WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, welcome back to AP Calculus.
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Today, we are going to talk about continuity.
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Let us jump right on in.
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Let me work in blue today.
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We mentioned several times already that the limit as x approaches a of f(x) and the value of the function at a,
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f(a), are independent of each other.
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We saw already that you can have a limit exist, as you approach a certain point.
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But that limit is not necessarily the same as the value of the function, at that point.
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It could be, it could not be.
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They are independent of each other.
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The limit as x approaches a of f(x) might or might not equal f(a).
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When it does, that is very special.
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For that, we say that the function is actually continuous there.
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Continuous means there is no gap.
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It is just if you take a pencil, drop it on a piece of paper.
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With one swing of your pencil, without lifting it up, you have the graph that means every single little point is accounted for.
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When it does, we say the function is continuous at that point.
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The function is continuous at a.
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Our definition, just to be reasonably formal about it.
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I just repeated what I just wrote.
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If the limit as x approaches a of f(x) = f(a), then f(x) is continuous at a.
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If we are talking about a particular interval, say from 0 to 5,
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and it is continuous at every point in that interval, we said it is continuous over that interval.
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That is it, pretty straightforward.
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Let us put some graphs to get a feel.
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Let us look at some graphs to get a feel for this notion of continuity.
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Graphs to get a feel what continuity is.
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Again, it is a very intuitive notion.
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Something that you know implicitly, you know intuitively.
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Let us look at some graphs to get a feel for what continuity is.
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Our first graph, let us go ahead and just do something like that.
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Let us say we have this and it goes like that.
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Let us call this 5 and let us say that this is 2.
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The limit of this function, this is our f(x), the graph.
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The limit as x approaches 2, which means from below and from above of f(x), it = 5.
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As we approach 2, the graph, the y value approaches 5.
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As we approach it from above, the y value approaches 5.
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The limit is actually equal to 5.
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Now f(2) is not defined.
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There is not even a value for f(2), that is why there is a hole there.
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This is not continuous.
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That is all continuity means.
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It is not connected.
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Let us do another graph here.
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Now we will just notice nice single sweep of the pencil.
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Let us say we have a point here.
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Let us say the y value of this point is 6.
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Let us say the x value is 5.
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Here the limit as x approaches 5, again when we do not specify, it is from both ends.
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The limit of f(x) as x approaches 5 is equal to 6.
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f(5), f at 5, is actually equal to 6.
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This is continuous because the left hand limit, the right hand limit, and the value of the function at the point equal each other.
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It is that simple.
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Let us look at another graph.
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Let us say we have got this function right here.
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Let us say we have got something that looks like that.
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Let us say this is the point 3, let us say this is 4.
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Let us say the y value of this one is 1.
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Now we have the limit as x approaches 3 from below, we are approaching it this way, of f(x), that is equal to 4.
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The limit as x approaches 3 from above, the value of the function, the limit is 1.
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f(3) = 4, that is the solid dot, right there.
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In this case, even though the value of the function, f(3) happens to equal one of the limits,
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in this case, the left hand limit, they are not all equal.
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The left hand limit has to equal the right hand limit, has to equal the value of the function.
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This is not continuous, we do say that it is left continuous.
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If the function were defined down here at 1, we would say it is right continuous.
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We do differentiate, just like we have left hand derivative, right hand derivative, left hand limit, right hand limit.
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We have left continuity and right continuity, but this function is not continuous.
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Here left hand limit does not equal the right hand limit.
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We know that the limit does not exist.
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I do not need to write all these, it is not a problem.
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You know what, I do not even need to write any of this.
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These do not agree.
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It is that simple, we do not need to label the point with silly formalities.
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Once again, the left hand limit does equal f,
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Let me write this out.
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The limit as x approaches 3 from below of f(x) does equal f(3).
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That is it, we just call this right, it is left continuous.
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f(x) is left continuous.
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That is it, it is continuous from the left, that is all that means.
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Over all, it is not continuous at 3.
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You can see this is not continuous.
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I have to do this then I have to stop.
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I have to pick my pen up, start again over here.
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That is the intuitive notion of continuity, one nice notion.
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It does not need to be a smooth curve motion.
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It can be something like this.
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This is still continuous because I have not lifted my pencil off the paper but these are sharp points.
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The graphs confirm your intuition or should I say your intuitive notion of continuity.
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In other words, without breaks in the graph.
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That is it, that is all it means, without breaks in the graph.
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In other words, when you put your pencil to paper to draw the graph of a continuous function,
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you can draw it without ever stopping and lifting your pencil.
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I should not say stopping because we can stop, without ever lifting your pencil.
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That is all continuity is.
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Once again, the limit as x approaches a of f(x) is equal to f(a).
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One, you find the limit if it exists.
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Two, the second thing you do is you find f(a).
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The third thing that you check is do they equal each other.
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If they equal each other, the function is continuous at that point.
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If they do not equal each other, the function is not continuous.
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You are doing two things.
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You are taking the limit, you are taking the two, and you are checking to see if they are equal.
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That is the question mark.
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This is the definition.
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When they give you a definition, when you see a definition that involves inequality,
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what that means is essentially what you have to do is you have to check the left side, check the right side.
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If the equality is satisfied, then the definition is satisfied.
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Then it is that thing, whatever the definition says.
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Those of you that go on in mathematics, when you take linear algebra,
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you are going to be talking about something called a linear function, which is not exactly what you think it is.
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There is a certain mathematical definition to it.
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A linear function is defined by equality.
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There is something equal to something else.
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What you are going to have to do is you are going to check the left hand side of the equality,
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check the right hand side of the equality, and then you have to confirm that they are equal.
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If they are equal, the function is linear.
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If they are not equal, the function is not linear.
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That is what equations in definitions mean.
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The laws of continuity correspond to the laws of limit.
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We will write them down.
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The laws of continuity correspond to the limit laws.
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They are entirely obvious.
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We will say let f(x) and g(x) both be continuous.
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They can be continuous over an interval or continuous at a given point.
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Then, the sum of the two functions is continuous.
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The product of two continuous functions is continuous.
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The quotient of functions is continuous, provided, of course, g does not equal 0.
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Any constant times a continuous function is continuous.
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Most functions you deal with are going to be continuous.
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If they have a discontinuity, it is going to be a very few places like for example,
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a place where the function is not defined.
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In other words, a vertical asymptote.
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It is going to be at a place of discontinuity.
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Those are really the only two cases where you are going to deal with something that is discontinuous.
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Most functions you will deal with are continuous.
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Functions are great, they behave very nicely.
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Let us do a few examples.
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We have an example, it says it is going to be example 1.
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We are going to draw a graph and we are going to ask you to tell us where the function is discontinuous.
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Example 1, give the points on the following graph. I decided to draw the graph by hand, where f fails to be continuous.
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There is your graph.
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Here is a vertical asymptote.
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The function is going off to infinity there.
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Where are the points of discontinuity, perfectly obvious.
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It is discontinuous here.
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At x = 1, it is discontinuous here.
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Clearly, at x = 2, it is continuous here.
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Even though it is a shell point, it is still continues.
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I did not have to lift up my pencil to continue again.
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It is discontinuous here.
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This goes off to infinity, never touches this, and never touches this.
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They are not connected.
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Those are your three points of discontinuity, 1, 2, and 6.
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Very intuitive, very obvious.
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Example 2, show that the following function is continuous at a given point.
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Also give the interval over which the function is continuous.
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Now we do not have a graph to help us out.
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If you have a graphing utility, if the particular test you are taking, you are allowed to use a calculator to,
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or graphing calculator or utility, to help you out, by all means use it.
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But again, the whole idea of the calculus is to learn to do things analytically so that we do not have to rely on the graph.
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If we have a graph to help us out, that is fine.
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But there are times you are not going to have a graph.
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We want to develop these notions that are always true.
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Not have to rely on geometric notion of a graph.
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Now we have to do this analytically, show that it is continuous at x = 2.
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We know what the definition of continuity is.
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The definition says the limit as x approaches in this case 2, of this f(x), has to equal f(2).
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That is what we are going to do.
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First, we are going to find the limit as x approaches 2 of this function.
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And then, we are going to evaluate the function.
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We are going to see if they are equal to each other.
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If they are, we are done, it is very simple.
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First, let us evaluate the limit.
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The limit as x approaches 2 of the function x³ - 14 - 4x.
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Again, do not let the symbolism intimidate you.
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It is just a basic limit.
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What you do with a basic limit, what you do to all limit, you put the value in first to see what happens.
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If you get an actual finite number, you can stop.
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You are done, that is your limit.
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If you get something that does not make sense, you try to manipulate the expression and take the limit again.
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That is all we have been doing, that is all you have to do.
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We put 2 in, this is going to be 2³ - √14 - 4 × 2.
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2³ is equal to 8 - the square root of,
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What is 14 – 8, it is 6.
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8 – √6, is 8 - √6 a finite real number?
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Of course it is, yes.
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That is our limit, very simple.
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Now we evaluate the function at 2.
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f(2) = same thing, 2³ - 14 - 4 × 2 = 8 -√6.
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They are equal to each other.
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Yes, this function is continuous.
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You might be saying to yourself, wait a minute, we just did the same thing twice.
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Yes, we do the same thing twice but under two different auspices.
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One, we are evaluating the function at.
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Here we are evaluating the limit but we know from our previous work with limits,
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that the way you evaluate a limit, in the case of a function like this, is to actually plug it in.
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Even though you are doing the same thing, you are doing it for different reasons.
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I know it seems a little weird but that is what is going on.
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It is actually two different processes.
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√14 - 4x, we see that this, our domain is going to be,
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Now I give the interval over which it is continuous.
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We know it is continuous at 2.
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Over which numbers is it actually continuous, besides 2?
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This implies that 14 - 4x has to be greater than or equal to 0, because you cannot take a square root of a negative number.
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Our domain is restricted.
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Therefore, 14 has to be greater than or equal to 4x.
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Therefore, 14/4 has to be greater than or equal to x.
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The domain of definition is, the domain of f is negative infinity to 14/4 exclusive.
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It is continuous over this entire interval.
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There is no point in this interval where it is not continuous.
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The function is defined and it is perfectly valid there.
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Is the following function continuous at a given point.
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f(x) = e ⁺2x, for x less than or equal to 0.
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It is equal to x², for x greater than 0.
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Is it continuous at x = 0?
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Here we have a piece wise continuous function.
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It is one function to the left of, 0 it is another function to the right of 0.
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We need the left hand limit to equal the right hand limit.
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We need it, in order for them to be continuity, we need it.
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First, the limit as x approaches 0 from below is this one.
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We use that function of e ⁺2x.
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As x goes to 0, 2 × 0 is going to be 0.
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It is going to be e⁰ and this can equal 1.
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Now the right hand limit.
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Now it is going to be the limit as x approaches 0 from above of x².
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When I put that in, that is just going to be 0.
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In this particular case, we can if we want to.
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Let us just do it for the heck of it.
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Let us just do f(0).
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f(0), here it is less than or equal to 0.
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We use this function.
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This is going to be e⁰ = 1.
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We see that the left hand limit and the value of the function are equal.
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The function is left continuous, but these three are not equal.
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It is not continuous at 0, it is not continuous at x = 0.
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In this particular case, you do not even have to do this.
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Basically, the left hand limit and the right hand limit are not the same, the function is discontinuous at the point.
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You did not really need to check the value of the function at the given point, at a.
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But in this case, we did, it is not a problem.
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It turns out to be left continuous.
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But overall, not continuous.
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Let us go ahead and do a theorem for composite functions.
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Again, this is mostly just a formality.
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Let us recall what we mean when we say composite function.
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If we are given f(x) and if we are given g(x), two functions of x.
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When we form f(g), that is equal to f of g(x).
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When you form the composite, that is what you are doing.
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If g(x) is continuous at point a and f(x) is continuous at g(a),
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then f(g) is continuous at a.
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Let us do an example.
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Let me go to red.
00:27:03.400 --> 00:27:07.000
I think this is example 4, if I’m not mistaken.
00:27:07.000 --> 00:27:22.900
The question here is, the cos of x³ + natlog of x is continuous at,
00:27:22.900 --> 00:27:27.200
I’m not going to write it all out, I’m just going to say cont.
00:27:27.200 --> 00:27:37.800
Is it continuous at x = π/2, that is our question.
00:27:37.800 --> 00:27:46.500
Using this theorem that we just did, here our f is equal to cos(x).
00:27:46.500 --> 00:27:53.400
Our g is equal to x³ + ln x.
00:27:53.400 --> 00:27:56.400
That is the composite function, it is going to be f(g).
00:27:56.400 --> 00:28:01.800
Cos of x³ + ln x.
00:28:01.800 --> 00:28:12.300
f(g) is equal to the cos of x³ + ln x.
00:28:12.300 --> 00:28:17.400
Let us see what we can do here.
00:28:17.400 --> 00:28:25.800
If g(x) is continuous at a and f is continuous at g(a), then f(g) is continuous at a.
00:28:25.800 --> 00:28:39.300
We ask ourselves, is g(x) continuous at π/2.
00:28:39.300 --> 00:29:13.000
The limit as x approaches π/2 of g(x) is equal to the limit as x approaches π/2 of x³ + ln of x = π/2³ + ln of π/2.
00:29:13.000 --> 00:29:15.100
That is the limit.
00:29:15.100 --> 00:29:19.000
Now let us check to see the actual value.
00:29:19.000 --> 00:29:30.800
g(π/2) = π/ 2³ + ln of π/2.
00:29:30.800 --> 00:29:32.300
This and this are equal.
00:29:32.300 --> 00:29:35.000
It is continuous.
00:29:35.000 --> 00:29:43.100
Yes, g(x) is continuous at our point of interest which is π/2.
00:29:43.100 --> 00:29:45.500
That takes care of the first one.
00:29:45.500 --> 00:29:51.500
The second part is, is f(x) continuous at g(a).
00:29:51.500 --> 00:30:11.600
Here is f(x) continuous at g(π/2).
00:30:11.600 --> 00:30:23.900
We said that g(π/2) is equal to this thing.
00:30:23.900 --> 00:30:28.400
Is f(x) continuous at g(π/2)?
00:30:28.400 --> 00:30:41.500
g(π/2) is equal to π/2³ + natlog of π/2.
00:30:41.500 --> 00:30:48.700
Therefore, what we want to check, for continuity we are checking to see that the limit = the value.
00:30:48.700 --> 00:30:55.600
We are going to calculate the limit as x approaches g.
00:30:55.600 --> 00:31:00.400
What we are saying is f(x) continuous at g(π/2).
00:31:00.400 --> 00:31:33.100
It is going to be x approaching g(π/2) of f(x) which = the limit as x approaches the value π/2³ + ln of π/2 of the cos(x).
00:31:33.100 --> 00:31:42.100
That is f, what we separated, that is why we wrote cos(x), instead of cos(x³) + ln x.
00:31:42.100 --> 00:31:45.700
We ask ourselves, is g(x) continuous? Yes.
00:31:45.700 --> 00:31:49.600
Now, is f(x) continuous at g(π/2)?
00:31:49.600 --> 00:31:51.100
We are going to do this.
00:31:51.100 --> 00:31:55.700
This limit =, we just put this here.
00:31:55.700 --> 00:32:05.300
It equals the cos of π/2³ + ln of π/2.
00:32:05.300 --> 00:32:07.400
This is a perfectly valid number.
00:32:07.400 --> 00:32:31.100
We also evaluate f(x) which is f of π/2³ + natlog of π/2
00:32:31.100 --> 00:32:41.900
= the cos of π/2³ + natlog of π/2.
00:32:41.900 --> 00:32:44.300
This and this are equal.
00:32:44.300 --> 00:32:49.100
Yes, that is true.
00:32:49.100 --> 00:33:08.000
Therefore, now we can conclude that fg which is equal to cos(x)³ + ln x is continuous at π/2.
00:33:08.000 --> 00:33:10.400
Those are long process.
00:33:10.400 --> 00:33:18.000
The basic idea is that a composite function is generally going to be continuous.
00:33:18.000 --> 00:33:31.500
If you have f of g(x), if g(x) is continuous and f is continuous at g(x), then your f(g) is going to be continuous.
00:33:31.500 --> 00:33:36.600
You can treat it individually like that or you can basically take a look at the whole function, f(g).
00:33:36.600 --> 00:33:42.000
You can form the function f(g) as a single function of x and check the continuity of that function.
00:33:42.000 --> 00:33:45.900
You can do it either way, it is not a problem.
00:33:45.900 --> 00:33:49.500
A lot of these things pretty straightforward, they are pretty intuitive.
00:33:49.500 --> 00:33:55.800
Sometimes you just break them down so much, that it tends to complicate them more than necessary.
00:33:55.800 --> 00:33:57.400
That is exactly what this problem was.
00:33:57.400 --> 00:34:09.100
It was just an excessive complication of something that is reasonably straightforward.
00:34:09.100 --> 00:34:24.100
Let us see, what value of a will make the following function continuous at every point in its domain?
00:34:24.100 --> 00:34:34.900
Here we have ax² + 7x, 4x less than 4, and x³ - ax for x greater than or equal to 4.
00:34:34.900 --> 00:34:39.100
4 is the dividing point.
00:34:39.100 --> 00:34:45.100
4 is our dividing point.
00:34:45.100 --> 00:34:54.100
We are basically going to be taking limits of the function as x approaches 4, from below and from above.
00:34:54.100 --> 00:35:05.600
In order for this to be continuous, for continuity,
00:35:05.600 --> 00:35:22.400
we need the limit as x approaches 4 from below of f(x) to equal the limit as x approaches 4 from above f(x).
00:35:22.400 --> 00:35:27.200
We need that equal to f(4).
00:35:27.200 --> 00:35:30.500
We need all three things to be equal.
00:35:30.500 --> 00:35:33.800
Let us calculate.
00:35:33.800 --> 00:35:40.100
The limit as x approaches 4 from below, that is this one.
00:35:40.100 --> 00:35:48.000
We are going to use this function of ax² + 7x =, we put x in for there.
00:35:48.000 --> 00:35:55.300
We get a × 4² + 7 × 4.
00:35:55.300 --> 00:36:01.600
You end up with 16a + 28.
00:36:01.600 --> 00:36:04.300
That is our left limit.
00:36:04.300 --> 00:36:07.300
Our right hand limit.
00:36:07.300 --> 00:36:10.000
What is the limit as x approaches 4 from above?
00:36:10.000 --> 00:36:14.200
From above, we are going to use that function.
00:36:14.200 --> 00:36:25.300
It is going to be x³ - ax = 4³ - a × 4.
00:36:25.300 --> 00:36:35.200
This one is going to give us 64 - 4a.
00:36:35.200 --> 00:36:42.200
f(4) less than or equal to, we are going to use that right there.
00:36:42.200 --> 00:36:54.800
That = x³ – ax, it is going to be 4³ – a × 4 = 64 – 4a.
00:36:54.800 --> 00:36:59.000
We see that this, that this is the same as this.
00:36:59.000 --> 00:37:01.700
It is right continuous.
00:37:01.700 --> 00:37:11.300
We want, again, we said that we want the left hand limit to equal the right hand limit to equal f(4).
00:37:11.300 --> 00:37:14.300
It already turns out that those two equal each other.
00:37:14.300 --> 00:37:16.700
We want this equal to that.
00:37:16.700 --> 00:37:26.000
We want this equal to these.
00:37:26.000 --> 00:37:45.800
We are going to set 16a + 28 is equal to 64 - 4a.
00:37:45.800 --> 00:37:55.100
Let us rewrite 16a + 28 = 64 - 4a.
00:37:55.100 --> 00:38:01.800
You are going to get 20a = 36.
00:38:01.800 --> 00:38:07.800
a is equal to 9/5.
00:38:07.800 --> 00:38:11.700
That is it, it is that simple.
00:38:11.700 --> 00:38:15.000
Whenever you need to set some piece wise continuous function,
00:38:15.000 --> 00:38:20.700
you need there to be continuity at the particular point where they are piece wise separate.
00:38:20.700 --> 00:38:24.000
You need to set the left hand limit and the right hand limit equal to each other,
00:38:24.000 --> 00:38:27.900
and it has to equal the value of the function there.
00:38:27.900 --> 00:38:55.500
Just use the definition of continuity which says that the limit as x approaches a of f(x) must equal f(a).
00:38:55.500 --> 00:39:18.900
This left hand limit must equal the right hand limit.
00:39:18.900 --> 00:39:24.400
A couple of nomenclature issues, just words that seem to come up.
00:39:24.400 --> 00:39:28.200
I will just say, by the way.
00:39:28.200 --> 00:39:31.600
Discontinuities are given different names.
00:39:31.600 --> 00:39:35.900
I do not why, but okay.
00:39:35.900 --> 00:39:37.100
That is just the way it is.
00:39:37.100 --> 00:39:39.800
What if we have a situation like this.
00:39:39.800 --> 00:39:43.000
We call this removable discontinuity.
00:39:43.000 --> 00:39:52.400
The reason I call it removable discontinuity is because basically at a point where it is discontinuous,
00:39:52.400 --> 00:39:56.300
whatever that is, let us say a, you are going to define it anyway you want.
00:39:56.300 --> 00:40:00.500
You can just say f(a) = either that point or another point.
00:40:00.500 --> 00:40:02.300
You can give any definition.
00:40:02.300 --> 00:40:06.800
It is a discontinuity that is removable, we can remove it.
00:40:06.800 --> 00:40:16.400
Here this type of discontinuity, we call this a jump discontinuity.
00:40:16.400 --> 00:40:21.500
Again, these names are completely ridiculous.
00:40:21.500 --> 00:40:23.400
They are completely meaningless.
00:40:23.400 --> 00:40:28.200
People use them but for all practical purposes, it is continuity and discontinuity.
00:40:28.200 --> 00:40:32.100
We do not need to know what type of continuity it is or discontinuity it is.
00:40:32.100 --> 00:40:40.000
Of course, there is this one, you have some sort of an asymptote and you have a function going this way and a function going this way.
00:40:40.000 --> 00:40:43.900
This is called an infinite discontinuity.
00:40:43.900 --> 00:40:53.200
Just to let you know that.
00:40:53.200 --> 00:40:58.900
You will see these terms, they do not mean anything.
00:40:58.900 --> 00:41:19.600
The most important and practical theorem regarding continuity,
00:41:19.600 --> 00:41:26.800
there are several of them but for our purposes,
00:41:26.800 --> 00:41:31.700
is something called the intermediate value theorem.
00:41:31.700 --> 00:41:42.800
The intermediate value theorem.
00:41:42.800 --> 00:41:48.500
Here are the hypotheses, the if section.
00:41:48.500 --> 00:41:53.300
If then, hypothesis, or hypotheses, conclusion.
00:41:53.300 --> 00:41:56.900
The hypotheses, there are few of them.
00:41:56.900 --> 00:42:10.100
Hypotheses are if a is less than b on the interval, if f(a) is different than f(b),
00:42:10.100 --> 00:42:23.300
if they are not equal, if f(x) is continuous on the closed interval ab.
00:42:23.300 --> 00:42:35.900
If all of these three are satisfied then the conclusion, the then statement, the conclusion is there exists,
00:42:35.900 --> 00:42:43.400
I will write it out, there exists, I will put it in parentheses.
00:42:43.400 --> 00:42:47.600
The symbol for there exists, the logical symbol is a reverse e.
00:42:47.600 --> 00:43:19.800
There exists a number c somewhere between a and b, such that f(c) is equal to some number a,
00:43:19.800 --> 00:43:27.000
or a is between f(a) and f(b).
00:43:27.000 --> 00:43:29.400
This is a formal statement of the theorem.
00:43:29.400 --> 00:43:36.300
We have the hypothesis which is your if clause and we have your conclusion which is your then clause.
00:43:36.300 --> 00:43:39.300
Let us go ahead do what this actually means.
00:43:39.300 --> 00:43:48.000
Pictorially, it looks like this.
00:43:48.000 --> 00:43:51.000
I have got something that looks.
00:43:51.000 --> 00:43:56.400
We have our axis.
00:43:56.400 --> 00:44:00.600
We have a nice continuous function.
00:44:00.600 --> 00:44:09.300
We have our point, let us say a is here that means f(a) is there.
00:44:09.300 --> 00:44:13.600
Let us say that b is over here.
00:44:13.600 --> 00:44:21.400
This is f(b) is over there.
00:44:21.400 --> 00:44:34.300
It says that this is some number c and this is the number a.
00:44:34.300 --> 00:44:42.100
Basically, what this is saying is that if a is less than b on the x axis, here we have a definitely less than b.
00:44:42.100 --> 00:44:48.700
If f(a) and f(b) are different, here f(a) and f(b) are definitely different.
00:44:48.700 --> 00:44:51.100
And if the function itself is continuous.
00:44:51.100 --> 00:44:55.900
This function f(x), yes, we can see that it is nice and continuous.
00:44:55.900 --> 00:45:01.000
If that is the case, then the conclusion I can draw is that there is some number, at least one, there may be more than one,
00:45:01.000 --> 00:45:14.200
that some number c between a and b such that the f of that number is actually going to fall between f(a) and f(b).
00:45:14.200 --> 00:45:17.400
That is what the intermediate value theorem says.
00:45:17.400 --> 00:45:22.500
Is that, if the three hypotheses are satisfied then there is some number between the two,
00:45:22.500 --> 00:45:29.400
such that when I take the f of that number, the y value is going to fall between f(a) and f(b).
00:45:29.400 --> 00:45:31.500
All three hypotheses have to be satisfied.
00:45:31.500 --> 00:45:36.700
a less than b, f(a) not equal to f(b), and it has to be a continuous function.
00:45:36.700 --> 00:45:38.800
That is the real important one.
00:45:38.800 --> 00:45:39.700
It has to be continuous.
00:45:39.700 --> 00:45:44.600
If it is not continuous then there is no guarantee that a value of a actually even exist at all.
00:45:44.600 --> 00:45:53.300
That is the whole idea because if we were to break the graph like that, if there is a discontinuity in the graph,
00:45:53.300 --> 00:45:57.800
you might have a number but it is not defined there.
00:45:57.800 --> 00:46:07.700
There is no guarantee that some number a in between f(a) and f(b) has some c value.
00:46:07.700 --> 00:46:14.300
Continuity of the graph is profoundly important.
00:46:14.300 --> 00:46:45.800
Let us write, since f is continuous as x goes from a to b, as x moves along from a towards b, f(x) hits every value.
00:46:45.800 --> 00:46:57.800
In other words, as we move from here to here, f actually hits every single value between here and here.
00:46:57.800 --> 00:46:59.900
It hits every single value.
00:46:59.900 --> 00:47:04.700
It might hit it once, we do not know what the graph looks like.
00:47:04.700 --> 00:47:07.400
But we know that it hit it at least once.
00:47:07.400 --> 00:47:28.700
Because f is continuous, as x goes from a to b, f(x) hits every value between f(a) and f(b).
00:47:28.700 --> 00:47:33.700
In another words, because the function itself is continuous, as we move from here to here,
00:47:33.700 --> 00:47:38.000
we move continuously from f of f(a) to f(b).
00:47:38.000 --> 00:47:41.600
There is always going to be some number between them.
00:47:41.600 --> 00:47:43.700
I hope that make sense.
00:47:43.700 --> 00:47:46.700
Again, I think intuitively, it absolutely does make sense.
00:47:46.700 --> 00:47:55.400
Let us do a problem, prove that the following function has at least one real root in the interval for 6.
00:47:55.400 --> 00:48:01.400
Here we have a function and they want us to prove it has at least one real root.
00:48:01.400 --> 00:48:07.400
What that means is that, you do not know what the function looks like.
00:48:07.400 --> 00:48:08.300
We can graph it, if we want to.
00:48:08.300 --> 00:48:10.400
But again, we do not have a graph at our disposal.
00:48:10.400 --> 00:48:13.200
We are going to use this theorem, analytical methods,
00:48:13.200 --> 00:48:22.800
to show that this function actually crosses the x axis at least once, between 4 and 6.
00:48:22.800 --> 00:48:31.200
It is either going to cross like this way or it is going to across this way, going from negative to positive or positive to negative.
00:48:31.200 --> 00:48:37.500
We do not know which one first but we want to show that at least it hits the x axis, at least one time.
00:48:37.500 --> 00:48:39.600
That is what this says, has at least one real root.
00:48:39.600 --> 00:48:43.600
That means it hits the real axis at least one time.
00:48:43.600 --> 00:48:48.100
The y value is going to be 0, that is what the root means, in the interval 4 to 6.
00:48:48.100 --> 00:48:52.000
Let us check our hypotheses.
00:48:52.000 --> 00:49:01.900
I think i will do this in blue actually.
00:49:01.900 --> 00:49:09.700
Our hypotheses, the first one is a less than b?
00:49:09.700 --> 00:49:16.600
Yes, 4 is definitely less than 6.
00:49:16.600 --> 00:49:38.900
Our second hypotheses, our second hypothesis is, is f(4) less than f(6)?
00:49:38.900 --> 00:49:41.000
We have to check that.
00:49:41.000 --> 00:49:50.000
f(4) = 1.643, when I put 4 into there, I get 1.643.
00:49:50.000 --> 00:49:56.000
Definitely, make sure that your calculators are in radian mode not degree mode.
00:49:56.000 --> 00:49:59.100
When you put in 4 in here, all the functions in calculus,
00:49:59.100 --> 00:50:05.100
when you use you calculator to solve trigonometric function, make sure you are in radian mode.
00:50:05.100 --> 00:50:07.900
You will get a number if you are in degree mode, but the numbers are going to be wrong.
00:50:07.900 --> 00:50:14.200
f(4) is that, f(6) = -1.497.
00:50:14.200 --> 00:50:17.300
Yes, this is true.
00:50:17.300 --> 00:50:21.200
f(4) is different than f(6).
00:50:21.200 --> 00:50:33.800
Hypothesis three, is f(x) continuous over the closed interval of 4 to 6.
00:50:33.800 --> 00:50:36.700
Yes, sin(2x) is a completely continuous function.
00:50:36.700 --> 00:50:39.500
Sin and cos are continuous everywhere.
00:50:39.500 --> 00:50:44.300
Cos is continuous and we know that the sum of the difference of a continuous function is continuous.
00:50:44.300 --> 00:50:48.200
Yes, that is continuous, that hypotheses are satisfied.
00:50:48.200 --> 00:51:28.700
Therefore, there exists at least one number between 4 and 6, such that f of this number, some number, let us call it c.
00:51:28.700 --> 00:51:32.600
There exists at least one number c between 4 and 6.
00:51:32.600 --> 00:51:59.300
There is some number here in that interval, such that f(c) is equal to 0, because 0 is between 1.643 and -1.497.
00:51:59.300 --> 00:52:03.500
f(4) is 1.643, let us just put it right there.
00:52:03.500 --> 00:52:08.600
f(6) is -0.497, let us just put it right over there.
00:52:08.600 --> 00:52:10.100
The function is continuous.
00:52:10.100 --> 00:52:15.500
Because it is continuous, when I'm going from here to here, I cannot lift my pencil.
00:52:15.500 --> 00:52:24.800
Therefore, somewhere, it does not matter what trajectory it takes, some path,
00:52:24.800 --> 00:52:29.300
whatever the function looks like, it has to pass through 0.
00:52:29.300 --> 00:52:30.500
That is what we are saying.
00:52:30.500 --> 00:52:33.900
That is what the intermediate value theorems says.
00:52:33.900 --> 00:52:42.000
If a is less than b, if the f values are different from each other and if it is continuous,
00:52:42.000 --> 00:52:54.600
that there is some number between the x values, such that the f of that number is between the f values of the two numbers.
00:52:54.600 --> 00:52:58.800
That is it, thank you so much for joining us here at www.educator.com.
00:52:58.800 --> 00:53:00.000
We will see you next time, bye.