all right welcome everyone to 1.8 this is continuity and this is the final section in chapter one so for this first video I want to talk a little bit well about continuity this idea that we've probably brought up before uh but we're going to be able to detect when a function is continuous and when it is not continuous or another fancier term is discontinuous uh and then we're going to use continuity and it turns out these have something to do with limits so we're going to be able to quickly evaluate out some limits so back in the day right we used to say something along the lines of a function is continuous if you can draw it without lifting your pencil and yeah this is true and a great intuitive definition but we can be more precise now that we know something about limits so in this example right here let's just get a little bit of a practice uh talking about continuity right so consider the graph of f ofx to the right on the domain from -2 to five so I can go all the way down to the x value -2 all the way up to the x value five but that's where the function stops determine where the function is continuous all right so it looks like I could take my pencil and I could draw all the way up to zero but then if I wanted to continue I'd have to lift my pencil right so it's not continuous at zero so I'm just going to go ahead and basically write some scratch notes Here X should not be equal to zero and then let's see what else we go on and then at four right I'd have to look my pencil to draw this dot down here so it's not continuous at four you can see there's a hole so X is not equal to 4 but besides those two points it seems to be continuous everywhere else so the question again is where is the function continuous well it would be all the way from -2 to 0o and then I'm using the parentheses to not include zero I want to jump over zero so I'm going to do a union from 0 to four not include four either Union from four to five so these are all the values between -2 including -2 and 0 but not including Zer then all the values from 0 to four not including either one of those and then from four to five not including four but including five so this is where the function is continuous where I could draw it without lifting my pencil more formally right so this is the old definition a nice intuitive definition if we have a graph but now that we know something about limits let me give you kind of the formal definition so a function f is continuous at a number a oops [Music] number a if and only if the limit and notice that this is a two-sided limit so as you approach from the left or from the right towards a for F ofx this should be equal to the function's value so note there's some additional subtleties here so this requires first of all that there is a value F of a right so this needs to be defined and then also it assumes that the two-sided limit exists so notice if we go back up here we can see that in this first case at zero the left hand limit and the right hand limit are not equal to each other so in this case the two-sided limit doesn't exist so it's not continuous in the second case the limit does exist as you approach from the left and from the right the limit exists the limit would be equal to probably three in this case right 1 two 3 however the function's value at four is not equal to three it's actually equal to here right this closed dot is at one height one so in this case the limit and the functions value are not equal to each other so you can kind of see how these limit actually kind of build on our intuitive idea of what is a continuous function all right so another definition right uh we often talk about continuity on an interval so a function is continuous on an interval if it is continuous at each point on that interval and then finally a nice theorem essentially all the functions we love pols rational functions root functions trigonometri functions are continuous everywhere in their domains let me give you an example so a couple here the limit as X goes to zero for this nice rational function right this is a polinomial over a polinomial well notice zero is in its domain so I can just go ahead and plug in zero everywhere I see an X so this is going to be 0 + 3 * 0 - 4 / 0^ 2 - 1 so -4 /1 is 4 now we used to call this in 1.6 the direct substitution uh property and this is kind of a little bit more formal right because the function is continuous that's why we can go ahead to in order to evaluate this limit we know that that limit is going to be the same thing as the function's value in this case at zero likewise cosine cosine is defined everywhere so if I want to figure out this limit this is going to be 5 and cosine of pi again just plug in pi everywhere I see a t in this case and cosine of pi is - 1 so this is going to give me the answer of5 so there are my two answers for these examples so we can really evaluate out lots of limits very quickly Now using something a little more formal than direct substitution it's actually because theun function is continuous all right so one kind of upgrade that you might want to make is you might be interested in what about combinations of these beloved functions right so rational functions root functions trig functions all these sorts of things so if I have this one right here right which we have a trig function and then inside of it we have a polom we have a root function on the bottom uh and another polinomial so the question is uh when we start combining these functions is it still uh continuous and does this theorem still apply and the answer is yes uh so let me go ahead and just plug this in so this is going to be S of 1^ 2ar that would be 1 + 1 2 / < tk1 * 5 - 1 so this is going to be S of 2 and then 1 * 4 so that's going to be four and that's going to be my final answer here okay so yes combinations do work uh and let me be a little bit more specific right the answer is yeah most of the time we are fine more formally if we have two functions f and g are continuous at a value a and then we have a constant C the following are also continu continuous at a so as you would guess if you were to sum two continuous functions you get a continuous function likewise if you were to take the difference of two continuous functions you would get a continuous function if you multiply a continuous function by a constant if you were to take the product of two continuous functions and finally if you were to take the quotient of two continuous functions F divided G of course so long as we don't want G of a to be equal to zero so so long as G of a is not equal to zero that's okay it'll be continuous otherwise you're dividing by zero that's an issue likewise if we have G is continuous at a value a and F is continuous at G's evaluation at a then the composite function right this kind of looks like fog here but it's F composed with G is given by right and that's just basically where you take G and you put it inside of F this is continuous a a so notice we use the majority of these theorems up here in this example without even knowing about it right because we have a composition of functions we have x^2 + 1 inside of a sign function we were using that hey s's a continuous function this polinomial is a continuous function this square root is a continuous function on its domain right we divided by stuff so we actually use quite a bit of this without even knowing it but these theorems right here I mean these can be proven and it's just to kind of help us sleep at night right so that we won't break math by any means all right for our last example I would like to determine where this function is continuous and you as you can see we have a composition right so we have 1 - x^2 in tied of a square root we have division by another polinomial in this case just X but from all of our theorems up above we know that this function should be continuous on its domain so the big question is really an algebra question and that is what is the domain of this function all right so I need to think back there were some kind of rules that we had when we were trying to think about domains functions one of them is that we don't like to divide by zero right so in this case in order for me to divide by zero I would have to have X is equal to Z so I don't want that to happen right I don't want X to equal Zer otherwise I'd be dividing by zero and would not be in the domain of this function a second big rule that we had when we were trying to figure out domains is that we don't want to have a negative inside of a square root so that is whatever I'm plugging into the square root in this case 1 - x^2 I want to make sure that that's not negative so that is that it's zero or bigger okay so I don't want X to be equal to zero and I need to know when is this true right when is this inequality true so I'm going to use a a uh technique in order to solve this inequality in which we find first where this is equal to zero those would be good points and then I'm going to kind of split up the real numbers into some intervals some sub intervals really and choose some test points so if you haven't seen this method before this is going to be uh a great introduction because we will use this a ton in this class all right so let me first figure out when this is equal to zero so I'm going to add x^2 to both sides sides take the square root and that'll give me plus or minus one remember when you take the square root you have the plus and the minus next up let me go ahead and draw the real number line here and I'm going to use the places when it was equal to zero to split up the number line so I'm going to go ahead and write Nega 1 here and one here so now you can see that we've split up the number line into three kind of sub intervals we have the numbers that are smaller than 1 we have the numbers that are between one and one and we have the numbers that are greater than number greater than one and so I'm going to choose test points in each one of these intervals and try to figure out is this inequality true on each one of those intervals for each one of those test points so let me go ahead and show you here's a test Point -2 that lives in the numbers that are less than1 if I was to go ahead and plug in -2 into this inequality I would have 1us -2 2^ 2 -2 2 will be 4 and is that greater than or equal to 0 so 1 - 4 would be -3 no this is not but next to this so it is not true at -2 and actually this test point will speak for the entire interval I'll go into more detail why this works uh in a later video next up let's choose a test point between netive 1 and0 my favorite has got to be zero so okay if I plugged in zero into my inequality 1 - 0^ 2 is that greater than or equal to 0 so 1 - 0 is 1 is that greater than equal to zero absolutely so this works out great and then next up one more I'll use green again I'll use black all right a number that's bigger than one so my test value will be in fact let me make a fun one right so this I would probably choose just like two right and that's kind of boring uh but let me choose a more interesting one maybe I'll choose 10 right that's a number that's bigger than one just to really show you that you can choose any number uh in this interval as a your test value and it'll work so okay 1 - 10^ 2 will be 100 is that greater than or equal to zero well 1 - 100 well that's going to be 99 is that greater than or equal to zero absolutely not so the only place that this in equality holds true is between -1 and 1 so if I was to go ahead and maybe you know draw the solutions to this right between 1 and one and then the question is what about at these end points here at negative 1 and one well we saw that was when actually this is equal to zero and this inequality is when it's greater than or equal to zero so I'm okay with equal to zero so yep let's go ahead and shade those in okay so the domain of this numerator bit is between -1 and 1 and we don't want X to be equal to zero so where is this function continuous well it would be everywhere from -1 up to 0o but not including zero then we can continue on our way to one so this will be our final answer this is where the function is continuous all right that is is the end of our first video on continuity here in Section 8 next time we're going to be talking about cases where functions are not continuous I'll see you then