WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, and welcome back to www.educator.com, and welcome back to AP Calculus.
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Today, we are going to continue our discussion of using the derivative to actually graph functions.
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Let us jump right on in.
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We have used the first derivative to talk about intervals of increase and decrease.
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We have also used it to discuss local maxes and mins.
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We have used the first derivative to find the intervals of increase and decrease,
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where the actual function itself is increasing and decreasing, as well as local maxes/mins.
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We know that if a function is from your perspective going from left to right, is decreasing then increasing,
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we know the function is a local minimum.
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If the derivative is increasing, it is positive, and then right after a certain point it becomes negative,
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we know that it hits a local min and local max.
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Now, let us go ahead and go a little further.
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Let us look at a couple of segments of a couple of graphs.
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Segments of the graph, these segments happen to be two increasing segments.
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Let us look at two increasing segments because we do not know what is happening to the left or right, or what it is that I’m drawing.
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Let us look at two increasing segments of two functions.
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One, we have our axis like that and we have something that looks like this.
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Over here, we have something that looks like this.
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In both cases, it is increasing, it is increasing.
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In other words, the derivative is going to be positive, the derivative is going to be positive.
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But clearly, there is a difference between these two.
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How do I differentiate, now is where we use the second derivative to differentiate between the two.
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In both cases, f’(x) is greater than 0, how do we differentiate between the two?
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One of them, let us call this one number 1 and let us call this one number 2.
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Number 1 is called concave up, in other words, the concavity is facing upward.
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Analogously, the other one is concave down.
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In other words, the inside of the curve, the inside of the concave portion is facing down.
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A nice definition, clearly, there is a tangent there, there is a tangent there, there is a tangent there.
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These are all increasing, they are positive tangents.
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Positive tangent, positive tangent, and positive tangent.
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A nice definition of concavity is the following.
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We fill it out here, not that we need it.
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I know we know what concave up and concave down means.
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These are pictorial motions, you know what concave is, it is the inside of the curve.
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The inside of the bowl, as opposed to the outside of the bowl which is called a convex portion of a bowl.
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A nice definition of concavity is as follows, just for the sake of having a definition.
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Concave up, I will just use cu, the graph of the function, the curve, the graph was above its tangents.
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Clearly, as you can see here, the tangents are all below the graph.
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The graph never falls below the tangents.
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Concave down is just the opposite.
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The graph lies below its tangents.
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In case you are not sure, in case it is really kind of subtle, draw a tangent and see whether the graph is actually below or above the tangent.
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And that will tell you whether it is concave up or concave down.
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Here we have a graph, let me go to red.
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That is your graph, it lies below.
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Let us go back to blue here.
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It lies below these tangent lines, it is concave down.
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Concavity can be very subtle.
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Clearly, something like this is really obvious.
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If you have something like this, it is obvious.
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Something like this, it is obviously concave up.
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You have something like this, ever so slight a curve, not so obvious.
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This is where we are going to use our analytical techniques.
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Now what does this mean in terms of the actual derivative?
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What does this mean in terms of the derivative?
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It means the following.
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If f”(x) is greater than 0 over a given interval, I will call it just a over a given interval a, then f is concave up over that interval.
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Remember, when we talk about intervals, we are talking about the x axis.
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Let us say from 1 to 2, every single x value in there, when I put in the second derivative,
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it happens to be positive then I know that in between 1 and 2, the graph is concave up over that interval.
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This gives us a numerical way of measuring concavity.
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In those cases where you have a slight very subtle curvature,
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where you cannot really decide or you cannot read it off the graph.
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Everything in math may have a geometric interpretation.
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We use geometry to help us along with the pictures.
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I mean, clearly, we are going to use all this to draw the graph of the functions.
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The graph is important but the algebra, the math, the analytics, that is was paramount.
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Number 2, if f”(x) is less than 0 over a given interval, then f is concave down over that interval.
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This is how you do it, you have a function, you take the first derivative that tells you whether the graph is increasing or decreasing.
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You take the second derivative and you check points to see where the second derivative is negative,
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where the second derivative is positive.
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Let us define something called a point of inflection.
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It is any point where f is, first of all continuous there and the graph changes from concave up to concave down.
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In other words, the second derivative goes from negative to positive or positive to negative.
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The second derivative passes through 0.
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Setting the second derivative equals 0 gives us an analytic way of actually finding these values of x
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where there is a change from concave up to concave down.
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In any point where f is continuous there, and the graph changes from concave up to concave down or concave down to concave up,
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either way, where it makes the transition.
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For a point of inflection, f” passes through 0, very important.
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In other words, f” at a point of inflection is going to equal 0.
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Let us see what is next here.
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Let us go ahead and draw a version of this.
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Here we have something like that and let us say we have a graph that looks like this.
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This right here, there is a point right here.
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That right there, from this point to this point over this interval, the graph is concave up.
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The second derivative is going to be positive.
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Over this interval, it is concave up which means that f” is greater than 0.
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From here onward, it is concave down.
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F”, if I take any x value in here, it is going to be less than 0.
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At this point, which is a point of inflection, it changes from concave up to concave down.
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In this particular case, this is a point of inflection.
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This point happens to be a point of inflection, the x value.
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If I want the y value, I put it into the original function to find the y value of where it actually flips.
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At this point, f”(x) is going to equal 0.
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This is a point of inflection, that is it.
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That is your geometric explanation of what a point of inflection is.
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Once again, over a certain interval, if the second derivative is greater than 0, the graph is concave up.
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It could be very clear concave up or it can be very subtle.
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If the second derivative over a certain interval is less than 0, it is concave down.
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At a point of inflection, it is going to equal 0, provided it is actually continuous there.
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Most of the functions that we are going to be dealing with are going to be continuous
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but there are going to be some instances where we have to be a little extra careful.
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We do not just take these rules and just apply them blindly.
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We always want to step back a little bit to use other resources at our disposal,
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to make sure we know exactly what is going on with the graph.
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That is really what mathematics is all about, is every class that you take teaches you a little bit more of something,
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a little bit more of something, gives you something else to put in your toolbox.
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You are going to apply some or all of these tools to a particular situation, depending on how complex or how easy the situation is.
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Let us do an example.
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Here says sketch the graph of a function that satisfies the following conditions.
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It is telling me that the f’(0) is equal to the f’(3), is equal to the f’(5), which is equal to 0.
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The first derivative of these points is equal to 0.
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That tells me that I’m hitting either a local max or a local min, or it is going to be not differentiable there.
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It tells me that the first derivative is greater than 0, when x is less than 0.
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Which means that when x is less than 0, the graph is actually rising, it is increasing.
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It is also rising between 3 and 5, which means between 0 and 3, it is actually falling.
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After 5, it is also falling.
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It tells me that f’(x) is less than 0, the first derivative is less than 0 between 0 and 3.
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It is decreasing from 0 and 3.
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It is decreasing when it is greater than 5.
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It tells me that the second derivative f” is positive between 2 and 4.
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It is negative to the left of 0 or when x is less than 2, or x is greater than 4.
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Let us see here, I think there might be a mistake in something here.
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But let us go ahead and see if we can work this out.
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Let us go ahead and draw it.
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I think I have might mis-worded something over here, but let us see.
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F’ at 0 and 3 and 5, I have 1, 2, 3, 4, and 5.
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It is a 5, 4, 3, 2, 1, and this is our origin 0.
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I’m going to hit a local max or local min at this point, this point, and this point.
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But I do not know whether it is going to be local max or local min yet, so I have to look at these.
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F’(x) is greater than 0 when x is less than 0.
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F’(0) is less than 0 between 0 and 3.
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The function is rising to the left of 0 and it is falling between 0 and 3.
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Therefore, I actually hit a local max at 0.
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It is rising over here and it is falling over here.
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F’(x) is greater than 0, the function is rising between 3 and 5.
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That means here, it actually hits a minimum.
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It is rising here, that means this comes this way and that is confirmed with this right here.
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The second derivative between 2 and 4, it is positive between 2 and 4, yes it is concave up.
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And then, it is less than 0 when x is bigger than 4, it is concave down.
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At 5, it looks like we are probably going to f’(x) is a greater than 0, when x is between 3 and 5.
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It is rising between 3 and 5, and is less than 0 when x is greater than 5.
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It hits another max over here and that is less than 0, for x greater than 5.
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It drops off like that.
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We have a local max, a local min, local max, and these corroborates.
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Let us make sure what we have here.
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F’(x) is greater than 0, when x is greater than 2 and less than 4.
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Greater than 2 and less than 4, yes, concave up.
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F’(x) is less than 0, when x is less than 2, yes, concave down.
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That is our concave down right here.
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This is concave down right here, or x is greater than 4.
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Yes, greater than 4, there was no mistake, it was exactly as written.
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Concave down, concave up.
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It certainly satisfies these conditions.
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You are just using the information given.
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Once again, when f’ = 0, it is a critical point.
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It may or may not be a local max or min.
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In this case, workout to be local maxes and mins.
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When f’, the first derivative is greater than 0, the function is increasing.
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When the first derivative is less than 0, the function is decreasing, the original function.
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When f” is greater than 0 over that interval, the graph is concave up.
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When f’ is less than 0, the graph is concave down, the original function.
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Let us do another example.
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Now we can bring all of our analytics to there.
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For the function x⁴ – 5x², discuss the intervals of increase and decrease,
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local maxes and mins, points of inflection, the intervals of concavity.
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And then, use this information to graph the function.
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There is no law that says you have to do anything at a given order.
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Generally, you are going to take the first derivative first, and then take the second derivative afterward.
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You do not have to wait until the end to actually draw your graph.
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You can draw it any time you want.
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If you have enough information and you feel comfortable enough
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to actually conclude the things that you can conclude from the mathematics given.
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You may use all the information, you may use only some of the information.
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You may use other information that you have, asymptotes, roots,
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whatever it is that you need, whatever things that you have in your toolbox.
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This first derivative and second derivative stuff, these are just two more tools in your toolbox.
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Let us start with the first derivative.
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Let us go back to blue here.
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F’(x) that is going to equal of 4x³ - 10x.
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We want to go ahead and set that equal to 0.
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I’m going to factor out the x.
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X × 4x² - 10 that is equal to 0.
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I get x is equal to 0 and I get 4x² - 10 = 0.
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That is going to be one of my possible points, one of my critical values.
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I’m going to have 4x² that is equal to 10, x² = 10/4 which is equal to 5/2.
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Therefore, x is equal to + or – √5/2 which is approximately equal to 1.58.
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I need to check my intervals of increase and decrease.
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Now that I find my critical values, 0, positive, this is positive or negative, +1.58 and -1.58.
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I have to draw my number line and I have to check points in those intervals to see where I have increase or decrease.
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Let us go ahead and do that.
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I have 0, I have +1.58 and I have -1.58.
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We are checking f’, I have to check points in this interval, this interval, this interval, this interval, to see,
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I’m going to put them back in the first derivative to see whether the first derivative is positive or negative, increasing or decreasing.
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F’(x), we said is equal to x × 4x² – 10.
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I’m going to try some points, I’m going to try a point here, here, here, and here.
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Over here, I’m going to try -2.
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When I put -2 into the first derivative, I do not really need to calculate the value.
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I just need to know whether it is positive or negative.
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-2 that means the x is negative.
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When I put -2 in here, I’m going to end up with a positive number.
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A positive × a negative is a negative.
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F’ is negative which means that the function is decreasing on that interval.
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I’m going to try the value here, I’m going to try -1.
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When I put -1 in for x into the first derivative, this becomes negative, this becomes negative, it is positive.
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The first derivative is positive in this interval which means the function is actually increasing.
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I'm going to try one over here, I get negative.
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1 is a positive number.
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If I put 1 in here, I get a negative number.
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I get a negative which means the function is decreasing on that interval.
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I have 2 which is going to be positive.
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If I put that there, 2 × 2 is 4, 4 - 10 is also positive.
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It is positive which means it is going to be increasing along that interval.
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The derivative = 0 at this point, this point, and this point.
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Those are our critical points.
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Our intervals of increase are this interval right here, let me go to red.
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This interval right here and this interval right here.
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Our intervals of increase are -1.58 to 0 union +1.58 to +infinity.
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Our intervals of decrease are going to be that and that.
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Our intervals of decrease are -infinity to -1.58 union 0 to 158, 0 to 1.58.
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We have also taken care of local maxes and mins here.
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We have decreasing/increasing, there is a local min right here.
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Increasing/decreasing, there is a local max, here, somewhere.
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There is a local min for x = 1.58.
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I have taken care of that as well.
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Let us go ahead and deal with those.
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We also know that -1.58 is a local min.
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We know that 0 is a local max and we know that +1.58 is a local min, as well.
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Let us go ahead and find the y values of these things, just for the sake of it.
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Let us find the y values of these critical points, these local maxes or mins.
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We just want to know where they are exactly, y values of these points.
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When I do f(-1.58), I get -6.25.
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I hope my arithmetic is correct.
00:26:07.600 --> 00:26:11.800
When I do f(0), that is going to equal 0.
00:26:11.800 --> 00:26:25.400
When I do f(1.58), I get a -6.25 again.
00:26:25.400 --> 00:26:43.900
We already have much of our graph.
00:26:43.900 --> 00:26:54.800
We do not yet know where it passes through the x axis yet.
00:26:54.800 --> 00:26:55.900
We do not know its roots yet.
00:26:55.900 --> 00:27:07.500
We can get those just by setting the original function equal to 0 and solving by with whatever means we have our disposal.
00:27:07.500 --> 00:27:15.000
In general for these graphing ones, whether you are doing them for calculus class or for the AP test itself,
00:27:15.000 --> 00:27:18.900
you may not necessarily need to know where it passes the x axis.
00:27:18.900 --> 00:27:24.800
What they want is a good solid graph showing increase/decrease things like that.
00:27:24.800 --> 00:27:27.800
But not necessarily if the function is too complicated and
00:27:27.800 --> 00:27:32.200
you do not have a calculator at your disposal or a computer algebra system at your disposal,
00:27:32.200 --> 00:27:36.500
it might be very hard to find a particular root.
00:27:36.500 --> 00:27:44.300
That is your secondary concern, where it passes through the axis.
00:27:44.300 --> 00:27:54.900
Let us go ahead and draw what we have so far.
00:27:54.900 --> 00:28:01.300
F(0) is 0.
00:28:01.300 --> 00:28:06.600
Let us do 1, 2, 3, 4, 5, 6, 7.
00:28:06.600 --> 00:28:14.200
Let us do 1, 2, 3, 1, 2, 3.
00:28:14.200 --> 00:28:19.000
At -1.58, at -6.25.
00:28:19.000 --> 00:28:23.400
1.58, -6.25, it is going to put us somewhere around there.
00:28:23.400 --> 00:28:26.500
It is also the same over here, somewhere around there.
00:28:26.500 --> 00:28:29.500
I know these are the local maxes and mins.
00:28:29.500 --> 00:28:35.100
It was a local max here, it is a local min here, it is a local min here.
00:28:35.100 --> 00:28:42.300
We have our graph, at least some of our graph.
00:28:42.300 --> 00:28:48.500
Of course we want to finish off with the other information which is points of inflection, intervals of concavity.
00:28:48.500 --> 00:28:53.800
We are looking for points where the concavity changes from positive to negative.
00:28:53.800 --> 00:28:56.300
We see here clearly this is concave up.
00:28:56.300 --> 00:29:02.100
But there are some points around here where the concavity goes from concave up,
00:29:02.100 --> 00:29:07.100
all of the sudden it is concave down, or from concave down to concave up.
00:29:07.100 --> 00:29:09.900
It is going to go up and pass through here.
00:29:09.900 --> 00:29:13.800
Let us go ahead and deal with those next.
00:29:13.800 --> 00:29:28.700
Intervals of concavity, let us go back to blue.
00:29:28.700 --> 00:29:41.500
Our points of inflection, we are going to take f” and we are going to set it equal to 0.
00:29:41.500 --> 00:29:44.800
Solve for x, that is going to give us our points of inflection.
00:29:44.800 --> 00:29:46.200
We are going to do the number line again.
00:29:46.200 --> 00:29:50.100
We are going to check points to the left and to the right of these points of inflection,
00:29:50.100 --> 00:29:59.300
to see where the second derivative is positive or negative, concave up, concave down, respectively.
00:29:59.300 --> 00:30:07.000
Let us rewrite, we have our original function f(x) is equal to x⁴ – 5x².
00:30:07.000 --> 00:30:15.400
We have f’(x) which is 4x³ - 10x, that makes our double prime.
00:30:15.400 --> 00:30:21.500
3 × 4 that is going to be 12x² – 10.
00:30:21.500 --> 00:30:26.800
We go ahead and we set that equal to 0.
00:30:26.800 --> 00:30:31.700
We get 12x² is equal to 10.
00:30:31.700 --> 00:30:37.800
We get x² is equal to 10/12 which equals 5/6.
00:30:37.800 --> 00:30:51.700
Therefore, x is equal to positive or negative √5/6 which is approximately equal to 0.913, + or -.
00:30:51.700 --> 00:30:54.500
Let us go ahead and do our little number line.
00:30:54.500 --> 00:31:01.700
We have +0.913 and we have -0.913.
00:31:01.700 --> 00:31:08.900
We are going to check points to the left in between and to the right, to see where it is concave up or down.
00:31:08.900 --> 00:31:15.000
We already know the answer, using the first derivative, you are able to get 90% of our graph.
00:31:15.000 --> 00:31:19.200
The only thing we do not really know is where it hits the x axis.
00:31:19.200 --> 00:31:22.200
We know where it is concave up and concave down.
00:31:22.200 --> 00:31:29.100
This procedure, this analytical procedure, gives us the actual numbers 0.913 to let us decide.
00:31:29.100 --> 00:31:38.300
We know that we are talking about something which is going to be concave up, concave down, concave up.
00:31:38.300 --> 00:31:40.200
Based on what we drew on the previous page.
00:31:40.200 --> 00:31:44.700
We just confirmed that here.
00:31:44.700 --> 00:31:46.400
This we are working with double prime.
00:31:46.400 --> 00:31:51.900
The hardest part of these problems is going to be, once you get the values of x and
00:31:51.900 --> 00:31:56.100
putting it back into the original function, the first derivative, the second derivative.
00:31:56.100 --> 00:32:00.500
You just have to be very careful which function you put them in.
00:32:00.500 --> 00:32:01.900
You are going to be dealing with 3 functions.
00:32:01.900 --> 00:32:04.100
The function of the first and second derivative.
00:32:04.100 --> 00:32:06.900
You have to make sure you know which one you are putting them in.
00:32:06.900 --> 00:32:14.100
Here, we are taking x values and we are putting them into the second derivative to see what sign the second derivative is.
00:32:14.100 --> 00:32:20.200
F”(x) is equal to 12x² – 10.
00:32:20.200 --> 00:32:25.000
I’m going to pick a point over on this side.
00:32:25.000 --> 00:32:27.500
I’m going to pick a point -1.
00:32:27.500 --> 00:32:34.500
When I put -1 into this, I'm going to get a positive.
00:32:34.500 --> 00:32:40.600
The second derivative, f” is positive here, this is concave up.
00:32:40.600 --> 00:32:44.500
When I check something in this region, I’m going to go ahead and check 0.
00:32:44.500 --> 00:32:48.600
I’m going to get -, 0 - 10 is negative.
00:32:48.600 --> 00:32:52.500
Negative means it is concave down.
00:32:52.500 --> 00:33:04.500
When I check +1 which is in this region over here, +1, I’m going to get positive which means it is concave up here.
00:33:04.500 --> 00:33:08.900
It went ahead and corroborated.
00:33:08.900 --> 00:33:23.300
Let us go ahead and re-graph our graph.
00:33:23.300 --> 00:33:36.700
We have this and we have something that looked like this, something like that.
00:33:36.700 --> 00:33:42.800
This point was 1.58, that is where it hit the local min.
00:33:42.800 --> 00:33:48.300
This was -1.58, that hit a local min.
00:33:48.300 --> 00:33:52.800
Let me draw this a little bit better.
00:33:52.800 --> 00:33:57.700
It hit the local min, the points of inflection are right here.
00:33:57.700 --> 00:34:02.500
These x values, that +0.913 and -0.913.
00:34:02.500 --> 00:34:07.500
That is a point of inflection, that is the point of inflection,
00:34:07.500 --> 00:34:18.300
it is where it changes from concave up, concave down, back to concave up, over that number line that we did in the previous page.
00:34:18.300 --> 00:34:19.900
Let us go ahead and find the y values.
00:34:19.900 --> 00:34:23.700
In other words, we found the x 0.913, -0.913.
00:34:23.700 --> 00:34:29.500
Let us go ahead and find the y values for these points.
00:34:29.500 --> 00:34:52.200
Let us find y values for the points of inflection.
00:34:52.200 --> 00:35:00.800
Remember, very important, if you are looking for y values, these x values of the points of inflection,
00:35:00.800 --> 00:35:14.700
they go back to the original function.
00:35:14.700 --> 00:35:18.500
Do not worry, we all make mistake like we all did.
00:35:18.500 --> 00:35:33.500
That is the process, the original function, in other words f(x).
00:35:33.500 --> 00:35:56.800
We have f(-0.913) = -3.4 and f(0.913) = -3.4.
00:35:56.800 --> 00:36:34.100
Our points of inflection are -0.913, -3.4, and 0.913, -3.4.
00:36:34.100 --> 00:36:40.500
Let us go ahead and write down the intervals of concavity, up and down.
00:36:40.500 --> 00:36:43.300
Sorry about that, I do not think I actually wrote them down formally.
00:36:43.300 --> 00:37:04.400
Intervals of concavity, it is going to be concave up from -infinity, we said to -0.913 union 0.913 to +infinity.
00:37:04.400 --> 00:37:13.300
It is concave down from -0.913 all the way to +0.913.
00:37:13.300 --> 00:37:16.900
Just to write them down formally.
00:37:16.900 --> 00:37:26.600
That takes care of that, the last thing we should do is actually find out where it crosses the x axis, just for good measure.
00:37:26.600 --> 00:37:46.000
The last thing we should do is find where f(x), the original function, crosses the x axis.
00:37:46.000 --> 00:37:53.200
In other words, the roots, the x axis.
00:37:53.200 --> 00:37:56.500
There is no guarantee that the function is actually going to cross the x axis.
00:37:56.500 --> 00:38:03.800
It could not cross at all, in which case you are going to set f(x) equal to 0 and you are going to get no solution.
00:38:03.800 --> 00:38:13.200
F(x), the original function is equal to x⁴ - 5x², we set that equal to 0.
00:38:13.200 --> 00:38:22.400
I’m going to go ahead and factor out the x² and I'm going to get x² – 5 and that is equal to 0.
00:38:22.400 --> 00:38:28.200
I get x² = 0 and I get x² - 5 = 0.
00:38:28.200 --> 00:38:32.100
Therefore, it crosses x = 0.
00:38:32.100 --> 00:38:35.500
It is a double root, it actually touches, we know that already from the graph.
00:38:35.500 --> 00:38:40.500
It touches the graph there, it does not cross.
00:38:40.500 --> 00:38:53.500
X² = 5, therefore, x = + or -√5 which is equal to + or -, roughly, approximately equal to 2.24.
00:38:53.500 --> 00:39:00.400
There you go, final graph.
00:39:00.400 --> 00:39:06.000
I will go ahead and put a point here, put a point there.
00:39:06.000 --> 00:39:18.800
They cross there, cross there.
00:39:18.800 --> 00:39:32.400
This is 2.240, this point is -2.240.
00:39:32.400 --> 00:39:38.800
This point is a point of inflection, this point is of course 0,0.
00:39:38.800 --> 00:39:46.800
This point is 0.913, -3.4.
00:39:46.800 --> 00:39:52.900
This point is -0.913, -3.4.
00:39:52.900 --> 00:39:59.800
This point, the local min, that is -1.58 and -6.25.
00:39:59.800 --> 00:40:02.300
If I’m not mistaken, if I remember properly.
00:40:02.300 --> 00:40:08.200
This one is 1.58 and -6.25.
00:40:08.200 --> 00:40:15.900
Let us see what a better version of the graph looks like.
00:40:15.900 --> 00:40:20.900
Local max, local min, local min.
00:40:20.900 --> 00:40:25.900
Where it crosses the x axis, where it crosses the x axis, those are two roots.
00:40:25.900 --> 00:40:31.500
Points of inflection are here and here.
00:40:31.500 --> 00:40:43.100
Concave up until this point, this interval all the way to -0.913, it is concave up then it is concave down until it hits +0.913.
00:40:43.100 --> 00:40:49.000
When it flips over and becomes concave up again, all the way towards infinity.
00:40:49.000 --> 00:40:54.800
They are you have it.
00:40:54.800 --> 00:41:19.800
The first derivative will tell us whether we have a local max or min.
00:41:19.800 --> 00:41:21.200
It does not tell us increasing/decreasing.
00:41:21.200 --> 00:41:27.100
It also tells us when it passes from increasing to decreasing or decreasing to increasing, we are going to hit a local max or min.
00:41:27.100 --> 00:41:32.900
The first derivative is all we need for local max or min.
00:41:32.900 --> 00:41:36.500
There is also a second derivative test, if we want it.
00:41:36.500 --> 00:41:57.100
There is also a second derivative test for local maxes and mins.
00:41:57.100 --> 00:42:04.000
It is generally unnecessary but we will go ahead and lay it out.
00:42:04.000 --> 00:42:20.100
Let f”(x) be continuous near a point a.
00:42:20.100 --> 00:42:23.400
To the left and right of it, it is going to be continuous.
00:42:23.400 --> 00:42:45.300
If f’(a) is equal to 0 and f” at a is greater than 0, f’(a) is equal to 0.
00:42:45.300 --> 00:42:47.000
It is either a local max or min.
00:42:47.000 --> 00:42:57.300
F” if a is greater than 0, which means it is concave up, it is a local min.
00:42:57.300 --> 00:43:08.100
Then f has a local min at a, and the opposite.
00:43:08.100 --> 00:43:24.700
If f’(a) = 0, f” at a is less than 0, which means concave down.
00:43:24.700 --> 00:43:37.800
Then, f has a local max at the point a.
00:43:37.800 --> 00:43:45.800
What you are looking at is this.
00:43:45.800 --> 00:43:49.400
You know this already, that point, that point.
00:43:49.400 --> 00:44:05.000
Here f’ is equal to 0, f” greater than 0, concave up, local min.
00:44:05.000 --> 00:44:12.500
Here you have got f’ is equal to 0, f” less than 0.
00:44:12.500 --> 00:44:16.900
This is concave down, this is your local max.
00:44:16.900 --> 00:44:19.200
That is it, second derivative test, in case you want to use it.
00:44:19.200 --> 00:44:29.300
You really do not need it but it appears in the calculus books, we are going to go ahead and give it to you.
00:44:29.300 --> 00:44:32.000
I’m going to go ahead and stop the lesson here.
00:44:32.000 --> 00:44:37.000
Do not worry that there is only been a couple of examples in this lesson and in the previous lesson.
00:44:37.000 --> 00:44:40.400
This was mostly just theory, something to get your feet wet.
00:44:40.400 --> 00:44:47.600
The next couple of lessons are going to be nothing but example problems, using what we just did in these last two lessons.
00:44:47.600 --> 00:44:52.000
Again, the next set of lessons is going to be just example problems.
00:44:52.000 --> 00:44:54.800
We are going to do plenty of these, no worries.
00:44:54.800 --> 00:44:57.000
Thank you so much for joining us here at www.educator.com.
00:44:57.000 --> 00:44:58.000
We will see you next time, bye.