the left, we do get our square root to cancel.
It's just 7x minus 33. But on the right, we have to FOIL this out. So a lot of students will miss that term in the middle, and it will create some problems for them later on. They won't get their answers to check out. You need to be sure that you are foiling there. Well, we can see the x squared and x terms starting to line up.
Let's go ahead and get this set equal to 0. It's a quadratic. So I'm going to subtract 7x from both sides, and I'm going to add 33 to both sides. Let's end up with this, which is factorable. So I end up with two possible solutions, x equaling 6 or x equaling 7, and I just want to check both of those to make sure that they will work.
On the left, I'll have the square root. Let's test 6 first. 7 times 6 is 42. 42 minus 33 is 9. So does the square root of 9, the positive square root of 9, as shown by the problem, does that equal 6 minus 3?
Yes, it does. Let's try 7. Square root of... 49 minus 33 is 16. Positive square root of 16, does that equal 7 minus 3?
Yes, it does. So both of our answers are acceptable here. But remember, if you think back to this section when we were solving, there are cases where only one solution will work, and there are cases where no solutions exist.
So we do want to be sure, especially when we're raising both sides to an even power, that we are checking those solutions, everybody. All right, we're closing in on the end. Next section, solve the inequality. Think about the domain, because for the most part, we're just going to solve this inequality like we would expect it to be solved, you know, treat it almost linearly. I like to have a little number line down here just to kind of sketch out my solutions.
It always helps me piece my answer together. So I'll go ahead and put those in now. For number 19, what is the domain? This is a square root.
It's an even root, so the domain has to be something that will keep the radicand, what's under that square root, positive or zero. So off to the side, I'll just kind of do this in a thought bubble. I know that 4t minus 4 has to be greater than or equal to zero. And then let's just solve this.
That means 4t has to be greater than or equal to 4. It means t has to be greater than or equal to 1. This will factor into my solution somehow. So I want to hang on to that. Now let's solve the original problem as we see it. Let's subtract 2 from both sides, and I've got the square root of 4t minus 4 has to be less than or equal to 4. We'll square both sides, add 4 to both sides, and divide both sides by 4. t is less than or equal to 5. So on my number line, I have two key values occurring here, one of them happening at 1. the other happening at 5. Out of my domain, I know I have to be greater than or equal to that 1, but I also have to satisfy this piece right here that says I have to be less than or equal to 5 as well. So I know my answer is all t values that are between 1 and 5. Not too bad, as long as we remember to check the domain.
And really, it's only if it's an even root. If it's an odd root... We don't have to worry about that thought bubble stuff.
We can just go right through and solve the inequality as we see it. Those domains accept any x value. Number 20, I do have another square root, another even root.
So that 5x minus 2 underneath, that has to be greater than or equal to 0. If I add 2 to both sides and divide by 5, x has to be greater than or equal to 2 fifths. I'm going to hang on to that. That's going to be part of my solution. Now let's go through and solve this as normal.
If I square both sides here, I find 5x minus 2 has to be greater than 9. We'll add 2 to both sides and divide both sides by 5. x has to be greater than 11 fifths. Well, now take a look at this. Our domain says x has to be greater than or equal to 2 fifths, but then we said that to actually solve the problem, x has to be greater than 11 fifths.
In order to make... both of those happy. Let's sketch those out here on our number line.
In order to make both of those happy, any of the values here aren't going to work. Yes, they work in the domain, but they won't output anything that's greater than three. So my answer is just all x values that are greater than 11 fifths.
And this is why that number line is so useful. We want to make both of those little inequalities happy. Next, combining functions together, performing operations on functions.
So here we've got f and g of x, and first thing we have to do, find f plus g of x. Remember, that just means I'm adding the two functions together. So 6x to the 3 fifths plus negative x to the 3 fifths. If I combine those together, hey, they're like terms. 5x to the 3 fifths.
Letter b, f minus g of x. Here I'm taking my function f and subtracting away negative x to the 3 fifths. Again, they're like terms.
I can combine those together. I get 7x to the 3 fifths. It's just like you would expect it to work. For letter c, f plus g, now they're giving me a number to plug in.
And we could go about this two ways. This is what we said in class you can use your calculator for. But if we already made f plus g of x right over here, let's just plug a negative 8 into that.
It's going to be 5 times the 5th root of negative 8 to the 3rd power. And we could get a decimal answer for that. It doesn't come out nice and clean, unfortunately. The only thing we can do is because it's an odd root and an odd power, we're able to free that negative sign. So negative 5 times the 5th root of 8 to the 3rd is as best as we can get.
or a decimal equivalent. For letter D, similarly, we're going to be doing a lot of the same work here. We're still plugging in a negative 8, but this time we're plugging it into what we set up for F minus G.
So that's 7 times, it would be the fifth root of negative 8 to the third power. It's going to look very similar to what we just did. We can free that negative, and we'll have the 7 times the fifth root of 8 to the third power for an exact answer.
Or again, we could get a decimal out of the calculator to work for that. Okay, and lastly, number 22. Here we have two more functions, 1 half x to the 3 fourths, and g of x is 8x. We want to find f times g of x. Notice that I believe on the test it's formatted a little differently.
It just says fg of x, and this is just my cursor when I captured the screen, so that's just an errant little line there. But the key is we don't want you to misinterpret this as meaning f of g of x. It means multiplication.
So I'm going to take f of x times g of x. And then I just want to multiply the coefficients together. 8 times 1 half out front is 4. And then x to the 3 fourths plus 1, that's 4 fourths, gives me a final answer of 4x to the...
7 fourths when I combine those terms together. Again, it's the same base of x, and we are multiplying the terms. That just means we add the exponents.
Closing in on the end here, last thing we covered in Chapter 5 was inverses of functions. So for 23, we have a function f of x. It's 2x plus 4, and we need to find its inverse. Notice it says to graph the function and its inverse.
I would graph this function right away. 2x plus 4. Welcome back to Algebra 1. We have a y-intercept of 4 and a slope of 2. I want to get this graph on paper as soon as possible. And of course, if I had a straight edge, I would use that here.
Remember that if we graph a function and its inverse at the same time, we should see perfect symmetry of those graphs across the line y equals x. so I should be able to tell pretty quickly if what I'm coming up with for an inverse makes sense or not. So if I plot out that original function, plot out y equals x, now we can find our inverse.
f of x means the same thing as y, our output, and our inverse flips the input and output. So we will do just that. Instead of y equals, we'll have x equals 2y plus 4. And we want our inverse to be a new function, so we'll solve it for y, just like all of our other functions have been expressed. We'll subtract 4 from both sides and divide everything by 2. So my inverse, we'll call that f inverse of x instead of y equals here.
It's going to be x minus 4, or x over 2 minus 4 over 2, which is just 2. Notice it's an acceptable answer to say the x minus 4 all over 2 that we arrived at here. That's acceptable. It's just much easier to graph. if we split this up into x over 2 minus 4 over 2. And that's what I did to arrive there. Another way of thinking this is 1 half x minus 2. It's 1x divided by 2. That's another way of looking at it.
So if I plot that out, another line, so at negative 2 is my y-intercept, and I chart out my positive 1 half slope, I get a pretty nice look here. Am I seeing the symmetry that I should be? And I sure am. There's perfect symmetry across that line y equals x, so I can be confident in my answer.
A calculator can help graph. There are two if needed. Finally, using compositions, this is your f of g of x, g of f of x, determine whether or not the functions are inverses. So the first thing I will do is try f of g of x, see what that produces.
So I'm working from the inside out. g of x, if I plug that into f, that's the same as taking f of 2x minus 3, which means instead of 2 times x, I have 2. times 2x minus 3 plus 3. And I just want to see, does that simplify back to the x that was the ultimate input? Did f undo g? Well, to get there, I'm just going to simplify. Distribute and combine like terms.
It did not simplify to x. It only simplified down to 4x minus 3. No further work is required. Because it did not simplify to x, we can say that those are not inverses.
And 25, let's see if this one works out. f of g of x, we're using compositions like the instructions say. Well, f of x says I have negative 1 third times my input.
That's all of g, negative 3x plus 9. And then... Plus 3, does that simplify to x? If I distribute my negative 1 third, I get positive 1x, negative 3, and then that plus 3 at the end. Hey, those cancel as well when I combine like terms.
That equals x, so it looks good so far. It has to work the other way as well, so g of f of x. What happens when I plug f into g? First thing to do is distribute that negative 3. I get a positive 1x and a negative 9. Then I have my plus 9 afterwards.
When I combine like terms, hey, I get back to x again. Because both of those brought me back to my ultimate input, that plain old x, these are inverses. Alright, so there you have it, the Chapter 5 review. If you know how to simplify and you can use the clues and the instructions to set up your problem solving, you're going to do just fine on this test. Much more procedural than what we've seen out of Chapter 4. So I wish you happy studying and, again, feel free to ask any questions along the way.
But, again, if this video made sense to you, you're going to be in very, very good shape.