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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In the last couple of lessons, we discussed the first derivative and the second derivative.
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And how they are used in order to decide what graph of a function looks like.
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In today's lesson and in the next lesson, we are going to do example problems using these techniques,
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getting progressively more complex as far as the functions that we are dealing with.
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Let us jump right on in.
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The following is the graph of the first derivative of some function,
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on which the original function increasing or decreasing and are there any local maxes or mins?
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With these particular problems, you have to be very careful to remember which function you are dealing with,
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which graph they actually gave you.
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In this case, they gave you the first derivative.
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This is not the function itself, this is not the second derivative, it is the first derivative.
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This is f’(x), we have to remember that.
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In the process of deciding whether it is increasing/decreasing, concave up, concave down,
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what is a local max and what is a local min.
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It is going to start to get confusing because you are going to try to describe the behavior of the function,
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but what you are given is the first derivative of the function.
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We know that increasing is where the first derivative is greater than 0.
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Decreasing is where the first derivative is less than 0.
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This is the first derivative.
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At this point, between this point and this point, the original function is increasing
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because the first derivative is positive, it is above the x axis.
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To the left of this point which is -1, the first derivative which is this graph is negative, it is below the x axis
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Pass the point 4, it is also negative.
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The function itself is decreasing on this interval, increasing on this interval, decreasing on this interval.
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Let me go ahead and work in blue here.
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I’m just going to write right on top of the graph.
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Our intervals of increase, it is when f’ is greater than 0.
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Our interval of increase is from, this is -1.
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I think I read the graph wrong.
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It looks like this is -1 and this looks like this 4, from -1 to 4.
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The interval of decrease is when the first derivative is actually less than 0.
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In this case, decrease happens on the interval from -infinity to -1 union 4, all the way to infinity.
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These points, where the first derivative actually equal 0
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because this is the first derivative of the graph, this one.
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This -1 and this 4, those are the local extrema.
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F’(x) = 0 at -1 and 4.
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These are the local extrema, in other words the local mins and local max.
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Let us see which one is which.
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At -1, we have decreasing, increasing.
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You have a local min at x = -1.
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Over here, it is increasing/decreasing.
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You have a local max at x = 4.
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That is it, that is all that is happening here.
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Just be very careful which graph they are giving you to read it.
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In the process, you are going to get confused.
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That is not a problem, that is the whole process.
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We have all gone through the process, we have all gotten confused.
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We have all make some minor mistakes and some gross mistakes, regarding this.
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This is the process, just be extra careful.
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You have to be very vigilant here.
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Let us do another example, same type.
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The following is a graph of the first derivative.
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This is f’ of a function, on which intervals is the function, the original function increasing/decreasing.
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Are there any local maxes or mins, you also want more information,
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on which interval is the function concave up and concave down?
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What are the x values of the points of inflection?
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A lot of information that they require here.
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Let us go through increasing/decreasing like we did before.
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Once again, increasing is where f’ is greater than 0, decreasing is where f’ is less than 0.
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This is the f’ graph, therefore here, between here and here, and from here onward, f’ is positive.
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Therefore, the function is increasing.
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Therefore, our increasing interval, I will write it over here, is going to be from, this looks roughly like -1.8 all the way to about here.
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We can read off as, let us just call it 0.9.
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This should be an open.
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This is roughly 2.2, all the way to +infinity.
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This is where the function, the original function is increasing.
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It is increasing where the first derivative is positive, positive above the x axis.
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Therefore, our interval of decrease, it is negative from -infinity all the way to this -1.8.
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It is also below between this 0.9 and this 2.2.
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These are our intervals of increase and decrease.
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What we have here is decreasing on this interval, increasing on this interval, decreasing on this interval, increasing on that interval.
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Therefore, that makes this point right here.
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I will say the local maxes are going to be this one which is at x = 0.9.
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Our local min, decreasing/increasing, decreasing/increasing.
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We have -1.8 and our 2.2.
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Remember, what we are talking about here is the actual original function.
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What we had here is our first derivative graph.
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This does not describe the graph.
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We are using the graph to describe the original function and we are doing it analytically, things that we know.
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Increasing/decreasing, positive, negative.
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Let us move on to our next page here.
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We have taken care of the local maxes and mins.
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We have taken care of the increasing/decreasing.
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Let us find some points of inflection and some intervals of concavity here.
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Once again, let us remind ourselves actually that this is f’ not f”.
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We want to talk about some inflection points.
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Inflection points, we said that inflection points are points where f” is equal to 0.
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F” is equal to 0, since this graph that we are looking at, since this graph is f’, f” is the slope of this graph.
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F” is the slope of this graph.
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I hope that make sense.
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We want the slope of this graph, where is it the slope of this graph equal 0?
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It equal 0 there, there, there, and there.
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What are these x values, it is going to be some place like right there, right there, right there, and right there.
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Therefore, at x equals -1.25, -0.6, 0, it looks like about 1.6.
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At these points, the second derivative equal 0.
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The second derivative is the slope of this graph which is the graph of the first derivative.
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These are points of inflection.
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We take those points of inflection, we put them on a number line.
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We evaluate whether the second derivative is positive or negative.
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To the left or right of those points.
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I have got -1.25, -0.6, 0, and 1.6.
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I need to check where the second derivative.
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We are dealing with the second derivative here.
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I need to check this region, this region, this interval, this interval, and this interval,
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to see whether it is concave up or concave down.
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I look to the left of -1.25, the slope here is positive, it is concave up.
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From -1.25 to -0.6, slope is negative, it is concave down.
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Not the graph, what we are talking about here is the original function.
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This is where the confusion lies in.
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This is not concave up but because this is the f’ graph, f” of the original function is the slope of this graph.
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The slope is positive, therefore, the original function is concave up.
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Let us write concave up, concave down.
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From 0.6 to 0, the slope is positive, this is concave up.
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From 0 to about 1.6, the slope is negative, we are looking at concave down.
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From here onward, the slope of this is positive.
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F’ is positive, this means it is concave up.
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Therefore, our intervals of concavity, concave up are, -infinity to -1.25 union -0.6 to 0 union 1.6 to +infinity.
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We are going to be concave down from -1.25 all the way to -0.6.
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And union where it is negative, it is going to be from 0 all the way to +1.6.
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I hope it makes sense what it is that we have down here.
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There you go, you got your points of inflection, you have your intervals of concavity
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We found our local maxes and mins, and we found out the intervals of increase and decrease of the actual function.
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It looks like we have everything.
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Let us try another one.
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The following is a graph of the first derivative.
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Let us remind ourselves, we are looking at f’ of a function.
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On which intervals of the function increasing/decreasing?
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Are there local maxes and mins, on which intervals of the function concave up and concave down?
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What are the x values of the points of inflection?
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The exact same thing as what we just did, we are going to do it again.
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Increasing/decreasing is where the first derivative is positive/negative respectively.
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Right about there, it looks like about 2.8, 2.7, 2.8, something like that.
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To the right of that, that is where the first root is positive.
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To the left of that, it is all below the x axis.
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F’ is negative, it is decreasing.
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Do not let this fool you, just read right off the graph.
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Trust the math, do not trust your instinct.
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Your instinct is going to want you to see this as the function.
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This is not the function, this is the derivative of the function.
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Our intervals of increase is 2.8 all the way to +infinity, 2.8 onward, that way.
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Our intervals of decrease, we have -infinity to 2.8.
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-infinity to 2.8.
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This is going to be, it is negative so it is decreasing.
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Here it is increasing, we are going to have a local min at 2.8.
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That is it, local min at x = 2.8.
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There are no local maxes because there is no place where it goes from increasing to decreasing.
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In this case, there is no local max.
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Let us go ahead and talk about some inflection points.
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I think I can go ahead and do this one in red.
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Inflection points, we said inflection points are points where the second derivative is equal to 0.
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Inflection points, there are places where f” is actually equal to 0.
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F” is f’ of f’, it is going to be the derivative of this graph.
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It is going to be the slope of that graph.
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F” which is the slope of this graph is equal to 0 at 2 points.
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That right there and about right there.
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Points of inflection are going to be x = -0.9 and x = roughly 1.5.
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Once again, we have our -0.9, we have our 1.5.
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Let us go ahead and do our f’ check.
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We have -0.9 and we have our 1.5.
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I need to check this region, this region, and this region.
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To the left of 0.9, the slope of this graph is positive that means the f” is positive.
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Therefore, the slope is concave up.
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From -0.9 to 1.5, from here to here, the slope is negative, this means it is concave down.
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The slope, remember this is f’.
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The slope is f” of the original function, concave down.
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From this point onward, we have a positive slope, positive slope, it is concave up.
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Therefore, our intervals of concavity are concave up from -infinity to -0.9 union at 1.5, all the way to +infinity.
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We have concave down from -0.9 all the way to 1.5.
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Again, be very careful, when you know you are dealing with f’.
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Now I’m going to show you the image of all three graphs right on top of each other, to see what is going on.
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This was the graph that we were given, this is f’.
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This is the graph of f”, in other words, it is the slope of this purple.
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This is actually f(x), this is the one whose behavior we listed and elucidated.
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Let us double check.
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Let us go back to blue here.
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We said the following, we said our interval of increase is 2.8 to +infinity.
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We look at our original function, it is this one the blue.
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Yes, decrease, decrease.
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Yes, from 2.8 onward.
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That checks out, very nice.
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We said that it decreased from -infinity all the way to this 2.8.
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Sure enough, the function from -infinity as we move from left to right,
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the function is decreasing, decreasing, decreasing, decreasing, until it hits 2.8.
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Yes, that checks out.
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We said that we had a local min at 2.8.
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Yes, there is our local min at 2.8, right there, that checks.
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We said we had inflection points, let us see what we have got.
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Inflection points, we said that we have an inflection point at x = -0.9.
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We also said we had one at x = 1.5.
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Let us go to -0.9, roughly right about there.
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Yes, there it is, our blue.
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There is our inflection point, it changes from concave up to concave down.
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And then, roughly around 1.5, right about there.
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There we go at 1.5, at this point.
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It goes from concave down to concave up.
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Yes, these two check out.
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And then, we had our intervals of concavity.
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We said that we have an interval of concavity from -infinity to -0.9 union at 1.5, all the way to +infinity.
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Let us double check.
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Concave up, concave up from -infinity to -0.9.
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From 1.5 all the way to +infinity, 1.5 to +infinity.
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Yes, that checks out.
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The last thing we want to double check is our concave down, from negative 0.9 all the way to 1.5.
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Yes, from -0.9 all the way to about 1.5.
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Yes, the graph is concave down, there we go.
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Here, we see them all together.
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The graph they gave us was this one.
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This is the graph they gave us.
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From that, I was able to elucidate all of this information that corroborated the actual function, which is this.
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Be very careful with this, that is the take home lesson, just be vigilant.
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As in all things with, when it comes down to higher math and higher science.
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You just have to be extra vigilant, there is a lot happening.
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You have to keep track of every little thing.
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Let us go ahead and do some analytical work here.
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For the function 10x²/ x² + 5, find the intervals of increase and decrease,
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the local maxes and mins, the points of inflection, and the intervals of concavity.
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Use this information to actually draw the graph.
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Let us go ahead and do it.
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I’m going to go back to blue here.
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I have got f’(x) is equal to, I have this × the derivative of that - that × the derivative of this/ that².
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I got x² + 5 × 20x - 10x².
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I really hope to God that I did my arithmetic correctly.
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All over x² + 5, I’m going to rely on you to double check that for me.
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All of that is equal to, when I multiply it out, I get 20x³ + 100x - 20x³/ x² + 5².
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Those cancel, I'm left with 100x/ x² + 5².
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This is my first derivative, I’m going to set that first derivative equal to 0.
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When I set it equal to 0, the denominator was irrelevant.
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It is only 0 when the numerator is 0.
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100x = 0, which means that x = 0, that is my critical value.
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I need to check values to the left of 0, values to the right of 0, to see whether I have a local max or min, and increasing/decreasing.
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Let us see here, I'm going to go ahead and check.
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Let us just go ahead and check -1.
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When we put -1 into the first derivative, you are going to end up with a negative number on top.
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It is going to be positive, this is going to be negative.
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This is increasing.
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If I check the number 1, which is to the right of 0, you are going to end up with a positive number on top.
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Positive, positive, this is going to be increasing.
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Therefore, we have our increasing interval from 0 to +infinity.
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We have our decreasing interval from -infinity to 0.
00:26:04.900 --> 00:26:12.400
That takes care of our increasing/decreasing of the actual function.
00:26:12.400 --> 00:26:15.700
We also know that this is decreasing, this is increasing.
00:26:15.700 --> 00:26:23.800
We know that there is a local min at x = 0.
00:26:23.800 --> 00:26:27.900
This critical value is a local min.
00:26:27.900 --> 00:26:30.400
Let us go ahead and find f”(x).
00:26:30.400 --> 00:26:33.800
Let us go back to blue.
00:26:33.800 --> 00:26:43.500
F”(x), we are looking for inflection points, in order to check intervals of concavity.
00:26:43.500 --> 00:26:44.600
We are left with this thing.
00:26:44.600 --> 00:26:49.300
It is going to be this × the derivative of that - that × the derivative of this/ this².
00:26:49.300 --> 00:27:13.200
We get x² + 5² × 100 - 100x × 2 × x² + 5 × 2x/ x² + 5⁴.
00:27:13.200 --> 00:27:29.300
I’m going to factor out an x² + 5 here, which leaves me with x² + 5 × 100 - 400x²/,
00:27:29.300 --> 00:27:36.200
I multiply this, this, this, to get 400²/,
00:27:36.200 --> 00:27:39.000
I’m sorry, I multiply this, this, and this.
00:27:39.000 --> 00:27:41.200
I have factored this out as here.
00:27:41.200 --> 00:27:50.100
I get x² + 5⁴, this cancels with one of these, leaving 3.
00:27:50.100 --> 00:27:57.100
I need that equal to 0.
00:27:57.100 --> 00:28:19.000
We have f”(x) is equal to 100x² + 500 – 400x²/ x² + 5³.
00:28:19.000 --> 00:28:31.700
Set that equal to 0 and I’m going to get -300x² + 500 = 0 because it is only the numerator that matters.
00:28:31.700 --> 00:28:36.700
I have got 300x² is equal to 500.
00:28:36.700 --> 00:28:51.500
I get x² is equal to 5/3 which gives me the x is equal to + or - 5/3, which is approximately equal to + or -1.3.
00:28:51.500 --> 00:28:59.200
I’m going to set up my -1.3, +1.3.
00:28:59.200 --> 00:29:07.600
I’m doing f”, I need to check points there, points there, and points there.
00:29:07.600 --> 00:29:09.800
Let us go ahead and actually do this one.
00:29:09.800 --> 00:29:26.100
F”(x)is equal to -300x² + 500/ x² + 5³.
00:29:26.100 --> 00:29:28.900
I’m going to check the point -2.
00:29:28.900 --> 00:29:39.200
When I check the point -2, I'm going to get a negative number/ a positive number
00:29:39.200 --> 00:29:43.600
which is a negative number, which is concave down.
00:29:43.600 --> 00:29:49.500
When I check 0, I’m going to end up with a positive number on top of a positive number.
00:29:49.500 --> 00:29:56.100
When I put 0 into the second derivative, which means positive, which means concave up.
00:29:56.100 --> 00:30:10.900
When I check 2, I'm going to get a negative/ a positive number that is going to be negative, this is going to be concave down.
00:30:10.900 --> 00:30:13.700
Let us move on to the next page here.
00:30:13.700 --> 00:30:26.100
I have got concave down from -infinity to -1.3 union at 1.3 to +infinity.
00:30:26.100 --> 00:30:35.600
I have got concave up from -1.3 all the way to 1.3.
00:30:35.600 --> 00:30:39.500
Let us see what happens.
00:30:39.500 --> 00:30:41.200
We are dealing with a rational function.
00:30:41.200 --> 00:30:43.400
We have to check the asymptotic behavior.
00:30:43.400 --> 00:30:50.000
We have to check to see what happens when x gets really big, in a positive or negative direction.
00:30:50.000 --> 00:31:05.800
Let us see what happens when x goes to + or -infinity.
00:31:05.800 --> 00:31:14.400
F(x) is equal to 10x²/ x² + 5.
00:31:14.400 --> 00:31:17.800
As x goes to infinity, this drops out.
00:31:17.800 --> 00:31:23.900
The x² cancel and you are left with f(x) ends up approaching 10.
00:31:23.900 --> 00:31:30.500
10 is the horizontal asymptote of this function.
00:31:30.500 --> 00:31:38.600
What do we have, let us go back and put it all together.
00:31:38.600 --> 00:31:45.000
We have a local min at 0.
00:31:45.000 --> 00:31:47.700
When we check f (0), it equal 0.
00:31:47.700 --> 00:31:52.200
We are looking at the point 0,0.
00:31:52.200 --> 00:31:58.300
We know that the function is increasing from 0 to +infinity.
00:31:58.300 --> 00:32:06.000
We know the function is actually decreasing from -infinity to 0.
00:32:06.000 --> 00:32:20.700
We know that we have points of inflection at x = -1.3.
00:32:20.700 --> 00:32:24.600
When I check the y value, I get 2.53.
00:32:24.600 --> 00:32:34.900
Therefore, at -1.3 and 2.53 is my actual point of inflection, both xy coordinate.
00:32:34.900 --> 00:32:42.700
My other one is at +1.3 and the y value is 2.53.
00:32:42.700 --> 00:32:49.600
1.3 and 2.53 is my other point of inflection.
00:32:49.600 --> 00:33:01.000
I am concave down from -infinity to -1.3 union 1.3 to +infinity.
00:33:01.000 --> 00:33:10.200
I am concave up from -1.3 all the way to +1.3.
00:33:10.200 --> 00:33:27.400
I have a horizontal asymptote at y is equal to 10.
00:33:27.400 --> 00:33:37.100
When I put all of this together, what I end up getting is the following graph.
00:33:37.100 --> 00:33:44.000
Here is my local min, here is one of my points of inflection.
00:33:44.000 --> 00:33:48.400
Here is my other point of inflection.
00:33:48.400 --> 00:33:55.100
It is decreasing all the way to 0, increasing past 0.
00:33:55.100 --> 00:34:04.300
It is concave down all the way to -1.3, concave down up to this point.
00:34:04.300 --> 00:34:07.300
It is concave up from this point to this point.
00:34:07.300 --> 00:34:14.800
It is concave down again, notice that it approaches 10.
00:34:14.800 --> 00:34:23.700
That is it, first derivative, second derivative, local maxes and mins, points of inflection, intervals of concavity.
00:34:23.700 --> 00:34:33.100
I need horizontal asymptotes, vertical asymptotes, everything that I need in order to graph this function.
00:34:33.100 --> 00:34:37.600
Let us go ahead and try another example here.
00:34:37.600 --> 00:34:41.600
For the function x² ln 1/3 x, find the intervals of increase/decrease,
00:34:41.600 --> 00:34:44.900
local maxes and mins, points of inflection, intervals of concavity.
00:34:44.900 --> 00:34:48.000
Use this information to draw the graph.
00:34:48.000 --> 00:34:51.600
Let us go ahead and do it.
00:34:51.600 --> 00:34:57.600
F’(x) is equal to x, this is a product function.
00:34:57.600 --> 00:35:00.800
This × the derivative of that + that × the derivative of this.
00:35:00.800 --> 00:35:16.500
X² × 1/ 1/3 x × 1/3 + ln of 1/3 x × 2x.
00:35:16.500 --> 00:35:28.700
1/3, 1/3, x, x, what I should end up with is x + 2x ln 1/3 (x).
00:35:28.700 --> 00:35:31.500
We want to set that equal to 0.
00:35:31.500 --> 00:35:34.200
I'm going to go ahead and factor out an x.
00:35:34.200 --> 00:35:53.400
I get x × 1 + 2 ln of 1/3 x that is equal to 0, that gives me x = 0.
00:35:53.400 --> 00:36:00.300
It gives me 1 + 2 ln of 1/3 x is equal to 0.
00:36:00.300 --> 00:36:05.300
0 is one of my critical points.
00:36:05.300 --> 00:36:08.700
When I solve this one, I will go ahead and solve it up here.
00:36:08.700 --> 00:36:17.300
I get 2 × ln of 1/3 x is equal to -1.
00:36:17.300 --> 00:36:21.400
Ln of 1/3 x is equal to -1/2.
00:36:21.400 --> 00:36:30.000
I exponentiate both sides, I get 1/3 x = e ^-½.
00:36:30.000 --> 00:36:34.500
I get x is equal to 3 × e⁻¹/2.
00:36:34.500 --> 00:36:42.000
X is approximately equal to 1.8.
00:36:42.000 --> 00:36:43.900
Let us go ahead and do it.
00:36:43.900 --> 00:36:46.400
These give me my critical points.
00:36:46.400 --> 00:36:49.700
I have 0 and I have got 1.8.
00:36:49.700 --> 00:36:53.600
I need to check a point here, check a point here, check a point here.
00:36:53.600 --> 00:37:03.600
Put them into my first derivative to see whether I get a positive or negative value.
00:37:03.600 --> 00:37:05.600
Let me go ahead and write it out.
00:37:05.600 --> 00:37:23.000
F’(x), I’m going to write out the multiplied form, = x × 1 + 2 ln of 1/3 x, that is my f’(x).
00:37:23.000 --> 00:37:25.800
I’m going to pick a point again here, here, here.
00:37:25.800 --> 00:37:27.500
Put it into this to see what I get.
00:37:27.500 --> 00:37:36.900
When I check the point, this here is not the domain.
00:37:36.900 --> 00:37:38.000
I do not have to check a point there.
00:37:38.000 --> 00:37:41.900
The reason is I cannot take the log of a negative number.
00:37:41.900 --> 00:37:44.700
That is not a problem, I do not have to check a point here and here.
00:37:44.700 --> 00:37:48.000
I'm going to go ahead and check 1.
00:37:48.000 --> 00:37:54.000
For 1, when I put 1 into here, this is going to be positive, this is going to be negative.
00:37:54.000 --> 00:37:56.700
Therefore, it is going to be decreasing.
00:37:56.700 --> 00:37:59.500
We are going to be decreasing on that interval.
00:37:59.500 --> 00:38:14.500
When I check the number 2, I get positive and positive which means it is increasing on that interval.
00:38:14.500 --> 00:38:20.100
I’m decreasing from 0 to 1.8, I’m increasing from 1.8 to +infinity.
00:38:20.100 --> 00:38:22.300
I have a local from decreasing to increasing.
00:38:22.300 --> 00:38:25.600
I have a local min at 1.8.
00:38:25.600 --> 00:38:27.300
That is what this information tells me.
00:38:27.300 --> 00:38:29.500
Let us write that down.
00:38:29.500 --> 00:38:43.900
I’m decreasing from 0 to 1.8, I am increasing from 1.8 to +infinity.
00:38:43.900 --> 00:38:55.300
I have a local min at x = 1.8.
00:38:55.300 --> 00:39:05.300
The y value at x = 1.8, that is just the original function.
00:39:05.300 --> 00:39:13.100
F(1.8), it gives me -1.7.
00:39:13.100 --> 00:39:23.900
My local min is going to be the point 1.8, -1.7.
00:39:23.900 --> 00:39:28.000
Notice that we have x = 0, what is the other critical point.
00:39:28.000 --> 00:39:33.900
But I cannot say that there is actually a local max there.
00:39:33.900 --> 00:39:38.600
The reason I cannot say that is because there is nothing to the left of 0, the domain.
00:39:38.600 --> 00:39:41.600
In order to have a local max or a local min, I have to have the point
00:39:41.600 --> 00:39:45.500
defined to the left on to the right of that particular critical point.
00:39:45.500 --> 00:39:49.100
Here it is only defined to the right.
00:39:49.100 --> 00:39:54.700
The positive and negatives do not count.
00:39:54.700 --> 00:40:00.500
Let me write that out.
00:40:00.500 --> 00:40:10.500
We cannot say that there is a local max.
00:40:10.500 --> 00:40:34.700
In some sense there is, but not by definition, that there is a local max at x = 0 because f is not defined for x less than 0.
00:40:34.700 --> 00:40:42.300
In other words, our function, we know there is a local min at 1.8 and -1.7.
00:40:42.300 --> 00:40:44.200
We know it is here.
00:40:44.200 --> 00:40:46.500
We know it is going to be something like this.
00:40:46.500 --> 00:40:51.800
We cannot necessarily say that this is a local max because it is not defined over to the left of it.
00:40:51.800 --> 00:40:54.300
That is all that is going on here.
00:40:54.300 --> 00:40:56.800
Let us do point of inflection.
00:40:56.800 --> 00:40:59.000
Let me go back to red.
00:40:59.000 --> 00:41:10.700
Points of inflection, we have f’(x) is equal to x + 2 ln 1/3 x.
00:41:10.700 --> 00:41:16.200
Therefore, f”(x), I’m going to take the derivative of this.
00:41:16.200 --> 00:41:28.400
It is going to be 1 + 2 ×, this is 2x, 1 + 2 × x × the derivative.
00:41:28.400 --> 00:41:45.600
X × 1/1/3 x × 1/3 + the nat-log of 1/3 x × 1.
00:41:45.600 --> 00:41:47.800
I just pulled out the 2, just for the hell of it.
00:41:47.800 --> 00:41:52.300
1/3, 1/3, x, x, this becomes 1.
00:41:52.300 --> 00:42:06.700
You end up with 1 + 2 × 1 is going to be 2 + 2 × the ln of 1/3 x.
00:42:06.700 --> 00:42:14.700
We are going to get 3 + 2 × the ln of 1/3 x.
00:42:14.700 --> 00:42:20.600
That is our second derivative.
00:42:20.600 --> 00:42:25.600
We need to set that equal to 0.
00:42:25.600 --> 00:42:34.200
F”(x)is equal to 3 + 2 × the ln of 1/3 x.
00:42:34.200 --> 00:42:35.800
We need to set that equal to 0.
00:42:35.800 --> 00:42:46.400
We get 2 ln 1/3 x is equal to -3, ln of 1/3 x = -3/2.
00:42:46.400 --> 00:43:01.100
We exponentiate both sides, we get 1/3 x is equal to e⁻³/2
00:43:01.100 --> 00:43:13.300
which gives us x is equal to 3 × e⁻³/2, which is approximately equal to 0.7.
00:43:13.300 --> 00:43:16.000
We have to check, this is a point of inflection.
00:43:16.000 --> 00:43:23.000
We need to check a point to the left of 0.7 to the right of 0.7, to see whether the second derivative is positive or negative.
00:43:23.000 --> 00:43:29.100
In other words, concave up or concave down, respectively.
00:43:29.100 --> 00:43:35.200
We have got 0.7 here, we are checking f”.
00:43:35.200 --> 00:43:45.700
F” is equal to 3 + 2 × the nat-log of 1/3 x.
00:43:45.700 --> 00:43:52.900
When I check the value, I will just check 0.
00:43:52.900 --> 00:44:09.300
I check the value 0, it is going to be concave down.
00:44:09.300 --> 00:44:12.400
I’m not going to check 0 actually.
00:44:12.400 --> 00:44:14.600
I will check my 0.5.
00:44:14.600 --> 00:44:26.300
0.5, you are going to end up getting a negative number which is going to be concave down.
00:44:26.300 --> 00:44:31.800
And then, I'm going to go ahead and check some other number 1, 2, 3, does not really matter.
00:44:31.800 --> 00:44:36.800
Let us just check 3, positive, that is going to be concave up.
00:44:36.800 --> 00:44:46.200
In other words, I’m putting these values into the second derivative to tell me whether something is positive or negative.
00:44:46.200 --> 00:44:52.100
This is going to be positive which is going to be concave up.
00:44:52.100 --> 00:45:06.000
That takes care of that.
00:45:06.000 --> 00:45:08.500
Now I have my intervals of concavity.
00:45:08.500 --> 00:45:24.600
It is going to be concave down from 0 all the way to this 0.7.
00:45:24.600 --> 00:45:28.200
Let me double check, yes.
00:45:28.200 --> 00:45:42.400
And then, concave up from 0.7 to +infinity.
00:45:42.400 --> 00:45:46.200
Let us find where f(x) actually equal 0.
00:45:46.200 --> 00:45:52.700
Let us see where it actually crosses the x axis, if in fact it actually does so.
00:45:52.700 --> 00:45:57.700
Let me go ahead and do that on the next page here.
00:45:57.700 --> 00:46:07.400
F(x) is equal to x² ln 1/3 x, we want to set that to equal to 0.
00:46:07.400 --> 00:46:16.000
That gives us x² is equal to 0 and it also gives us ln of 1/3 x is equal to 0.
00:46:16.000 --> 00:46:19.300
This is 0, it attaches at 0,0.
00:46:19.300 --> 00:46:28.000
This one you get 1/3 x e⁰ is 1, that means x = 3.
00:46:28.000 --> 00:46:34.300
It touches the x axis at 0 and 3.
00:46:34.300 --> 00:46:39.900
Let us list what we have got.
00:46:39.900 --> 00:46:48.800
Our root x = 0, x = 3.
00:46:48.800 --> 00:46:59.600
We have a local min at the point .1.8, -1.7.
00:46:59.600 --> 00:47:07.700
We have a point of inflection at 0.7, -0.7.
00:47:07.700 --> 00:47:20.700
We are concave down from 0 to 1.4.
00:47:20.700 --> 00:47:24.300
I’m sorry not 1.4, it is going to be 0.7.
00:47:24.300 --> 00:47:27.600
Concave down from 0 to 0.7.
00:47:27.600 --> 00:47:35.700
We are concave up from 0.7 to +infinity.
00:47:35.700 --> 00:47:42.100
The function is decreasing from 0 to 1.8.
00:47:42.100 --> 00:47:55.100
The function is increasing from 1.8 to +infinity.
00:47:55.100 --> 00:48:03.700
When we put all of that together, local min at 1.8, -1.7.
00:48:03.700 --> 00:48:07.300
1.8, -1.7 probably puts us right there.
00:48:07.300 --> 00:48:14.200
Point of inflection at 0.7, -0.7, 0.7, -0.7, somewhere around there.
00:48:14.200 --> 00:48:16.200
Our graph goes something like this.
00:48:16.200 --> 00:48:26.700
It is decreasing from 0 to 1.8, decreasing, concave down from 0 to 0.7.
00:48:26.700 --> 00:48:35.200
It is concave down here, concave up from 0.7 to infinity.
00:48:35.200 --> 00:48:41.000
That is our graph, let us see a better version of it.
00:48:41.000 --> 00:48:45.500
Here is our root, here is our root.
00:48:45.500 --> 00:48:50.700
Points of inflection is somewhere around there.
00:48:50.700 --> 00:48:52.700
Local min somewhere around there.
00:48:52.700 --> 00:48:57.100
As you can see, we have concave down from here to here.
00:48:57.100 --> 00:49:01.500
Concave up from here to here, and continuously concave up.
00:49:01.500 --> 00:49:05.700
The graph goes, passes through 3.
00:49:05.700 --> 00:49:09.600
There you go, that is it, wonderful.
00:49:09.600 --> 00:49:12.100
Thank you so much for joining us here at www.educator.com
00:49:12.100 --> 00:49:18.700
We will see you next time for a continuation of example problems on using the derivative to graph functions.
00:49:18.700 --> 00:49:19.000
Take care, bye.