 In this video I want to talk about a very important theorem known as the intermediate value theorem. Now in a math textbook you often see numbers attached to theorems like this is theorem 2.5.17 which this number might be relevant for this lecture series because this is the 17th result inside of section 2.5 in our lecture series which is about continuity 2.5 is. But if I was to go ask someone like oh you remember theorem 2.5.17 no one's gonna know what I'm talking about and that's because if a theorem is really important enough to we want to talk about it you know outside of a classroom out you know without reference to a textbook we need to attach a name to it and so when a theorem like the intermediate value theorem has a name attached to it that means it's kind of a big deal and as a student of calculus you're going to want to know this theorem by name you need to know the assumptions of the theorem you need to know what the conclusions of the theorem are so you can use it correctly so let's read what the theorem says and talk about what the interpretation should be suppose that a function f is continuous on some closed interval a to b closed interval is important here and let n be any value that sits in between f of a and f of b assuming f of a and f of b are not the same y-coordinate otherwise there's nothing between them and this is actually where it gets the name intermediate value theorem because if n is between f of a and f of b then it's an intermediate value one rule of thumb I want to mention to you when it comes to calculus we often use the words value and number differently in regular English these are these are considered synonyms right but in calculus to help make the language a little bit more articulate whenever we talk about a number we're talking about an x coordinate that is we're talking about something in the domain on the other hand if you ever talk about a value we're talking about a y-coordinate and so that's something we try to distinguish between a value is a y-coordinate that is it's something with respect to the indirect variable a number is referring to an x-coordinate or something to do with the direct variable and if I ever talk about a point a point will always reference a ordered pair we have an x-coordinate and a y-coordinate together so we have the intermediate value theorem I want you to talk about it's the intermediate y-coordinate theorem that's another way of thinking about it and so getting back to it if f is continuous on some closed interval a to b and you have some intermediate value n between the y-coordinates f of a and f of b then there exists a number so there's guaranteed a number in the domain that is there's some number c that sits inside the interval a to b this notation here just means that c is greater than a but less than b so there's some number there's some intermediate number between a and b that guarantees that f of c equals n so to the right here on the screen you see an illustration that's that's well illustrating the intermediate value theorem if you have some function f like you see right here it's continuous right so there's no breaks holes or rips inside of our function here if we pick specific numbers in the domain so like x or x equals a x equals b you're gonna see there's these points right here on the graph and so we look at the values f of a and f of b if we were to pick any into any intermediate value between f of a and f of b if we draw the line eventually it'll intersect the graph somewhere and then there's some coordinate in the domain somewhere between a and b that produce that value so if we pick any intermediate value between f of a and f of b there's guaranteed to be something in the domain something between a and b that'll produce that value so this point right here is c comma f of c where f of c is equal to n and so let me show you some applications well one in particular of the intermediate value theorem we are going to use the intermediate value theorem to show that there is a root to the equation 4x cubed minus 6x squared plus 3x minus 2 equals 0 now i deliberately chose this polynomial because this is one you're going to have a hard time factoring if you sort of do it using techniques you might have learned in a pre-calculus setting we're going to show that there's actually a root to this polynomial there's a solution to the equation between the numbers 1 and 2 so the first thing to do is you have to first mention what is the function in play here what is f of x and so i'm going to take f of x to be the left hand side of this equation right here 4x cubed minus 6x squared plus 3x minus 2 notice that this function is a polynomial function therefore it is continuous on its domain which is all real numbers so f of x is continuous everywhere so that's the first part whenever you do a proof like this you always need to verify the assumptions of the intermediate value theorem because if you take a function that's not continuous then the result is not guaranteed like if we take a function which has some type of jump discontinuity like so so here's our function f and then we pick values like okay whoops here is my f of a we'll pick this right here to be like my f of b and then let's choose my intermediate value to be this one right here this is my n is there some place on the graph of f that intersects this line and no there's not because there's a big old stinking hole inside the function it's kind of like if i throw a rock at your house right you know it's like people who live in glass houses shouldn't throw stones well they can if the window is open if you throw the rock it just goes through the window there's a hole in the wall it doesn't break anything no big deal so continuity is important so it is important to declare in your in your argument here that f is continuous because if f was discontinuous the intermediate value theorem doesn't apply here so once we've once we've identified that the function is continuous because again the types of the ones we'll do will be like polynomials or trigonometric functions things that will be obviously continuous based upon what we've already talked about in this lecture in the previous lecture the next thing to do is then to verify that i'm going to get rid of this picture right here we then need to verify that there's some place where the function is below the x-axis and somewhere where it's above it right so think of this in our think of this as our picture here so we're going to take the x-axis right here compute the function f of one okay well f of one just by regular calculation right you're going to take four times one cubed minus six times one squared plus three times one minus two simplify that of course all powers of one are themselves just one so you end up with four minus six plus three minus two that equals negative one negative one notice is a negative number so what we're saying here if you take the value x equals one we're going to do x equals two in a moment if you take the value x equals one we're going to get a point something like this at x equals one the y coordinate turned out to be negative one what about at f of two if we were to do f of two well again just evaluate the function you're going to get four times two cubed uh minus six times two squared plus three times two minus two like so two cubed uh is equal to eight times five four is 32 two squared is four times six is 24 three times two is six and minus two and so we combine those together 32 minus 24 plus six minus two that adds up to be 12 which is positive and so you're going to get something like this do not worry if it's drawn to scale because in the middle there's no x there's no y coordinate no y-axis for it to be drawn to scale at all anyways there's no scale but if you take the point two comma 12 you see that there's a point below the x-axis and you have a point that's above the x-axis because the function's continuous there must be somewhere the only way we can get from point a to point b is somewhere we have to cross this line there's got to be an x intercept between x equals one and x equals two and so that's how the intermediate value theorem applies the intermediate value theorem using the intermediate value zero you'll notice zero is between negative one and 12 because the function is continuous the function is somewhere negative and somewhere positive there must be an x intercept somewhere between one and two so we can guarantee the solution to this equation by the intermediate value theorem even though we don't actually know what the solution is now the intermediate value theorem does actually provide a technique for which we can approximate the solution we know there's a solution between one and two well if we got a little bit closer notice that f of 1.2 is negative 0.128 that's negative if you take f of 1.3 that's 0.548 that's positive and therefore we can even do better not just you know we knew there was a solution between one and two but now we actually know there's a solution between 1.2 and 1.3 um we can also do better right if we take f of 1.22 that's gonna be negative 0.007008 and we could also do f of 1.23 which is 0.056083 which is positive so even doing better we know c is between 1.23 and 1.22 and we could we could kind of start zeroing in on the root of this polynomial more and more and more so by trial and error we could approximate the solution as a consequence of the intermediate value theorem now in general we're not going to do this to solve equations because there's a lot of guessing and checking and there are more efficient ways like Newton's method which we'll talk about later in this lecture series but that does conclude for us lecture 13 about continuity thanks for watching everyone if you like this video feel free to give it a like click right if you want to see more videos like this feel free to subscribe so you get updates about new math videos that'll be created in the future if you have any questions whatsoever when you watch any of these videos in this lecture series feel free to post them in the comments i'm glad to answer them for you and try to help you learn some more about mathematics bye everyone