hello everyone its mr. Spinelli today I'm gonna give you a little rundown of the key topics in AP calculus a B so one of the first things you see is limits now remember the limit exists at a point C if the limit from the left equals the limit as X approaches from the right very basic definition limit from the left has to equal a limit from the right now we can use limits to discuss some asymptotic behavior in particular horizontal asymptotes are when we take the limit of a function as X goes to positive or negative infinity now you can have situations where the numerator and denominator have equal degrees and in that case your limit will just be the ratio of those coefficients on the leading term and you can also see this result if you apply l'hopital a number of times and then you also have situations where the numerator has a degree that is greater and in those cases you will go to positive or negative infinity you got to be careful with there's square roots or odd powers involved when you look at these kinds of limits and then you've got situations where the denominator has a greater power so then these situations your limit will go to zero when the denominator has a greater degree ace then then we can also look at limits to address vertical asymptotes and in this case we're approaching a number so X is approaching some number and the limit either doesn't exist or it approaches positive or negative infinity so I just got a simple example rational function 1 over X minus 3 from the left that goes to negative infinity from the right it goes to positive infinity therefore the limit does not exist now we'll never say that an infinity equals an infinity so if this were like 1 over the absolute value of x minus 3 you could still say the limit doesn't exist or you could say the limit is positive infinity both those would be acceptable the limit doesn't exist but the function tends towards positive infinity now limits are also very useful in continuity now we're only going to add one little thing to our equation so typically the limit has to equal be the same from the left and the right but for a function to be continuous we also need that limit to equal the value of the function so the limit must equal the value of the function at a particular point then there's removable discontinuities that you might see this is where you'll have a hole so an example of that might be X minus 3 over X / minus 9 now you can factor the x squared minus 9 is X plus 3 and X minus 3 and that X minus 3 component simplifies to 1 so you have a hole at x equals 3 the limit will still exist there but the function will not be continuous because the functions not defined there then you run into the situation where you take limits and more often than not you're gonna want to apply l'hopital's opie tall basically just says that if you see 0 over 0 or infinity over infinity that's an indeterminate form then you apply l'hopital's I live ative of the top and the bottom separately you don't apply quotient rule you just take the derivative of both the numerator and the denominator separately and then plug in your value that you're approaching if that returns another indeterminate form you keep applying low feet all until you get a constant 0 or infinity some kind of number or thing you're used to seeing when you take a limit now if you don't get an indeterminate form you may need to rewrite it so if you can't apply l'hopital because you don't get 0 over 0 or infinity over infinity then you need to rewrite it or apply some other algebraic manipulation to the problem and now if you have graphs same rules apply the limit exists if the function approaches the same value from the left and the right limits can't exist at holes so here we've got you know a nice little graph as we go to negative infinity you see the little arrow so it's going to go to negative infinity the limit of this function as we go to negative three from the left that approaches two from the right it approaches negative four so the limit as X goes to negative three doesn't exist now as X goes to one it does exist even though the function is defined at a different value it's positive two when X is one but the function approaches negative 2 as X goes to one and that's that hole down below the x-axis now as X approaches four the limit doesn't that exist it goes to positive infinity from the left and from the right it doesn't even approach anything at all then the limit as X goes to positive infinity is simply positive infinity because we've got an arrow pointing up tables very much the same thing as graphs you're looking for the limit to be the same on both sides and if necessary make a little sketch so here we've got a table we're gonna look again at that limit as we go to negative infinity which it appears it's approaching negative 1 so we probably have a horizontal asymptote of y equals negative 1 now as we go to negative 5 it appears some more interesting things are happening we approach negative 2 then as X goes to positive 2 from the left it looks like it's negative infinity because we've got that negative 19,000 large number but as we come from the right x equals 2 point 0 0 0 1 we're approaching 17,000 positive so from the left it's negative infinity from the right it's approaching positive infinity therefore the limit doesn't exist and then as X goes to infinity it appears the values going towards 3 so for this particular function that we've been given a table 4 has a horizontal asymptote at y equals negative 1 and y equals 3 and it appears it also has a vertical asymptote at x equals 2 so then we get into derivatives and initially derivatives are defined as a limit because you start with secant lines which are just average rates of change so you can see in this little Jif that as those two points they're on either side of that black dot when they're far away we're approximating the average rate of change at that point x equals 1 and we do that by just computing the slope slope old school way change in Y over change in X then we get to instantaneous rate of change so we want those two points so you can see right now they squeeze close to the black dot that is the so-called tangent line as they get right on to that dot we've got a tangent line and you first see that using what's called the difference quotient so the derivative is defined as in limit so the limit as H goes to zero of f of X plus h minus f of X all over H now if I wanted to find the derivative at that particular black dot I can apply it by using the limit as X goes to a and in this case I would say a is 1 so f of X minus F of a over X minus a so these are the derivatives in the context of limits which is how you first see derivatives and then you talk about differentiability very much like continuity so first you need to check that a function is continuous so you check that the limit from the left equals the limit from the right equals the actual functions value so you can't have a hole it's got to be filled in and the functions got to approach that filled in dot from both sides then you check the continuity of f prime of X so you take the derivative and you can do this using limits or you can just take the derivative if you have the function then you're checking that the derivative is the same on the left and the right and that it is defined at the value C once you verify both of those things then the function is differentiable remember a differentiable function is continuous so oftentimes the AP test will say function f is differentiable therefore you know it is also continuous if you're going to apply a theorem like intermediate value theorem you need to specify that because it's differentiable it's also continuous there for intermediate value theorem applies you need to be able to clearly state how you know the things you know so here's just a little example of why you need to check the continuity of the function so this particular function has a slope of 1 on both sides so you might think it's differentiable but it's not continuous it's got to be continued to be differentiable so then here speaking of your theorems intermediate value theorem remember the word intermediate just means in between so if we've got a continuous function on some closed interval from A to B then there is some x value between a and B such that will hit the values in between so here's a nice little jiff to show you we're hitting every value between those two blue dots your function may go above as you see ours does but you know for a fact that it's gonna hit every value between those two blue dots again your function has to be continuous and then we've got the mean value theorem mean value theorem we need continuous and differentiable and again it can be differentiable on just the open interval and then that just means that we can find some point that's gonna have the slope see that blue line in this chip that's the slope between our two endpoints and our functions continuous and differentiable between those two endpoints and you're gonna see here as that blue dot moves along in just a moment that red line is going to turn green because there's a point where the tangent line right there is equal to the average slope between those endpoints in this particular particular function there's two points you see in just a moment this red line is gonna turn green right near the top right about there and again we see that there's exist at least one point where the instantaneous rate of change is the same as the average rate of change over that interval now you also have rolls theorem which is just the same exact thing except f of a is equal to f of B so those two blue dots would be perfectly in line perpendicular to the x axis having the same Y value which means we know the slope has to be 0 at least once between those two points speaking of derivatives you got all your different rules you've got power rule you've got quotient rule but the big one that you've got to remember is chain rule chain rule can be it appears everywhere so just remember you work your way from the outside in then you've got inverse functions remember function is an inverse of one another if you compose them and you get X as a result so F of G of x equals x and G of f of X equals x and I've just written in here two different ways so you're aware of the ways it can be presented on a test and then remember that you've got some fancy formula for remembering the rules now I always just remember that I need to solve for the function value given so if I'm given F or G depends on which one I'm given if i plug an x value into one of the functions let's say I plug a into G and it gives me B that means a B is a point on G which means B a is a point on F so if I want to find the derivative of G at a I need to do the inverse the reciprocal I should say of the derivative at B of F so G prime of a is equal to the reciprocal of F prime of B again it helps to write all that you know about F on one side and all that you know about G on the other side and remember that the x and y values are going to flip-flop they're gonna switch roles and the derivative at one point is the reciprocal of the derivative at the corresponding point for the other function and you've got an implicit differentiation this is where you need to remember that Y is a function of X so every time you take the derivative of Y you're gonna get a dy/dx so for example this function my first term is gonna give me 2x my second term I've got x times y both of those things involve X cuz Y is a function of X so I've got to use product rule and then for the sine term Y is inside a sign so I'm gonna have chain rule same with e to the Y so I should get a result that looks like this the 2y plus 2 X dy/dx that second and third term on the left side both of those come from 2xy using product rule and then you're gonna want to group all the dy DX's together and then ultimately divide by what why DX is x to get your derivative and remember when you have to take a second derivative it's gonna involve dy/dx so you'll need to be sure to plug this back in to your second derivative if it asks you to evaluate just plug numbers in and you don't have to simplify but if it asks for what is the second derivative you need to be sure to plug this 2x plus 2y at yadda-yadda into your second derivative result applications derivatives are very useful for finding extremeties and mins so the first derivative test you're gonna take the derivative set it equal to zero and make a sign chart remember that the sign chart is not your justification you have to explicitly say I know the function is switching from increasing to decreasing because f prime goes from positive to negative okay and this should say explain work okay so when F prime of X changes from positive to negative that is when the function f of X changes from increasing to decreasing I must explain my sign chart second derivative test once I know my critical numbers where the derivative is zero I simply need to compute the second derivative at those points if it's positive I know it's concave up which I mean that means it's smiling if you will so I know I'm sitting at a relative minimum if it's the second derivative is negative then I know the function is concave down so I'm frowning and I'm sitting at the top of a hill at a relative max to find absolute Max and mins you have to check the end points as well so not just the critical numbers where you had the derivative was 0 or undefined you need to also check the endpoints key thing in calculus first semester is remembering how to connect a function to its derivatives and vice versa so whenever you're given a function you know that the first derivative is positive when your function is increasing and it's negative when your function is decreasing you know the second derivative is positive when the function f of X is concave up or smiling you know the second derivative is negative when the original function is concave down or frowning when you're given the derivative F prime you know that the original function is increasing or decreasing when F prime is positive or negative so above or below the x-axis and you know that f of X has local extrema when the sign changes for that first derivative so it goes from positive to negative or negative to positive and then you know when the function f of X is concave up or concave down because the second derivative is the derivative of F prime so when F prime is increasing going up then you know f of X is concave up when F prime of X is decreasing going down then you know f of X is concave down now when and that's just this is this next bullet point is the same exact thing I just said because you know that the second derivative is positive when the function f of X is concave up so that's just the same statement but sometimes you've got to consider when this involves position velocity acceleration you've got to make all these connections now you know the function f of X has inflection points when F prime of X has local extrema because what's happening is the concavity is changing because f prime of X is going from positive to negative that means the second derivative F double prime is going from positive increasing to decreasing so when the F prime goes from increasing to decreasing f double prime is going from positive to negative when you're given the second derivative there's not as much that you can tell but you do know when the function f of X is concave up or down based on when F double prime is positive or negative and you know when f prime of X is increasing or decreasing and that's simply based on when it's concave up or concave down so when F double prime is positive or negative then you've got position velocity acceleration which very much ties into what we just discussed so position is often described by X of T tells you where an object is velocity is the derivative of position it tells you the speed and direction speed is the magnitude of velocity and direction when your function V of T is positive your object is moving in the positive direction which is often to the right or forward and when the velocity is negative its moving left or backward in the negative direction acceleration is the derivative of velocity which means it's the second derivative of acceleration and it measures the change in an object's velocity which remember involves speed and direction so acceleration can be a change in speed it can be a change in direction or it can be a change in both so when acceleration velocity are the same sign then the speed is increasing when they're opposite signs the speed is decreasing when acceleration is positive that means the velocity is increasing when it's negative that means the velocity is decreasing so there's a key distinction between the difference in velocity and speed then you've got function behavior so you can tell when a function is increasing or decreasing and again that's based on knowing where the function has derivatives where the points are zero or do not exist and you're gonna make a sign chart you know when it's concave up or down based on taking the second derivative setting it equal to zero solving for those points to determine where the function has concave up concave down intervals by using a sign chart for it as well inflection points the second derivative must be zero and the second derivative must change sign at that particular point and more importantly it's got to actually be a point on the function so you've got to be sure that if you want to determine if something there's an inflection point it has to be defined for that point linearization key thing is you need to write a linear equation you need a point you need a slope and you need to write an equation and you often do that in point-slope form optimization you're just looking for Max and min local extrema so the goal is to often write a function as a single variable usually this requires using the volume or area or perimeter you have some perimeter of fence you have a certain amount of material to make a box so that's a surface area constraint so you're gonna need to solve for X as a function of Y or Y is a function of X or you may decide to use different letters to represent the base and height width of a box but the goal is still the same find the extremley using the first and/or second derivative test and check your endpoints so sometimes you are explicitly given an endpoint think of the classic problem where you got to cut a corner from a piece of paper you can only cut so much paper before you can't even make a box and you've got to cut something or you can't make a box so there's usually some constraints on what our so called impassable values think about the context of the problem to determine what endpoints you have to check related rates very much like implicit differentiation everything is a function of whatever variable you're dealing with unless that particular quantity is a constant pi is a constant e is a constant if the cone is restricted to some particular dimension the height of the water is going to certainly change the radius is also going to change but if we switch that cone to a cylinder the radius is not going to change as we fill it up with water this particular example is when you've got some kind of triangle so you're gonna connect things using Pythagorean theorem maybe one car is moving in one direction to the east the other one's moving to the north what's the rate of change between those two cars you're looking for DC DT then you've got your classic volume of cylinder PI R squared H so here's an example where that dr/dt is likely zero because odds are the radius of your cylinder is not changing but I've just written out what that derivative would look like using related rates everything is a function time except for pi and then you may have something that involves angles so this might have involved some height floating into the air or something along those lines and remember whenever dealing with problems on the AP test all angles are in radians and less otherwise specified next big topic is semester two you're jumping into integrals first thing you often do is approximate integrals and you do that with Riemann sums you need to remember the area of a rectangle simply the length times the width or base times the height you've got left Riemann sums right Riemann sums and midpoint Riemann sums remember that a left Riemann sum is going to be an over approximation when the function is increasing and the right will be an over approximation when the function is decreasing so you've got your rules to remember that trapezoidal sum you got to remember the area formula for a trapezoid it's going to be an over approximation when the function is concave up and it's gonna be under when it's concave down if you do not remember this which I don't bother remembering it I always draw a picture the picture will help you determine if you are sketching things above if your rectangles or trapezoids are above the function it's gonna be an over approximation otherwise it's gonna be under midpoint is gonna be somewhere in between so you're often not asked to assess whether the estimate would be over or under for a midpoint then you get into summation notation this is how we then extend the Riemann sum to an infinite sum you need to be able to go both directions using this these formulas remember that the integral can be written as an infinite sum so out of all of those rectangles of the infinitesimally wide rectangles and you will get the exact area so here's just an example you're integrating from negative 5 to 8 that's a distance of 13 so you see that 13 over N as the Delta X the initial point a is negative five so that's wherever you see an X x squared and the cosine of X it's replaced with negative five plus that Delta X times I I is just a counter to move you along to more and more rectangles and then recall you've got to take the limit as n goes to infinity so you gotta be able to go both directions left to right and right to left then you get into the fundamental theorem of calculus which basically just says that the integral from A to B is whatever that antiderivative is that function evaluated at B minus the value evaluated at a and I've written just two versions of it oftentimes you're given a graph of f prime let's say and it tells you an initial condition like at x equals to f of X is equal to 3 so you know 2 comma 3 is a point and let's say it wants you to find f of 7 what you're gonna want to do is use this bottom equation integrate from x equals 2 to x equals 7 because you know the value of the function f at 2 so you'll know F of a on that right side and you're probably given a picture that you can determine the area between x equals 2 & 7 so you can determine the left side of this equation which allows you to solve for f of B which is f of 7 that point that you're looking for now the other thing to remember is you need to remember that the area is associated with this integral it's positive when it's above the x-axis and it's negative when it's below the x-axis and then remember too that you can split it into pieces so if I want to break down the integral from negative 1 to 3 I can go from negative 1 to 1 1 to 2 and then 2 to 3 and then also if I switch the order of integration I've got to remember that it becomes the opposite value so if I integrate from A to B that's the opposite sign of if I integrate from B to a this is very useful when I was talking about those top two equations when you're connect a function given to you with its area in some initial condition to find the value of the antiderivative at some other point so oftentimes you've got to remember what direction you're going then we got the FTC part two which is really just taking the derivative of the previous fundamental theorem key things to note in these two equations the top one is integrating from a constant to X that one is nice and easy the bottom one is integrating from some function of X to some other function of X so I've got to remember to apply chain rule do not forget chain rule would apply in FTC part two techniques you've seen a bunch of them but you've got u sub u sub you're usually just looking for some function it's inside of another function and whose derivative is outside of that said function and it may be some multiple of that function so you got to deal with some constants remember to change the limits of your integration if you've got a definite integral otherwise just focus on looking for some function whose derivative is elsewhere other techniques are long division and completing the square these are a little less common so I'm not gonna spend a whole lot of time talking about those then you've got applications so remember that integrals are accumulation of change so you've got your average value you're integrating from A to B and you're dividing by the width between a and B position velocity acceleration total distance travelled you're integrating the absolute value of velocity whereas with displacement you're just integrating velocity from two endpoints then you often run into the fr Q's where you see units involved so let's say a function has units of cubic feet per hour so it gave me f of X and then it's gonna ask me a series of questions well I often like to write what's f Prime's units gonna be what are the units of f double-prime going to be what are the units going to be if I have to integrate so as you notice that independent variable time hours ends up getting we end up dividing our units by a time as we take derivatives and we multiply by time as we integrate as we go up if you will so these are just this just an example of the units that you would encounter so if it wanted me to say how many cubic feet are gonna be in the pool if f of X is defining the rate at which water is entering the pool I'm gonna have to integrate to get cubic feet if it's asking me how is the rate at which the water is entering the pool changing I'm gonna have to take the derivative and my units are gonna be feet per cubed per hour per hour which some of us struggle to wrap our heads around what that means in this particular case that top unit feet cubed per hour it makes a lot more sense if you deal with units of man-hours so oftentimes if you're working in the business world you may see man-hours otherwise those of you that take physics you know that Newton's times meters quantifies energy so then just extending on that let's say f of X still has units of cubic feet per hour if we were to integrate we know we get feet cubed so if we integrate from A to B then we know between let's say time x equals a and time x equals B we've accumulated so an X amount of feet cubed X amount of cubic feet during that time frame you may also be asked what's the average across that interval well now you're dividing that cubic feet by the time B minus a so now your units are back in feet cubed per hour we found the average rate of change f of X and then here's just showing you well what if you're looking at f prime of X so notice that all the results on the left just got divided by time then lastly you've got differential equations so often you may be seeing a slope field you're going to look for lines of constant slope don't draw anything where the slope is undefined because you don't know what's going on with you original function then lastly if you have to draw a particular solution through a point go with the flow sometimes you might have to verify solutions that just means if a particular differential equation says three y double prime minus y prime equals zero and it gives you that a potential solution is is y equals two x well you're gonna take the first and the second derivative plug it into that original equation to see if it actually does equal zero and then lastly you've got separation of variables so you're gonna want to get the X's on the side with DX and the Y's on the side with dy and then integrate both sides respectively when you're looking for a general solution you don't have to solve for C when you're looking for a particular solution you do need to solve for C using some initial condition which is going to be at point so thank you very much for watching and good luck on your tests