all right in this video I'm going to be working a full practice AP pre-calculus exam and if you're in my class this is the full mock exam homework and this kind of what the front page looks like so the first section is multiple choice no calculator and the first one's talking about which of these statements about really the nature of the behavior of f is correct all right so 1 to three in increasing at an increasing rate that doesn't seem right okay from x = 1 to x = 3 okay the graph of f is decreasing and concave up so it's decreasing at an increasing rate hey don't think too deep on this like you are going to trick yourself into the wrong answer right if f is decreasing it has a negative rate of change and if the graph of f is smiling then f is concave up which is equivalent to having an increasing rate of change so it's decreasing at an increasing rate the slope is less negative it's closer to zero it has increased as I move left to right I think that's enough of that decreasing and an increasing rate this one looks like more of the same graph of the polinomial function G is concave down from 0 to 10 which is the following could be a table of functions right well concave down is going to mean uh the rate of change is decreasing uh and we could see that if we were looking at you know tables with evenly spaced X values if the second differences were all negative okay so up here I am yeah only say concave down so 1 to two that's going to be plus 1 2 to 4 is + 2 4 to 8 is + 4 and 8 to 16 is + 8 okay and I'm kind of now recognizing that's an exponential function with exponential growth that's going to be concave up okay so that one's out okay but we could also see that looking at the second order differences these are all positive right + one + 2+ four okay then for B I'm going to do the same thing I'm going to add 10 here add five add 10 and add five okay so from adding 10 to adding five that's a decrease of five it looks like the graph would probably be concave down on that interval okay but then 5 to 10 is an increase of five so that's a positive second order difference which is feeling that concave up okay so it's not concave down the entire time from 0 to 10 because we're seeing X between one and five have a positive second order difference all right so then G of X I'm looking up here 1 to 5 that's + 4 + 2 - 2 - 4 this feels right okay so I'm decreasing by two decreasing by four decreasing by two all three are negative numbers yeah I feel good about that okay what's Happening Here -4 -2 yeah that's uh that's increasing by two okay so that's not necessarily going to be concave down all right we go that's number two the graph of H is a transform of G where H is 2G + 3 okay which of the following points is on the graph of H inverse the inverse of H well I think I'm going to start by knowing the one point that must be on the graph of H and then I'll think about the graph of H inverse all right so um 3 -2 is on the graph of G so that means that g of 3 = -2 then I'm going to say h of yeah I know how to plug in three so h of three = 2 * G of 3 which we know is 2 then + three did the work to know that that was going to be -2 so 2 * -2 is going to be -4 + 3 is going to equal -1 okay then that tells me that on the graph of H I've got 31 okay so that you know if you like to do that one thing I I don't like you get to say okay it's going to be answer Choice a but if I know that h of 3 = -1 then I know that H inverse of1 is going to equal three just works in Reverse so that's showing me that answer Choice a is definitely going to be right in a way that I like a little better okay so what are all values of X for is natural log of 3x plus natural log of 2al 4 okay when we have one of these with like kind of two different logs we're going to want to combine them with log property um because we can't say that e to the this side equals e to the that and then say it's 3x + 2 = 4 that's what they want you to do and that is not going to be right that's going to probably give you answer Choice a so we're not going to do that we're going to combine these two logs first with a log property okay so we have two logs added together that's going to be like multiplication on the inside of the argument so 3x multili by 2 that's going to equal four so that means natural log of 6 X is equaling 4 now I can take e to both sides power and that's why we always you know when we were doing this we worked to isolate the exponential or isolate the log so that we didn't run into any sort of like you know exponent property or log property issues where we assuming things that weren't true okay then that's going to be 6 x = e to 4th power I'm going to divide both sides by 6 and I would get 1 over 6 * e 4 so that's going to be answer Choice B okay yes this one all right so we've got a cone filling up with water a situation you might become more familiar with if you continue on and take AP Calculus next year um T equals 0 we're going to start pouring in the rate that we pour into the tank varies in such a way that the depth of the water inside the tank increases at a constant rate with respect to time this is very unrealistic right when we usually fill up a paper cup of or a paper cone of water uh from like one of those you know water coolers uh the water's coming out at a constant rate now if you think about it if as the water comes out at a constant rate the water level is going to rise a lot at first because there's not that much space for water to go in and then it's going to slow down later because there's more cone and that's the same idea that we need to be thinking about here but this is just not the same situation we're having the height increase at a constant rate okay now volume of water versus depth in feet actually you know the rate doesn't even matter we're kind of just comparing for the cone okay now as the depth of the water increases we're going to have more water in there so I can eliminate C and D immediately CU these are decreasing um we know the volume is going to increase as the depth of the water increases so I'm going to go back to these two graphs say well um kind of what I would expect to happen is uh is B like right for the height to increase at an increasing rate but that's not what we're looking at at all the volume versus the depth and so what I'm you know suggesting is that oh maybe no pardon me a is what you would expect because the water level is rising fast at first and then slowing slowing down I just said B um but may as I'm you know the water level is going up along the red portion I think we're not really adding that much volume you know we're adding relatively less volume then as the water level goes up to here and then eventually goes up even higher that's really what I'm talking about here in the blue area there's more space for water to you know be so as the depth increases the volume is going to increase at an increasing rate so that's going to be answer Choice B and that one is very tricky I would say like significantly trickier than the other like filling bottle type problems that I've seen in AP preal but you know for a practice exam I think that's a good thing all right the binomial theorem can be used to expand an expression of the form a plus b to the nth power that is true which of the following is equivalent to this binomial to the fifth power okay what we're going to want to do especially because this is like relatively unsimplified we're not going to want to grind through this by hand we're going to want to use the triangle so the triangle starts with like ones here and then we kind of just add in the middle one two one 1 3 3 3 1 and we're going to go until we hit this five 1 4 6 4 1 and then 5 10 10 5 1 okay so now that I've got the numbers I need I like to kind of arrange them vertically so I'm going to just know do that there and then go back and copy this in right here one five 10 10 5 and 1 okay then I'm going to have the powers of X and the powers of -4 Y and I think that's really what we're going to have to be worried about is the the negative here so I'm going to start off with X to the 5 and then I'm going to decrease the power of X all the way down to X the 0o so it's going to be x 4 x to the 3 x to the 2 and x to the 1 okay and then X to 0 I know that's equal to one so I'm just going to replace that with a one okay going be taking -4 y to various Powers okay so as the power of X decreases the power of4 y will increase so4 y the 04 y to the 1 okay and so I'm they all start with x to the 5th a is just clearly too good to be true I would not ask you the question if that's how it worked and then I'm going to look at the answer choices and you know I'm seeing kind of some similar I'm looking for one where all three are different um and I'm looking at that third power of X term all three answer choices that left that are left or different even emphasize that I'm kind of looking here here oops that was okay that's not going to snap too but you see what I'm saying like all three of those terms are different so I should be able to make that determination just based on that term okay so I'm going to look at that over here which is X 3 y the 2 it's right here right and then we're going to have all right so 10 x 3-4 y^2 should be positive okay so that's going to be answer Choice C all right then number seven rational function K is here what are all intervals on which K of X is greater than or equal to zero okay it's been a while since we've done this at least in my class I remember that we're going to identify our key numbers put those on a number line make a sign chart and you know hope for the best all right so my key numbers are the zeros of the numerator and of the denominator uh but a hole in the graph did not necessarily count as a key number I could simplify first and if that made my life any easier so that's going to be x = 3 x = -2 not because of that leading coefficient because of that I really don't like I don't see why they would do that um have that -2 out front that's just like needlessly confusing or you could be right for the wrong reason but anyway x = 5 is the other one know could just be like A4 on the outside that be totally different um but then okay I've got these three key numbers I'm going to make a number line -2 three is here and five is over here all right I'm going to choose a representative from each sub interval so a number less than -2 like I don't know maybe -10 K of 10 is going to equal -2 * -10 - 3^ 2 is a positive number * -10 + 2 is a negative number all divided by -10 - 5 is also a negative I've got three total negatives that's an odd number of negatives that'll end up being negative all right and something between -2 and 3 that would be 0o K of 0 equaling -2 is a negative number 3^ 2 is positive pos2 / x - 5 over a negative is going to be positive okay K of X is greater than or equal to Zer from -2 to 3 what did I do wrong here greater than or equal to zero oh maybe it just goes all the way to five maybe there's yeah okay there's not maybe no sign change at three okay I can't figure it out based on the workout I thought maybe after just two you know sub intervals I could stop but I'm going to check K of four I'll equal -2 * a squared thing which is always going to be positive 4 + 2 is positive and 4 - 5 is negative yeah it's going to stay positive okay and then continuing beyond that -2 to 5 okay I mean I think a is the only one that has that option um I'm going to finish out the problem just like so you understand what's going on here but K of six something Beyond five would be negative positive positive divided positive now since we went bigger than five and then that would be negative okay so we know that we're definitely interested in this area and that area then we would need to think about these kind of edge points are they included in our solution Reg or no when x = -2 the numerator is 0 K of x equals 0 that is greater than or equal to Z okay that's n when xal 3 same story it's part of the numerator but when x equal 5 that makes the denominator zero that's a vertical ASM toote it's definitely not defined so it's not going to be able to be greater than or equal to zero if it's undefined and that's why it's -2 to 5 including -2 but not including five all right function J is given by J of x = 55 * 1/3 x which these describes the N behavior of J all right well 1/3 to the X oh that's a little rough I I do a little better than that okay that's better so an exponential growth function is where we've got you know the base of the exponential being greater than one 1/3 is not greater than one so each time x gets bigger we're going to multiply by a third it's going to get closer to zero um and so so I'm going to say okay when x equals 0 J of X is 5/3 so graph is going to be kind of down here and then as X increases I'm going to keep multiplying by a third it's going to get much smaller much closer to zero okay that works too and then when X goes negative I'll end up you know like kind of flipping the fraction and multiplying by larger powers of three uh times it's ne5 so it's going to be very negative so I'm going to say as X approaches positive Infinity maybe this first limit right here as X approaches positive Infinity J of X is approaching zero and then over here I'm seeing it's going to negative Infinity as I go to the left so I'm going to say the limit as X approaches Nega Infinity of J of X that's going to equal negative Infinity so looking for where is it that's going to be answer Choice B yeah okay here we go now function K is given by K of theta is 2 sin Theta what are all values of theta between 0 and 2 pi where K of theta equal 1 okay we'll solve a trick equation that somewhat recently 2 sin of theta = -1 honestly this is best case scenario for us this is one I think we probably did pretty good on so --1 / 2 okay looking for where where on the unit circle is s equaling half that's really high up but it'll be fine umga half not very far down at least in the unit circle World an exaggerated version that's kind of like the first stop after zeros that's going to be here um down below the xais sign is the vertical displacement away from the x-axis on the unit circle soga half that's going to going be Theta = 7 piun / 6 just bigger than < by < / 6 and just less than 2 piun by pi over 6 11 pi over 6 so go 7 pi over 6 11 pi over 6 there we go keep going which of these is equivalent when X is a positive number okay so we're going to undo all the log properties okay so division you know I see that division right here that's something I'm going to be able to kind of manipulate on it's going to be a difference of two logs so I'm going to have natural log of 2 e 4 minus natural log of x to 3 okay then I might uh you know point out that this these two things are connected with multiplication so there's going to be addition there and I'm going to say that that's natural log of 2 plus natural log of e 4 okay now you need to know that this thing here that's just going to equal four because the natural log of a number is the special exponent we put on E to get that number so if I put that you know ask you what exponent do I put on E to get e to the four you would tell me four and that's why that log is four minus natural log of x to the 3 okay now x to the 3 right there that's a power on the inside of a log and the argument we can swing that around the front so like kind of a multiplier okay so that's where I'm going to have log of 2 plus a 4 - 3 natural log X so 4 + log 2 - 3 log X that's a right there all right now function J is given by J of x = 8 - 7 tangent of X over 4 these gives the vertical ASM tootes of J he well I just remind you that a vertical ASM toote that comes from division by zero number divided by 0o and remember tangent is s over cosine so what we really need to be doing is thinking on the unit circle like where does tangent have its vertical ASM tootes just like maybe tangent Theta yal tangent Theta okay and since tangent Theta equal sin thet / cosine Theta it would be wherever cosine is equal to zero so I'm thinking where is cosine of theta equal to Z okay that's where there's no displacement from the Y AIS horizontally that would be at Pi / 2 and also at 3 pi over 2 but that's not what we've got we're not looking at yal tangent Theta we're looking at J of x = 8 - 7 tangent of X over4 but really what matters is that this thing here it's going into tangent and think about when that equals pi over 2 or 3 Pi over2 that's going to cause vertical ASM toote so I'm going to say pi/ 2 = x / 4 and then I'd multiply both sides by 4 and I'd get x = 4 piun / 2 which is going to be 2 pi and then okay I kind of know that it's going to be probably this one uh but I'll figure out what's going on with the other situation here um 3 pi over 2 is going to equal x over 4 I know that there's going to be other you know infinitude of solutions but I figur just figure it out based on having two of them and looking at the answer choices since this is multiple choice so I'm going to multiply this both sides by four I have 12 pi/ 2 is x = 6 Pi yeah that I guess is plus 4 Pi K uh the next one would be I guess I'd have to add a pi so it' be a 10 pi okay I see what they're doing 2 pi 6 Pi 10 pi they're all separated by four Pi okay that's great okay now they're going to give us the end behavior of a polinomial you're going to ask which of these could be an expression for K all right okay I I I don't know why I wasn't seeing the nature of the question all right so the limit as X approaches negative Infinity of K of X is positive Infinity all right so that means the graph goes up like that and then the limit as X approaches positive Infinity if K of x equals negative Infinity so we don't really know what kind of polinomial we've got here but um we know know that eventually it's going down to negative infinity and it was coming from positive Infinity so it's going to look about like that okay based on that I know that it has to be an odd degree with a negative leading coefficient okay that's that's something that we learned in the polinomial unit um odd because the in behavior is the opposite and negative leading coefficient because as X approaches positive Infinity K of X approaches negative Infinity so the highest power term has to be multiplied by a negative leading coefficient okay we're putting in large positive numbers and getting huge negative outputs so that's going to be answer Choice okay not B that's an even degree not D that one's also even degree it's going to be C because it's even it's odd degree negative leading coefficient there we go rational function here is f ofx describing the output values of f as input values of f get arbitrarily close to one that means they're talking about a limit where X is greater than one okay so we're thinking about what happens as we approach there okay that idea of one-sided limit we're going to be saying you know really what we're doing here and this is not really what the answer Choice you're getting at but we're doing limit is of f ofx as X approaches one from the positive side so what I'm going to think about is you know what's happening when f ofx equals you know 1.1 let's say F of 1.1 which is a little bit bigger than one and notice that there is going to be definitely like a vertical Asm toote on the graph of F at xal 1 and that's what I usually do when I work this type of I forgot to label it there F of 1.1 is going to equal -2 * 1.1 + 3 which would have been 43 divided by 0.1 I'm dividing by 01 one that's like multiplying by 10 so 8.6 that's going to be 86 now if you wanted to you could probably you know add you could do 1.01 and see that it's even more negative um but we're getting a large negative number here um you know maybe this is 1.1 right there and that's very negative because this thing is drawn very out of scale um but that means you know as we approach this vertical ASM toote we're not going to have a sign change between x = 1.1 x = 1 we're approaching negative Infinity output values decrease without bounding okay so I say the limit as X approaches one for the positive side of f ofx is negative infinity or maybe that's what we say extra in AP calculus but all right the graph of the rational function K has a hole at xal -3 it's known that limit as X approaches -3 if K of X is equal to 4 okay so there's a hole at -3 and the limit of of K is 4 all right I think the fact that there's a whole and I mean really maybe I'll draw you the picture after okay x = -3 I need to be seeing x + 3 over X+ 3 so no no and I'm down to two Okay so when I do that actually you know reduce I'm G to have 4 * x - 4 over x -1 then if I do the same thing for D I'm going to have x - 5 / x + 1 all right when they say that the limit of k ofx equal 4 yeah what we're going to do is we're going to kind of plug into what's left over and see all right four * and we're plugging in x = -3 to see what would have happened had the you know we not had that those redundant factors3 - 4 is -7 / -3 -1 okay that's a -4 so be of -7 = 7 that's not right so it must be D okay but we know we'll know that's true for sure if we do3 3 - 5 is -8 / -3 + 1 is -282 equals pos4 yeah that's I like that okay and that's going to be answer Choice D now this graph shows a polinomial function P one of the zeros is 2 - 3 I what's the least possible degree of P well okay least possible I know that that right here looks like a root okay it's [Music] bouncing so it's an even multiplicity roote um I guess the minimum it could be would be two and then I'm going to have a single root right here so I'm already sitting at degree three because I've counted three Roots but then remember the complex Roots come in conjugate pairs so that's going to be at least two more so the least possible degree would be five um so just write that in and just okay so I'm counting two + 1 is three two more would be the very least because we know that um and the by the pairs they're conjugate Pairs and by conjugate if it's 2 minus 3 I that its conjugate is going to be 2+ 3 I all right but you don't need to know too much about complex numbers for this course just they come in complex pair or conjugate pairs to when they're roots of polinomial equations and also what's the other thing oh that we can express their coordinates polarly if we want but that's hardly ever actually about imaginary numbers it's more about TR we got F of theta = -3 sin of 3 Theta okay uh with these AP pre-cal polar graph questions my strong strong recommendation is just plug in the major points check and eliminate okay so Theta and F of theta or R okay I'm going to plug in theta equals 0 and see what happens okay we need to know that s of 3 * 0 is s of 0 we need to know that's zero and I would really prefer for you to know that like really mentally off the top of your head um but if you don't we're drawing a circle okay theta equals z is right there so s of Z is the vertical displacement from the xaxis which is going to be zero okay so Theta equal 0 r 0 okay when theta equals let's say Pi / 2 okay -3 * the S of 3 * < over 2 okay 3 piun / 2 is right there s is -1 so -3 * -1 is 3 and I think that should probably be enough yeah we need to be seeing Theta = PK /2 R = 3 that's only this point right here and then we're not seeing Thal Z or we're not seeing like any point on theta equals z elsewhere but really this only only one has pi over 2 and 3 so if we just went through and my next one was going to be Theta equal Pi I would have gotten r equal Z so that could have meant that like this would have been like bad here and that one would have been bad there because I knew that at both zero and Pi for input values F of theta was going to equal zero um but really pi over two was the one that showed me then let's see f is an odd function remember when they say an odd function they don't necessarily mean an odd deegree polom they mean that f ofx it has that symmetry across the origin it's the negative of f ofx okay and then the table gives values of F at selected values of X where A and B are constants it's the value of a plus b okay well if we know that F of -2 = pos1 okay you could either use the equation if you're more algebraically minded or you could draw a picture if you're more graphically minded -2 10 up here okay so that means I'm going to have to have a point that corresponds over there it's going be where X is positive2 a is going to equal -10 okay get rid of those okay what kind of corresponds with b and -7 well probably something with an output value of postive 7 okay so -4 and 7 so -4 POS 7 okay so that would mean pos4 -7 okay so B = pos4 -10 + 4 is -6 all right now we've got a figure showing a graph of a trigonometric function G which of the following could be an expression for G okay 4 S of 2x okay I think with these has been that long we should start with the period and amplitude I think or no but no amplitude and midline is going to be the way for us to go okay so the minimum value is yal -2 the maximum value is six that means the midline would be halfway in between we would average them so it' be y equal 2 okay and I am seeing that that does look believable okay but these all have a midline of y equals 2 so that was not helpful at all okay now I'm going to think about from two how far do I go up to the maximum and how far down do I go to the minimum that would be four okay yeah that's four so four these all have a multiplier of four okay I did not read the question all that carefully I still think that was probably the thing to do all right so if we are going to start off at with no phase shift okay because these all not what I'm looking at it they all have the same like the multiplier for the period and everything it's just a question of face shift sh so if we phasee shift nothing and we start at the you know xal 0 uh we're starting in the middle going down that needs to be negative sign okay so this is positive sign that's out that's at least how I think about it okay if we started at theta equals Pi over4 which we would if we were shifting right by pi over4 was a phase shift that's at the very bottom of the sine wave that's negative cosine okay but I'm looking at a positive 4 cosine so that one's out too 4 sin x - < / 2 okay here I'm in the middle going up okay that would be positive sign okay so that was wrong also so it must be D what's going on here 4 cosine of 2 * x - 3 piun over 4 okay the phase shift we to kind of like start our perspective here that is at the top that is positive cosine that's going to be answer Choice D all right now we've got an exponential function which the following is an equivalent form for J of x okay well all right what we're going to need to do is use you know kind of an exponent property on that four to the X plus 2 because there's nothing you can really do to combine the three and the four so I'm going to say J of x = 3 * 4 x + 2 well 4 x + 2 can be Rewritten as 4 x * 4 2 and that's an exponent property you need to be very comfortable with an AP preal um the fact that when we multiply two numbers with the same base in different exponents we're going to add the exponent okay when I always when to ask students like how many X's is this they're very confidently able to tell me that this is X 5al to five this is five powers of X um but sometimes you know splitting something like 4 x + 2 up into 4 to the x 4 2 is not as natural okay but I'm telling you they are exactly the same thing all right then I could say all right 3 * 4 the 2 um well I'm just going to do one step at a time here 4 to the x 4^ 2 is 16 3 * 16 is 48 I still have a 4 to the X here and then that's going to be answer Choice B now we've got H ofx equaling f ofx divid g ofx we're thinking about rational functions which this is a slanted ASM toote all right now if they say slant ASM toote um we know that we probably have a situation where the degree of the numerator exceeds the degree of the denominator by exactly one F has to be one power higher than G yes -2X ^2 versus x what we can often times do is take the leading Co or the leading terms and say all right -2 x^2 ided x = -2X and slanted ASM toote is going to be parallel to that okay well that doesn't really help us all that much that only helps us eliminate answer Choice D if we want the actual equation we have to do long division okay now this one is set up for synthetic division because we're just dividing by xus 3 but evidently synthetic division is not supposed to be part of the AP preall course so I'll just do it with long division but if you want to do this with syn synthetic be my guess you know like that's just as good it's all the same numbers and you get the same results so I would not care okay that was x - 3 I think I should probably make sure of that nope it was x + 3 right there we go okay so what do I multiply X by to get -2X ^2 be a -2 and an X I'm going to distribute the -2X across the X and the + three 6X I'm going to subtract that and I'm going to be left with all right 0 3 - -6 is POS 9 and all right what do I multiply X by to get 9x that would be a positive 9 -2x + 9 that'll be enough you know there's going to be a remainder and whatever but that's you know what makes it an interesting rational function uh we're just interested in the slanted ASM toote and that'll just be we can read off right there next one is got yeah I remember this one so angle Theta is all the way out here right so Point p i mean what you're supposed to see is that the coordinates of Point Q are -4 and 3 I guess you can know that by the fact that they're sketched in the right angles there and stuff but um that's what we need to know is that coordinates of Point q are4 and 3 so what is cosine Theta okay well if and then the radius of the circle is five if I just drew a representative triangle okay a reference triangle okay okay over here it looks like this leg would be -4 that leg would be three hypotenuse would be five and that angle will be equivalent to Theta at least it's cosine will be all right so cosine Theta is adjacent over hypotenuse -4 over 5 of these is an equivalent form worked one very similar to this but that's all right okay when we've got power of a power that's where we're going to multiply the exponent so I could say that's four to the -2 let write it like that to the X power now 4 the -2 remember that that's 1 over 4 the 2 power the negative exponent doesn't make it negative or anything it just turns it into a fraction okay so 1 over 4 the 2 power to the X power so that's 116th to the X number 23 g of X is log base 3 of X what is the value of G of 1 9th okay well that's just going to be plug in one nth for X this is just like do you know how to take a log yeah it's actually good question um the type of thing I really expected to be in AP pre-cal okay so when you're taking a log with base three and argument 1 nth you're asking yourself what do I put in this box what power do I put on three to get 1 nth well to turn it into a fraction like something is less than one it's going to have to be a negative value so I know it needs to be negative in there and 3^2 = 9 so I think it should be -2 the 1/2 power remember the 1/2 power is the square root okay and the square root of three is not nine three is the square root of n that's something different so that thing that number that needs to go in the box that's -2 so that's it's going to be G of 1 9th all right now what are all values of X for which log base 7 + 2al okay this another one we're going to have to use log property this is the second time this has happened in this problem set so it shouldn't be a big surprise all right so we're going to combine the logs uh by getting them first to the same side of the equation so log base 7 of 4x + 6 and then I'm subtracting that other log which was log base 7 of x -1 okay here where I'm div dividing pardon me where I'm subtracting two logs that is going to turn into division in the argument to write it as a single log so log base 7 of 4x + 6 / x -1 then I'm going to take seven to both sides Powers so 7 the 2 is 49 7 to the log base 7 last is going to be 4x + 6 / by x -1 now I do feel like this is the type of situation because uh we might need to make sure that we are not looking at some sort of like extraneous root or something I just feel like this is the type of situation that could bring one of those up multiply both sides by x - one so I'm going to multiply 49 * the X and the 49 * the -1 it's equal 4x + 6 then I'm going to I guess subtract 4X from both sides so 45x equals and add 49 to both sides so that' be 55 all right now I might divide both sides first you know I mean if you're feeling like you are going to need help simplifying this fraction this answer because I don't see a 55 or a 45 anyway uh I see both of these are divisible by five so 9 * 5 is 45 let divide both sides by 5 so 9x = 11 x = 11 / 9 right now I like that but let me just make sure that that's not going to cause me to take the log of a negative number or the log of Zero no that's not going to happen no not there either okay I feel good happy moving on all right now we're going to think about these two exponential functions and F is the ratio of them okay what is the x coordinate of the point where f ofx = 1 interesting well if the ratio of them equals one then they're just the same thing why would they they often times ask you where the two graphs intersect and this is equivalent to that but it's not exactly the same all right whatever so f ofx = 1 1 is going to equal G ofx 4 to the x + 1 / h of X which is 2 3x X I'll multiply both sides by 2 3x okay now if I want to take a log of both sides um really I think it would be most effective if they had the same base or if the these two expressions had the same base we could just kind of clearly equate the ex the exponents if they were if that was just like exponential equals exponential so I notice that four is 2^ squar so I'm going to just kind of like work on that relationship that's 2 the 2 power and I'm taking that to the x + 1 that's going to equal 2 to 3x all right then I'll have 2 3x 2 to the 2x + 2 right I'm going to have to double the exponent multiply the 2 * x + 1 I'll distribute the two over the X and the one and then I'll say all right well you know kind of this point I can see 3x is going to need to equal 2x + 2 then I'll subtract 2x from both sides and I'll realize ah X needs to equal two and then what you could do is you could go and just kind of check this really don't think it would be all that hard x = 2 4 3 / 2 3 * six okay yeah both of these are 64 um or you could think about 2 the 6 is 2 2^ 2 cubed it all works right we're 26 this one's only 28 questions so the table shows values for f it's selected values of X which of the following claims and explanations is going to fit the data all right okay what are we noticing on the table okay it kind of looks backwards right f ofx is going 2 4 6 8 10 and X is really going downhill this looks kind of backwards from what we're expecting usually X is evenly spaced and F ofx has action okay so really I mean if you see that like really inverse relationship and they're like oh exponential versus log oh it must be log right because log is kind of like by Nature an inverse type function okay so I'm thinking it's log because X values uh form like a proportional geometric sequence um an exponential pattern uh while the output values form a linear pattern um increase and equally intervals um yeah so output values change you input values change proportionally output values increase in equal length intervals that's the answer to C always got to read through that twice make sure you're choosing the right one all right which of these is the range of H okay 2 cosecant of 4X what we're need to do to think about that is kind of like the underlying partner function to h of X okay 2 cant uh is going to be very related to 2 sin okay so I'm going to draw the sketch of Y = 2 * the S of theta I don't even need to worry about 4X is kind of my point right so 2 * the S of Thea is going to look like this there we go okay now if we wanted to think about h of X okay we' just kind of be thinking about Y = 2 * 1 / s well when s is equal to 1 and the actual s of theta okay cosecant is also going to equal one so that's going to be the point that they share and then when s should equal zero um that's where we're going to have a vertical ASM toote for cosecant I'm saying is this is kind of what the graph this H ofx is going to look like in Black so let copy like that that was terrible she obsessed over it so much there we go and then this one is going to be like that I do that so much quicker and it looked better better um you know we can even get rid of the graph of y = 2 sinx but if we think about what was actually going on there uh the midline was zero and the amplitude was two so it should be you know uh B yeah just like that okay and you probably could have figured that out sooner I was just wanted to give you know full explanation to all these you get 80 minutes I think for this section so I think I'm still way under time um so the function f is given by three + 2 e x when is FX equaling 11 right so 3 + 2 e thex = 11 I'm going to isolate the exponential by subtracting three dividing by two going to take the natural log wait second that's get rid of that um e to thex equal 4 I'm going to take the natural log of both sides log of e thex is just going to Bex equals natural log 4 and then we're going to multiply both sides by - 1 and we're going to get X = the negative of natural log 4 now I'm kind of interested that they left that that way as answer Choice C I would point out that that negative multiplier could be applied up there as an exponent and we could call that the natural log of 4 to the -1 power which would be the natur log one and fourth um I would not be the least bit surprised if you know on your AP exam if you got a problem like this you might see um you might see the correct answer choice you know with a with a log properly applied to it you will not see that anymore in AP Calculus though okay because that's now very firmly a pre-calculus topic all right now we're going to move on to section one part B which is multiple choice calculator there's only 12 questions and we get 40 minutes and it will not take us 40 minutes now uh think about expectations for this one we should not expect to use the calculator on all 12 questions that would not uh that's just not how this exam is going to work um you know it's going to maybe be a couple before we even need it and when we need it I'll bring it in otherwise I'm not going to be you know having a calculator sitting off to the side not using it all right so the figure shows the graph of H in the XY plane with four labeled points it's known the relative maximum of H occurs at xal C I can see that um which of the following pairs of points is the average rate of change of H the least okay so remember average rate of change is the slope of the cant line which is the line that connects the two points between a and C all right so I'm going to try to get that there okay that's approximately it um but between points B and D that looks a little bit more negative to me that was awful um let's try that again okay okay that's closer yeah that is more negative because this one was a to c was actually probably more negative than it should have been okay so that looks most negative okay that's the least and then B and C I didn't do B and C B and C is positive so that's not going to be the least uh C and D is very negative though that's going to definitely be the least slope because it's the most negative all right CN D yeah all right let GB the pie wise defined or function defined by g ofx equals this stuff on that domain what's the ative change of G over the closed interval -2 to 4 so I just need to think about for average rate of change G of 4 minus G of -2 all ided by 4 - -2 so that's going to equal G of 4 is okay four is between two and 9 so that's 3 + 2 fours um just going to leave it like that and deal with the arithmetic after um I've got a calculator in case I need it I don't think I'll be needing that um 3 + 8 minus G of -2 G of -2 okay well -2 is between 5 and positive2 so I'm going to do 1 minus the square of -2 1 - 4 is -3 now if I'd written it as 1 - 4 I'd have needed to put in parenthesis and that's seems a little risky um for y'all so I'm just going to do the computation mentally 11 + 3 is 14 over 6 okay those are both even numbers I can at least divide them in half 7 over three that's answer Choice d right there all right now 78 the graphs of the linear function L and the exponential function e both pass through 02 and 16 f is given by L of xus e of X what is the maximum value of f okay well um near function L and exponential function e okay yeah I'm going to have to use a calculator for this one all right if they're telling me that I need I've got two different functions running through both of these points I should probably just edit the lists and uh run two regressions I think it's going to be the way for me to go and clear that one out okay the points were 02 and 1 six I know those are covered up but I think it's all right I can just tell you you saw it earlier okay what is the maximum value of f okay but first I need to do linear function l so I'm going to go over here do a stat calculate a linear regression um I'm going to store L into y1 calculate um even though we could have easily computed that right we could have gotten 4X plus 2 off those two points Maybe actually faster than we did that um now that we need to do an exponential though doing two of these by hand would probably take considerably longer than I've spent on this I'm going to store the exponential in y 2 and that might be a good thing to like make note of on your paper like hey I did L and y1 and E and Y2 now we should never do that in free response right because the we don't use calculator notation in our answer for free response more over you know reader doesn't know what we mean really when we use calculator speak like that okay so we've got them loaded up in here we did those and really what we're interested in is f ofx = l minus E so Y3 is going to equal y1 minus y 2 it's going to be a course of action here now I want to find the maximum value of these things I'm going to go up and I'm going to turn off the graph of these because I don't really want to have the calculator Wast oh wait I didn't turn it off okay Enter okay I want to have it waste time on graphing y1 and Y2 before graphing the thing I'm actually interested in this is not going to be the window I want so I'm going to act like I'm turning the calculator off and zoom back to standard okay here we go I'm just going to hope for the best on the standard window I think there we go yeah okay yeah we can find the maximum so we're going to use that maximum command it's still graphing and let it think for a second all right calculate a maximum okay that is to the left of the maximum looks like xal 1 is to the right of the maximum okay it's going to give me the coordinates there now 545 and 540 okay I think it would have been nice for them to have 545 as an answer Choice um because remember what they're looking for is the value of f like the output value of f the y coordinate there so that's going to be 0.540 all right now looking at this one I'll take that away for a second whoops there we go certain hobby shop wants to attract customers on slow sales days oh we've seen this question already this year if we make purchases on successive Wednesdays discounts doubled from the previous Wednesday all right so 1.5 % on week 1 3% on week 2 6 point 6 or 6% on week three based on this pattern what is the first successive week that this customers's discount would exceed the purchase price of an item okay I'm trying to think how I want to do this um it's because 6 seven and 8 and 11 are like all very close to one another I should probably like write an equation and be like really certain about it so we' see a a doubling pattern that's very exponential so I'm just going to say the discount rate um R oft is feels like something times 1.5 to the T okay a * 1.5 to the T now really what I want is uh for it to be 1.5 when T equals 1 so I think what I want to do is make that a T minus one and um then think about it that way oh wait whoa whoa whoa whoa whoa whoa this is not right because I'm not multiplying by 1.5 each time I'm multiplying by two each time I just didn't multiply by two the first time yeah okay now I can put the 1.5 out here and then I'll have something that'll be just right yeah now if you wanted it to be 2 to the T you could figure out okay when T equals 1 2 * what is 1.5 and you get you know I think 75% okay but let's just um wrote it like this I'm going to go with it okay so R of 1 is equal to 1 and A2 R of 2 equal 1.5 * 2 the 1st is 3 R of 3 is 1.5 * 2^ 2 which is going to be six this makes a lot of sense I'm going to go back over to the calculator and use it um I don't know if you really need it for this but um I guess it depends on your arithmetic skills I'm going to go over here kind of clear out um oh whoops uh is that going to no that definitely didn't do it I don't know what the button is on the keyboard word for clear out um okay so in y1 I'm going to do 1.5 * 2 to the power of T -1 I'm just going to use the table happens all right there we go and I'm going to use the table um I'm actually going to have it ask me for the independent variable um I really prefer to use the table that way but if you wanted to start at zero and just scroll through it I do not want to graph I want to table I'm just going to type in six okay sure I didn't like that x = 6 x = 7 okay that's a 96% discount x = 8 192% that's the first time that the discount exceeded the price purchase price cuz that would be 100% he when tal 11 I'm not sure why we would want that but okay there we go tal 8 all right now number 80 function G is constructed by applying three transforms to F in this exact order horizontal dilation by a factor of three okay so we replace x with 1/3 of X okay so I start with f of x and then that horizontal dilation by a factor of three is going to be F of 13x a vertical dilation by a factor of five okay I'm going to multiply this whole thing by five okay and then a vertical translation by -7 units okay I'm going to just subtract seven after the fact so it's going to look like that and so 5 F of 1/3x - 7 okay 13x - 7 there it is we got that we can do those right number 81 let H be a function where when the output values increase without bound where they are increasing without bound as the input values increase without bound okay so that means the limit as X approaches positive Infinity if h of X is positive Infinity I know that limits are something that at least the students in my class this year are feeling not all that confident on so I'm just going to write it out every time okay so output values increase without bound H is approaching positive Infinity as the input values increase without bound as X approaches infinity and output values of H decrease without bound found as the input values decrease arbitrarily close to zero okay so that means limit as X is approaching zero and it's decreasing towards zero so it's approaching from the positive side of H of X that's going to go to negative Infinity so I need to see a vertical ASM toote at x equals 0 and okay that's going to be and then approaching positive Infinity that's going to be answer Choice B there okay all right sin Theta and S of theta + 2 theta plus pi / 2 0 to 2 pi how many solutions exist okay well I can I don't know if you know the hint but you can kind of see it's in a different font it probably is a calculator question um you're allowed to use a calculator on this one might as well try um now the thing is like it kind of looks like a polar situation but you actually don't want to graph this on Polar you just want to graph this on rectangular and count the intersections uh to find the solutions so I'm going to go over here clear this out I don't know what the button would be but clear it out that way all right so F of theta is sin Theta and as soon as we hit one of these trig buttons we need to make sure we're that we're in radiant mode okay we are in radian mode so we're good to go you need to always be in radiant mode in this class so I'm going to graph a sign of of X and then also going to graph s of x + < / 2 I believe in pimos so I know it's going to know what I mean when I do PI ID two that way if you were unsure use the fraction bar right no shame in that 0 to 2 pi I'm going to set the window there 0 to 6.28 or 6.3 or something like that you can type in 2 pi if you like but one intersection two intersections just two all right so on the interval 0 to 2 pi I counted two times where the graphs ran into each other so that's how many solutions there are to f equals g oh goodness right now um average daily temperature in degrees Fahrenheit in a certain city can be modeled by function f given by that business Okay C is a constant January 1 corresponds to tal 1 all right tal 10 the average temperature was 41.7 based on the model what was the predicted average temperature on February 28th okay so really what we need to do is find the value of C so if they tell me that F of 10 is 41.7 I know that a 41.7 is going to equal 63 .6 + 21.9 sine of 2 pi over 365 time T minus C now I do know the value of T so I'm going to substitute on that um that's going to be T was equal to 10 right and so I'm just going to try to solve this equation for C using the calculator okay so on uh you know 41.7 is bigger than 10 which is kind of my standard uh y window I'm going to actually subtract 41.7 from both sides but I'm not going to try to do that without a calculator right I'm going to go over here bring back the calculator oh whoops wrong thing there um okay get rid of that um don't need these anymore I'm going to find the zero of a related function so I'm going to do 63.6 + 21.9 * the sign in radian mode of okay call for the fraction bar because there's just like kind of a lot of drama in this equation two pi over 365 times no T was 10 minus C and that's the thing we're solving for for that's like the variable okay we're interested in where that's equal to 41.7 so the difference between that and 41.7 should equal zero and that's a calculator trick we're going to employ quite a bit in AP Calculus now um I'm going to zoom to the standard window and just hope for the best uh just figure out what C is great okay what is going on here how am I supposed to figure this out um think about this for a second all right now I'm about to do something I I am certain I have never done this on my YouTube channel I'm going to use the equation solver I hardly ever do this and I have not had a lot of success with this in the past um but I feel like this is probably going to be the way to go um I want to solve y1 equals z so in equation one I put y1 in equation two I put zero I press okay iner um x = 200 was kind of like my right bound I have not pressed solve yet okay when it says solve really what it's talking about is this F5 button so I press F5 and it's thinking select variable press solve okay and then I get this 101.2 499 okay I think that that's going to be that's what I need okay and so that's going to be C equaling 101 2499 you know we've been through all this is a calculator multiple choice question let's just use 10 1.25 because these answer outputs are o they are kind of I've got two that are really close together I should probably go back and look at that again I pardon me this is be using more than my allotted time for this problem that's for sure okay I'm going do yes yes let's please solve and then okay 101.2 I don't know maybe I'm just being paranoid but I'm going to copy all the decimals 2 4 99459 that has got to be enough all right so I'm going to go back in and I'm going to kind of just edit this equation right here so it's 10 no so it's really it's T minus C so it's going to be x minus I'm going to be inserting over this in front of this 101 yes I'm going to need to delete that 2499 I think that was 4 59 I think I copied down one too many um too many nines there and then I will need to get rid of that minus 41 because that was just when I was trying to solve an equation there all right so now I'm going to want to plug in t equals it looks like 59 and hopefully this did not test this out resumed the recording 49.0 396 okay good 49.4 that problem was a bit of a journey all right now a new special attraction open at a museum and I remember this one too all right so this is going to be a rate graph um and this is patently not pre-calculus but that's just I guess not my place to say um I think if you've got a graph of a rate of change that that makes it the graph of a derivative which is calculus but okay I've said my piece so right they m models the number of people that visited but R is the rate of change in people per day um for each day after the attraction opened R is given by this and which of the values of D does m have a point of inflection okay M has a point of inflection when it goes from concave up to concave down or concave down to concave up when the rate of change change of M goes changes signs but maybe yeah goes from incre no pardon me it doesn't change signs the rate of change of the rate of change changes signs okay so it goes from increasing to decreasing or decreasing to increasing um okay now the rate of change of M is R so that's we're going to graph R of T no R of d okay goes increasing to decreasing or maybe decreasing to increasing okay so let's get the calculator let's just get to it uh it's going to take me probably longer to type this in 1 over 200 okay 1 over 20000 times the negative of D to 4+ 35 D to the 3 - 411 D to the 2 plus 1845 D to the 1 just like what is the benefit of the numbers being this bad 26 86.5 we're going to graph it and we're going to look at the answer choices for sure we want to make sure that these values of D which are represented by X in our graph are something we're seeing so we're going to set the window to be 0 to 15 and hopefully -10 to 10 is enough to see the action here okay up down okay yes up okay these are changes in decrease to increase okay so there's three of them um I only see one place where there's three answer choices maybe I'll make sure that these local extrema like these points here and there and over here are actually those three x values that I'm seeing in the answer choices so I'm just going to calculate a maximum uh it looks like it's between let's say two and six all right so left bound of two a right bound of six I guess all right 3.8 94 yeah that's the first one answer Choice D I'm happy with that let's move on that one's that one's pretty tough for you as a pre-calculus student that really doesn't know all that much about rates of change besides that they're the kind of some slopes and I don't know fearing a new computer virus a security company performs a simulation to predict the number of computers might be affected by the virus okay number of infected computers can be modeled by a geometric sequence okay geometric sequence that's a fancy name for an exponential function that only takes in whole numbers okay first day of the simulation is day one on day 6 virus had affected 750 and on day 10 virus infected 3200 how many virus how many on day 14 all right so okay on day six we' done 750 Let's do an exponential regression we got a calculator anyway right okay so I'm just going to say on Dat day six edit the lists they told us about day six and then day 10200 on day 10 and it said 750 on day six going to run an exponential regression on this one I'm going to store it in y1 he's got some one calculate okay now on day 14 I'm going to call for y1 of 14 whoops do that and 13,656 actually there we have it all right I think exponential regression was definitely the way to go there all right table gives values of polinomial function G it's selected of X which of these statements of G could be true okay or G is an odd function we already talked about that earlier um I probably need to tell you about even remind you about that so even function is where F ofx = NE F ofx Well that has to do with even degree polinomial is that a polinomial will be even if it only has even power terms okay so f ofx = f of negx that's obviously not happening because F of 2 and -2 are negatives of each other all right and G of 4 equals the negative of G of4 this thing is looking like it's odd okay so even is out odd is in and G of1 would be the negative of -6 it needs to be positive so there we go kind of weird that we had odd function twice and even function no times but that's why this is just a practice exam kind of cobbled together at the end of course all right here we've got table giving the population in millions of a certain region for selected years exponential regression is calculated for the five data points great in what year will the population first read 35 million in okay so we're going to go into the stat menu bring back the calculator stat menu edit the lists okay xal 4 represents 2004 okay so 4 8 12 plus 16 and 20 then okay 22.4 24.3 26.0 [Music] 27.9 and 29.2 we're do that exponential regression store that in y1 yeah cuz we're going to need to to go ahead and graph this and find out where it reached a population of 35 million okay so we're going to be interested in where this thing hits 35 so what I'm going to do is I'm going to go up and I'm going to turn on plot one so that I can zoom to stat I don't know that that actually turned it on okay now now I think it's on yeah now I'll go over here to zoom zoom to stat is option nine and there we go I've got this exponent growth pattern um if 35 is not in the window then I can just expand it a little bit looks like it's not okay that's kind of a shame but that's a good place to start right yeah the Y Max was 30 we should make that 40 so we can comfortably see the intersection oh no we're going to need to look out in the longer term I can already see that so let's just kind of double the X window make it 2 to 40 that should do it for us I think yeah that'll give us a full picture see the intersection it'll be great graphing here comes the horizontal line they intersected let's calculate the intersection between blue and red give it a guess it's going to be 30149 okay so what year it's going to be 2030 okay there we go all right okay and that's the end of the multiple choice section okay so we're going to move on to free response now um and the first part of the free response allows the use of a calculator wait what second plus 712 I'm just going to clear out the RAM and to start fresh and all right here we go so let's work on the free response questions for this practice exam okay we're starting off with the first two which allow the calculator we already know what's going to be on them because the free response is fully prescribed in AP pre-cal right so the first one we're need kind of some room to write here okay we don't need to see that I'm realizing I need to get this thing ready to write right so number one is going to going to give us a graph or a table it's going to do some composition of functions and you know a little bit of solving we're going to be able to use calculator um kind of all over the place and this is another one I kind of wish we' brought out a little sooner for you to to Grapple with but y'all are doing really well or at least kids in my class this year yall are doing really well on it um hopefully if you're watching this in the future you're also doing well with free response question one because it's it's not too hard um h of xal G of f ofx which is can also be written with the more common notation G of f ofx okay um find the value of H of three so this is going to be A1 so that would be h of3 = g of f of three we're looking at a picture of the graph of f f of three is looking like two so this is is going to equal G of 2 and okay find the value of H of3 as a decimal approximation G of x = 2 + 3 log X so that's going to be 2 + 3 natural log of two pardon me I said log I meant natural log in general when I say log I actually do mean natural log but I'm going to you know get that calculator out and do two okay now 2 + 3 natural log of two and I'm going to get 4.07 n so I'll kind of report that there don't need the calculator any longer that 4079 remember always three decimals always um so find all real zeros of f or indicate that none exist okay we're looking at a picture of the graph of f and well they told us that the points - 1 0 and 32 are on the graph of F I feel like that was pretty helpful um part A2 I'm just going to say x = -1 is the only one I'm seeing um decreasing function f on its domain of all real numbers um except where X is ne1 okay here we go so next for Part B we're going to be finding all values of X as decimal approximations so we need use calculator where G of x equals e all right so 2 + 3 natural log x = e I think it would be tempting to try to solve this one by hand we don't need to right we've got a calculator so we're going to go over here we're going to type in y1 2 + 3 natural log X and then Y 2 e we're going to see where they intersect okay so going over here to y equals 2 + 3 natural log X and then I'm going to type in E down there okay where even is that that's over here by that but if you couldn't find that you know it's we know e to the x is above natural log you could do either the one that would also work right and okay we're kind of on the standard window yeah we should be able to find that okay we see them intersecting we're going to calculate the intersection between blue okay and if it gave you an error we need to make sure we're actually on the blue curve I don't know if that would have happened there but I fig that might happen sometimes okay I've got what I need I'm seeing that value that's what I'm going to be taking right there oh wow you can't actually see that let's that's not quite right didn't realize that my calculator was I don't know if I can make that okay you see I'm grabbing that number right down there it's hopefully you're working Along on your own calculator that's 1.27 7 1 or 1.27 depending on if you like to round or truncate so that's going to be x equal no I can't see it 1.27 one okay now um I think I'm not sure about this because you know we've not actually seen the scoring notes for the pre- response for a real AP pre-cal exam yet as of when I'm recording this video but I think you might be able to get away in AP pre-cal with just like reporting X = 1. 1271 like a bald answer but I really don't like that cuz I teach AP Calculus and that would not fly on the free response in that class anytime we solve an equation an AP Calculus with a calculator we do have to show the equation that we solved so I think it's going to be you know definitely best for the students in my class for you know almost all of which are going on to calculus next year um you need to show what equation you're solving using the calculator it's just the best practice so okay there we go determine the end behavior of G as X increases without bound okay that means we need to be thinking about the limit as X approaches positive Infinity okay and it's going to say express your answer using the mathematical notation of a limit that means you cannot write a sentence as X approaches Infinity y approaches that that is not the mathematical notation of a limit okay the mathematical notation of a limit we start off with a limb okay we say what x is approaching right below the limit I don't know where we went wrong this year in showing you how to like the notation but so many of my students we're on free response we're putting the X approaches stuff somewhere else besides underneath the limb and that's not where you're supposed to put it and like we're going for standard mathematical notation here so we need to need to like kind of do it the right way so the limit of G of X as X approaches Infinity put positive Infinity because it's increasing without bound okay well 2 + 3 log X we could look at it at the graph I mean could bring it back up for you and here uh look it's the blue curve as X approaches Infinity Y is going to be approaching positive Infinity yes see here we go um log does not have a horizontal ASM toote what would it be approaching you know what is what would you think the horizontal ASM toote would be there's no you know upper limit to the exponent we can put on E so there's no limit to log so this limit is positive Infinity we're just going to write it like that and move on okay now part C determine if f is is invertible okay now I'm going to say then this is a yes no question um yes f is invertible okay there's a variety of reasons that work but you know our reason has to be accurate I'm going to write a variety of reasons for you that that could potentially you know earn the credit um one reason uh I mean based on the definition of a function in the graph of yal F ofx I know I didn't notice that okay okay I don't know that yeah that's not going to I was going to say because f is strictly decreasing on its domain a function that's monotone is going to be invertible um a function that is you know that's an injection which this one is is going to be invertible but that doesn't necessarily have to do with the definition of a function in the graph of yal f ofx okay um now the graph of y f ofx basically yes because each output value each of the y-coordinates is only hit by One X value or at most One X value but each output value corresponds to a unique input variable a unique input value pardon me and that's going to have to do with the definition of a function right um because when we think about the inverse function that's taking in y values and returning us the X's that got us there and if it's going to be a function it can only return Us One X corresponds to a unique input value okay but um I think another valid answer would be all right if um f is oh we don't have an equation for f but if we did somehow have happen to know that F ofx equal is that going to be 2x - 1 + 1 is that part possible okay 2 over 3 - 1 would be 2 yeah that one works when x = -1 that does seem to be an equation that works I guess that's kind of taking a lot um on faith because they just told you it was a decrease function where X is not equal one uh this seems to work it fits the points I don't I wouldn't do this on mine I would definitely write the sentence that had to do with the definition of a function uh but if y equals that we could solve for the inverse right we could say y - one = 2 overx -1 we can multiply both sides by x -1 this would be a little bit unpleasant um and then we could solve for x eventually I'm really running out of room here um r on the sheet I guess I could just keep going and then I could say x = 1 + 2 over y - 1 that's a formula for the inverse I could say F inverse of X is = to 1 + 2x -1 that's a function in its own right so I think you could say hey I solved for it if you had an equation you could say hey I solved for the inverse and it's a function uh but we didn't didn't really have that that formula so I don't think that was valid I'm going to move on okay so that was free response question one okay free response question two still allowing the use of a calculator oh no what's going on here okay no that's not good all right fore response question two this is going to be modeling a non-periodic context so possibly exponential possibly logarithmic possibly quadratic here we've got a piece-wise model okay so initially seven students knew about the rumor after 2 hours 15 had heard the rumor after 6 67 had heard it here we've got a pie wise function R use the given data to write two equations that can be used to find the values for constants A and B in the expression for R of T all right well um oh yeah this one's a little bit different um I think we will no we're not actually running aggression on this one and but usually we do okay but we don't necessarily know that we're always going to for number two all right so for two part A1 they told us that at T = 2 15 had heard the rumor yeah so we're going to need 15 equal 7 * a to the tal 2 at that point 2 over 2 and then the other thing we know is after 6 hours 67 had heard it so 67 equals and 6 is okay that's where T is greater than or equal to 6 so I have to use that NE - 213.29 plus b natural log of in this case six okay so we're going to be able to figure that out um I think we're just going to have to um you know actually do the algebra though okay so 15 = 7 * a a = not again think 15 over 7 so a is 15 / 7 B is going to but they want a decimal approximation so I can give them that here in a second um B natural log of 6 okay maybe I'm just going to kind of do some work here 67 + 213.29 and then to solve for b I'm going to be dividing by the natural log six and I guess with my handwriting that kind of looks the same natural log of six all right so now I'm going to need the calculator for sure I don't I'm not that quick with mental arithmetic here we go all right that's okay what is going on okay quit out 15 / 7 that's the easy one 2.43 and then this one uh let's bring out the fraction bar we want to be careful there 67 + 213.29 divided by the natural log of 6 that's 156.43 3 okay now I am going to go up and I'm going to store these values because I'm going to be using them later in the problem probably and I do not want to be using rounded values wait no I don't Stow to a because that was the value of a and I'm going to go up to this 156 number I'm going to store that into B in case I need to use those later which I'm sure I will all right now after that I've got okay so that's part two right here okay then Part B One is using the G using the given data to find the average rate of change the number of students that have heard the rumor between two and 6 hours okay so that's going to be R of 6 - R of 2 / 6 - 2 which is equal to 67 - 15 over 6 - 2 now it's I think again I think this might be a situation where you are allowed to what am I saying um give a bald answer but I really don't like the sound of that um I think we should be showing the setup that leads to our calculations um 67 - 15 over 6 - 2 and then we would get 13 you can put 13.00 but you don't have to and again I'm going to uh give the units it's not going to ask me to but it's just a good habit the units are going to be number of RS students per number of t days no hours so students per hour and okay so I've got students per hour as my units and oh it's very laggy but it's all right interpret your meaning from part one in the context of the problem okay well we can do that now that we have um you know got the units also um from T = 2 to T = 6 hours the number of students who heard the rumor increased 13 students per hour just kind of leave it at that um consider the values that result from using the average rate of change found in part one to estimate the number of students who have heard the rumor for times P hours where p is between 0 and six okay well we found that a was equal to 2.43 okay and it's an exponential function with a growth factor bigger than one so it's exponential growth I just draw myself a little picture to you know convince myself of that if that's 0 to six okay just draw it like that and say okay we use the average rate of change maybe from 2 to six very frustrating um and that's going to be a straight line the average rate of change if we yeah to estimate the number of students yeah yeah are these estimates less than or greater than the number of students PR by model R okay well I would say these estimates estimates from average rate of change uh they're going to be greater than the number of students predicted by R because the graph of r is concave up right it's that curvature that's keeping it below the straight line because the graph of r is concave up on the interval 0 to 6 okay think that will be enough okay there we go that's B3 and remember my advice on B3 we're just going to do the best we can right okay it's very laggy unsure of what it's doing on the left side but um just do the best you can it's always going to be kind of a toss up and same story for C the model R is valid for 0 to 12 hours explain how the range values of the function R should be limited by the context of the problem okay well um the range values it's the output values of the function it's the number of students who have heard the rumor okay it's valid for 0 12 hours um there's only so many students to hear the rumor at the school right um there's not like a million students at the school that could possibly have heard this rumor that you know know who Mr pwater is and are going to be like you know hearing that he's won the lottery and moved to Japan so range values should be limited to the number of students at the school okay I think my tablet is tired as well yeah range should be limited to the number of students at the school and I don't even know if that would work okay if you want to see more examples of me doing problems like this and being uncertain about the answer for c um I've got a video of it you know there's other I've got a few of these examples of free response question two that you can see and you know see they're all kind of like that a little bit all right now questions three and four do not allowed the use of a calculator so I'm going to I'm going to actually close that out that's making my computer go slower or anything but it's definitely in my way all right so a windshield wiper blade on a car window rotates back and forth okay so we got lag on the right side as well okay when the wiper blade is farthest to the left or farthest to the right the measure of the angle formed by the wiper blade in the vertical Center Line is you can't see it but it's 0.75 radians um and I tried to show you in that okay there you go that was probably a mistake let's see if I can get that all in there and it's probably going to be too blurry for you to raid all right at time T equals 1 second uh the wiper blade is farthest to the right for the first time okay so I'm going to say farthest to the right is let's see is that positive or negative a positive value of H of T indicates the wiper rate is to the right okay so farthest to the right T equals 1 maximum value of H of T so I'm just going to make that note right here at Point F that's going to be tal 1 okay um and that's the first thing that I see okay then they say that the maximum number of radians is 0.75 so I would think that yeah it's going to be 0.75 up here and netive 0.7 7 5 down here meaning zero is the center okay so we want the possible coordinates for T and H of T okay now it should probably have given us okay then the wiper blade rotates left passes the center vertical or the vertical Center Line and is farthest to the left again at tal 2 okay so p is going to be tal 2 so J would be 1 and a half G is going to be one and a quarter and J is going to be one and 3/4 okay so a coordinates of f could be T = 1 h of T = 0.75 G is going to be T equals 1.25 t or h of t equal 0 J comes next fgj so that's going to be 1.5 and 0.75 K it's 175 and zero and then P the last point will be two and 0.75 because is back to the farthest left spot okay so I've got possible coordinates for each of the Five Points okay function H can be written in that nice form a sign of B * t + C + D now if we are going to write it like that with a positive value of a and it's sign okay we need to be coming up through the middle so we need to kind of probably be looking for that spot for our phase shift um Point K so I'm going to say I think that would be like probably the easiest way so for us we're going to go over here and say all right Part B is find the values for a b c and d all right a is the amplitude that's going to be 0.75 I think we can see that from the graph that's how far we go up or down D is the midline that'll be zero okay if we want positive sign that's why I'm going to say okay we're just going to shift right T equals to 1.75 oh whoops still on highlighter over here t equal 1.75 okay I think we should just shift right 1.75 units so that we've got what looks like you know the start of a positive sign graph and so if I'm shifting right 1.75 I need to have T minus 1.75 C is -175 and then B is going to be something I need to think about the period okay so the period equals one as I'm seeing there from top to top so period equals 2 pi over B so that means B must be 2 pi okay it's coming there it is all right now part C is is um something that right I think we're doing all right G is T1 and J is is T2 so we're kind of interested in uh that portion of the graph that I've I've kind of emphasized in red and so since we're looking here where the graph is decreasing and concave up but that's actually no it's just asking about H H is negative right cu is below zero and it is decreasing H is negative and decreasing I'm going to clearly I I don't know if you're going to I really don't know if you're going to be bubbling your answer that's crazy if you're going to be d uh bubbling your answer just writing a letter D that seems kind of risky because it's like you know need a handwriting sample to be sure about what your letter is so some of yall Like A and D might even look the same who even knows have interesting pin manip but I'm just going to say answer Choice D maybe they will give you a okay it's going to be all right we are almost done here so um and then C2 okay describe how the rate of change of H is changing on the interval T1 to T2 okay well at first the rate of change is very negative and then the rate of change is less negative it's closer to zero so the rate of change is increased um you could say I think you can say increasing and that'll be enough like to describe what is going on with the rate of change of H the rate of change is increasing because the graph is concave up less is more I do not want you rambling going off saying things that are untrue and talking yourself out of the point so we're just going to be really direct with our answer not even give a reason because they didn't request one the rate of change is increasing let's think about the slopes slopes work first very negative and they were less negative as I went to the right all right so that is an increasing rate of change because the slopes are increasing that's number three right now we're going to move on to number four remember number four is the one we have to be very careful about simplifying our answer I tried to zoom in on it some um but you've you've seen the Box before you've read it um you know what the deal is he we must simplify our answer so functions G and H are given here like an exponential and something with tri all right so solve and solve that's that's going to be pretty typical for four it's going to be two times where you're going to solve two times where you're going to rewrite and then you're going to do something else totally crazy in part C okay but really if you just kind of like don't freak out at the beginning it's going to be all right you know like just have faith that you're going to get weird numbers for these but it's okay G of X = 27 so 3 2x * 3 x + 4 = 27 G of X = 27 okay we can do this okay when we're multiplying two things with the same base in different exponents we're going to add the exponents it's going to equal 27 and I think this can actually you know you need to take a log really um because we know that 27 is a power of 3 so 3 3x + 4 = 27 which is 3 cubed so I think it'll be all right for us to equate the exponents that didn't go well okay here equal to that and I get 3x + 4 = 3 3x = 1 nope that's going to be - 1 and then x = - 13 okay then solve H ofx = 5 and I like I'm not even sure if they're I really don't think they're grading really the work um but may i' seen one where they did grade the work kind of partially on the way on part C so I don't know this is May Prett thing it's a little weird to me um for part two we want to know where H ofx equals 5 so 2 tangent SAR X subtract one is equal to five okay now I'm going to isolate the tangent so I'm going to be adding one to both sides of this equation if that five ever like shows up there it is okay so 2 tangent 2 x = six and it looks like I can get about 10 seconds thing before it does that now divide both sides by two okay take the square root of both sides so that's going to be tangent x = + or minus the < TK 3 and okay that's going to be kind of you something that we have to solve uh we have to know how to find that on the unit circle okay and if we're unsure you know we see that square root of three so we know it's going to have to do with 30 60 90 triangles that's that's one thing we should definitely know from AP preal so if we think you know maybe my hypothesis is pi over 3 because that's what it's going to be okay go up here and I'm just going to think about you know the standard diagram we draw where the hypotenuse is one and then the short leg is a half and long leg is root3 over 2 yeah that's going to work okay so tangent X is equal to positive < tk3 there would also equal positive < tk3 down here okay and then negative < tk3 would be these two points also so that's going to be X equaling pi over 3 2 pi over 3 and they only ask in the interval 0 to 2 pi so that would be 4 pi over 3 and also 5 pi over 3 and we don't need the 2 pi in because they only asked about the ones in the interal 0 to 2 pi all right so now Part B let's just keep going Part B here J and K are given by these two things right J is a single logarithm with base 10 all right so 2 log base 10 don't really need the 10 of x x + 3 - log base 10 of x - log base 10 of 3 okay let's just take this one step at a time here um I'm going to deal with that two that can become an exponent log of x + 3^ 2 and if you look at the actual rules you know that you see um in the Box you don't have to like Square out a binomial so minus the log of x - the log of 3 okay now we're going to do log of x + 3^ 2 over X okay subtraction becomes a uh division on the inside of the log and then I'm going to be subtracting log of three okay so that's going to be divided again and so that's just going to go into the denominator so x + 3^ 2ar / 3x X could write it base 10 like that but you don't need the base 10 that's implied okay um but the result should be of the form log base 10 of an expression so you maybe I'll just do what they say all right K of X involving secant and nothing else okay okay we know K of X is equaling tangent 2 x coent X okay well why would they even do that already reciprocals divided by cant X all right so I'm honestly I'm of a mind I think I I don't even think I need to use Pagan identi for this one um tangent X over 1 time tangent X over 1 hey that's my tangent squared the cotangent is a one over tangent okay and then cosecant in the denominator I'll just leave that there for now okay just deal with these tangents okay now I have tangent X over cosecant maybe I should just write everything in terms of s and cosine and then s over cosine is tangent and then 1 over secant is s over one okay need more paper I guess where even am I okay so that's sin^2 x / cosine it said cant and nothing else okay all right s squ x I do need a Pythagorean identity and almost two hours into this video I'm about to go nuts I'm really really tired of this practice exam to be honest but it's okay it's good that we have one full practice exam video you know on this channel I think so we know that sin^2 thet + cine 2 thet = 1 so sin^2 thet is going to equal 1 - cine 2 Theta in our case where I've got this sin^2 x I'm going to replace that with 1 - cine s x divide that all by cine X okay now I'm going to look at this as like 1 over cine X you know kind of distribute the 1 / cosine X to the one and the cosine squ X oh my goodness um over cosine x 1 over cosine is secant x minus this is just one cine X but they wanted it in terms of secant only um I'm not very impressed with this um with this type of problem here I don't really see the benefit to your like overall understanding of pre-calculus or later on calculus by being able to do this um but it's something we have to do for the AP exam so it's something we learn how to do and we perform it and we get our credit and we'll move on with our lives so now we're going to lastly in part C find where m ofx equal 1 over6 and that is going to bring this video to a nice close hopefully just short of two hours so 2 5x + 3 divided by 2 x - 2 cubed equals 1 over 16 I think I should simplify what I've got going on on the left hand side of the equation first before even dealing with the 1 over 16 I think it's very tempting to cross multiply and we could do that but I just don't and and you could even write 16 is 2 the 4 and do it that way uh I'm just going to go with the exponent properties um and I'm going to reorganize this to see that anymore so I'm going to be just going on exponent property level 5x + 3 deal with that power of power we're multiplying those exponents multiply the 3 over the X and the -2 and then um if I'm dividing two things with the same base in different exponents I'm subtracting the exponents 5x - 3x is 2x 3 - -6 is POS 9 okay 1 over 16 is 2 The -4th Power um CU 16 is 2 4th in the denominator that would be 2 the4 power now I'm going to equate the exponents 2x + 9 = -4 subtract 9 IDE by 2 get -13 / two that's going to be all for this one um that's all for this video uh I'm done thanks for watching