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 do some more example problems using derivatives to graph functions.
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Let us jump right on in.
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Our first examples says, for the function f(x) = x²/ x - 4²,
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find the intervals of increase and decrease, local maxes and mins, intervals of concavity and inflection points, and asymptotes.
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Use this information to graph the function.
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In general problems, we will just say graph the function.
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In these set of problems, we are laying out specifically what we want.
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In general, all of these things are things that are going to do anyway.
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You may do some of them, you may do most of them,
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but we want to get you into the habit of actually doing a nice systematic procedure.
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Let us see what we can find.
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This is a rational function, notice this mentioned asymptote.
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We may or may not be dealing with asymptotes, whenever we deal with rational functions.
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Let us start off by doing the first derivative, it is always a great place to start.
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The first derivative gives us a lot of information.
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I think I’m going to work in blue today.
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F’(x), we know this is going to be quotient rule.
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We are going to have x - 4² × the derivative of the top which is going to be 2x - x² × the derivative of the bottom
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which is going to be 2 × x – 4¹ × the derivative of what is inside, 1.
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All of that is going to be / x – 4⁴.
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I’m going to go ahead and pull out the x – 4.
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The numerator, I’m going to write it as 2 x × x - 4 - 2x²/ x – 4⁴.
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I end up getting, this cancels this, turning it into a 3.
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When I multiply that, I get 2x² - 8x.
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I get 2x² - 8x - 2x²/ x - 4³.
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The 2x² cancel leaving me -8x/ x – 4³.
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This is our first derivative, this is the one we set equal to 0, to find our critical points.
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We set that equal to 0 and what we end up getting is -8x = 0.
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Therefore, x is equal to 0.
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0 is our critical point.
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Recall that a critical point is not only where the first derivative equal 0,
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it is also where the function might not be differentiable.
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We have to watch out for those.
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This procedure, the first derivative setting equal to 0, it is only going to give us those that are.
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We still have to look at the function and see if there is some place where the graph is not going to be differentiable.
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In this particular case, we look at the denominator.
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The denominator was x – 4.
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Clearly, we cannot have a denominator equal to 0.
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Therefore, at the point 4, we are going to see some sort of asymptote.
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The asymptote, it is not going to be differentiable there.
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4 is also going to be a critical point.
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A critical point that this first derivative procedure will not give us.
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Something we have to elucidate by other means.
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Let us write that out, recall that the critical point, that a critical point, is also where f’ fails to exist.
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In other words, does not exist, fails to exist.
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Here we have a vertical asymptote at x = 4.
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x = 4 is also a critical point.
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In other words, it is one of the points to the left of which and to the right of which,
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we are going to have to check to see what the derivative is equal to, positive or negative.
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In order to decide whether it is increasing or decreasing to the left or to the right of that.
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We are dealing with f’ and we have our two critical points, we have 0 and we have 4.
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We are going to be checking points in this interval, in this interval, in this interval, to check increasing and decreasing behavior of the function.
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F’(x) we said is equal to -8x/ x – 4³.
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We are going to take points in these intervals, put them in here to see whether we get something that is positive or negative.
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Let us try -1, when I put -1 into this, I’m going to get something which is a positive number/ a negative number.
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This is a negative number which means decreasing.
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I know that to the left of 0 is actually decreasing.
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Let us try 1, when I put 1, I’m going to get a negative number/ a negative number.
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This is also a negative number.
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Definitely, negative divided by a negative is positive, sorry about that.
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Positive, we have an increasing.
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It is going to be increasing between 0 and 4.
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Let us go ahead and check a point to the right of 4.
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I’m going to go ahead and try 5.
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When I do 5, I get a negative number on top and I’m going to get a positive number which is negative.
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It is going to be decreasing on that interval.
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We said that 0 was a critical point that we got from the first derivative test.
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We also have a decreasing to the left and an increasing to right of it.
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That means that 0 is a local min.
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All of this gives us.
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There is a local min at x = 0, we have that information.
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There is an asymptote, a vertical asymptote at x = 4.
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Even though we have an increasing and a decreasing, this is not a local max.
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At 4, the function is not differentiable and we see that we have a rational function.
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This is clearly a vertical asymptote.
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Basically what is happening is it is going that way and it is coming back down that way.
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It is going to the same place and we knew that from pre-calculus because it was the original function had a squared term.
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Whenever it is an even multiplicity on that, it is going to go to the same place.
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When it is odd multiplicity, it is going to go to different places, the opposite ends.
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Let us go ahead and see what f(0) is.
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F(0) is equal to 0, the local min, the actual point is, the local min occurs at 0,0.
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We have partial graph.
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You are welcome to graph it as you go along, even if you do not have all the information.
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We know that we have a local min at 0,0.
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We know the graph looks something like this.
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We also know that the value 4, you know we have a vertical asymptote.
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We know we are looking at something like this and we know we are working at something like that.
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That is what this first derivative tells me, really nice.
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It gives me a lot of information.
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Not done yet but it is a good place to start.
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Before we take the second derivative to check points of concavity and inflection points,
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let us see if there are any horizontal asymptotes here.
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Before we take f”(x), let us see at that horizontal asymptotes.
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F(x) that is equal to x²/ x - 4².
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If I expand the bottom, I get x²/ x² - 8x + 16.
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We caught an asymptote behavior means what happens when x really big, positive infinity.
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x gets really small, negative infinity.
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That is it, that is all that it is saying.
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In this particular case, we notice from pre-calculus, if we remember, the degrees are the same.
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When x gets really big, these other terms do not even matter.
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What you end up with is x²/ x² which is just 1.
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Let me write this down because I have got a polynomial written down here, we might as well be complete.
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Recall that asymptotic behavior means what happens to f(x), when x goes to + or – infinity, it is really huge.
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In this particular case, I’m going to run through the process, that is fine.
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F(x), we said is equal to x²/ x² - 8x + 16.
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I’m going to multiply the top and bottom by 1/ x².
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In other words, I’m going to divide the bottom and the top by the highest degree that occurs.
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I’m going to end up with 1/1 - 8/x + 16/ x².
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As I take x to + or –infinity, this goes to 0, this term goes to 0, and what I’m left with is 1.
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This is my horizontal asymptote.
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You remember this from previous lessons, when we were talking about limits earlier in the course.
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Horizontal asymptote at 1, at y = 1, horizontal asymptote.
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Let us go ahead and draw out real quickly what is it we are looking at here.
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We know that we have a 1, 2, 3, 4.
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We have a vertical asymptote at 4.
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We know at 1 we have a horizontal asymptote.
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What is probably happening is this, we know that we have that.
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We know that it comes probably like this.
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And this one probably goes like that.
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More than likely that is what is happening.
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Again, the function is getting close to 1, when x gets big or when x gets big in a negative direction.
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That is what is happening.
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Let us go ahead and find f” and see what sort of information that gives us.
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Let us find f”(x) and determine the points of inflection and the intervals of concavity.
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We said that f’(x) was equal to -8x/ x – 4³.
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Therefore, f”(x) is equal to x – 4³ × -8 - 8x × 3 × x - 4² × 1/ x – 4⁶.
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I’m going to go ahead and factor out an x - 4².
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I’m going to get -8 × x – 4.
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When I take all of these out, I’m going to get +24x.
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All of that is going to be over x – 4⁶.
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I’m sorry, I know that this is a lengthy procedure.
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I apologize for actually going through this process.
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I just figured it might as well be as complete as possible.
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This knocks out that, turning that into 4.
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It leaves me with -8x + 32 + 24x/ x – 4⁴,
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which gives me -8x + 24 is going to be 16x.
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If my arithmetic is correct, + 32/ x – 4⁴.
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This is what I’m going to set equal to 0 and this gives me 16x + 32 = 0, 16x = -32.
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I get x = -2.
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This is my inflection point.
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We have already seen the graph of this thing.
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We have already seen that it looks, this is our horizontal asymptote.
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This is our vertical asymptote.
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We know we are dealing with the function that looks like this.
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Clearly, this point right here at -2 is the inflection point.
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From the graph, we already know the intervals of concavity.
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-infinity to -2, it is going to be concave down.
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From -2 all the way to 4, it is going to be concave up.
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From 4 onward, it is also going to be concave up.
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We know these already, but let us go through the process analytically for the second derivative.
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From the graph, we already know the intervals of concavity.
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But, let us run through analytically.
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We went ahead and we found that this point of inflection, one of the points of inflection is -2.
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We also have to include the 4, it is an asymptote.
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There is something going on to the left and right of that.
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Even though, again, it does not show up, we have to include it.
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It is a point that is not differentiable.
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Something happens to the left of that point.
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Something happens to the right of that point.
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I have to check 3 regions.
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I’m checking the second derivative.
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The second derivative is equal to 16x + 32/ x – 4⁴.
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Let us go ahead and find f”(-3).
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When I put that in, I’m going to get a negative number/ a positive number.
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This is negative which means this is concave down, which I already knew.
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When I do f” of something between -2 and 4, let us say -1, I end up with a positive/ a positive number.
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This means it is concave up, second derivative.
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And then, when I checked f” of let us say 5, something over here to the right of 4,
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I’m going to end up with a positive number/ a positive number.
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This means concave up.
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My intervals of concavity, concave down, like I said –infinity to -2.
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Concave up from -2 to 4 union.
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I do not just say -2 to infinity because the 4 separates those two.
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I do have to break it up into regions, -4 to +infinity, that take care of that.
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Let us go ahead and find, now I have got my graph.
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I have got 1, 2, 3, 4.
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I have my vertical asymptote, I have my horizontal asymptote.
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I know my graph goes something like this.
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I know it goes like this.
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Let us find what that point is.
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We said that the point of inflection, we said that x = -2.
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When I take f(-2), I end up with 1/9.
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This point right here is the point -2, 1/9.
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The actual coordinates of that point.
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When I put it all together, let graphing function actually take care of this.
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This is what the function actually looks like.
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You can see we have a vertical asymptote at 4.
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We have a horizontal asymptote at 1.
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We have our point of inflection from concave down to concave up, still concave up.
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There you go, all of the information.
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You are using what you already know, in addition to the tools that you are learning from calculus.
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Notice, the first derivative did not give me a critical value of 4.
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But I included it because there is a vertical asymptote there.
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All of the bits of information that I have at my disposal, I want to be able to bring to there to be able to graph the function.
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That is my goal to be able to graph the function and to find important points of reference for the graph.
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Let us go ahead and try another one here.
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For the function f(x) = ln, 2 - ln x, find the intervals of the normal stuff,
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increase/decrease, maxes and mins, concavity, inflection points, asymptotes, graph the function.
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Let us do the asymptotes first.
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In this case, the first thing we are going to look at, first let us look at the domain.
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We are looking at a logarithm function.
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A logarithm function is going to have a restricted domain.
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We want to see what that domain is first, and more than likely some thing is going to be happening at the endpoints of that particular interval.
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Asymptotic behavior, domain, they are going to be mixed up here.
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Let us take a look at the domain.
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Let us look at the domain.
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We have the function ln of 2 – ln x.
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This part, the argument of the logarithm function, this has to be greater than 0.
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Let us set it greater than 0.
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We have 2 – ln x greater than 0, let us solve for x.
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I have 2 greater than ln x.
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I'm going to exponentiate both sides.
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I get x is less than e².
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Or if I like to write it this way, x is less than e².
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The domain is from 0 to e² or 0 to about 7.4.
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This is not 1, this is a comma, 0 to about 7.4.
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Now we have our domain.
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Now that we have the domain, let us ask ourselves what happens as x goes to 0 and as x gets close to 7.4.
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We ask ourselves, now we work out the asymptotic behavior.
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We ask ourselves, one, what happens to f(x) as x approaches 0 from the left.
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Because there is nothing less than 0, we are approaching 0 from the right, from the positive side.
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Two, what happens to f(x), as x approaches e² from the left.
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Number 1, as x gets close to 0 from the right, the ln(x), it is going to go to –infinity.
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Negative of the ln(x) is going to go to +infinity.
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2 – ln(x) is going to go to +infinity so that makes 0 a vertical asymptote.
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Remember, my function here, my f(x) is equal to, 2 – ln(x).
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Ln(2) – ln(x) goes to +infinity.
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This is ln(2) – ln(x).
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I'm asking myself, what happens to the function as x gets close to 0.
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As x gets close to 0, the ln(x) part, that goes to –infinity.
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The –ln(x) is +infinity.
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2 – ln x goes to +infinity.
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As this goes to positive infinity, the ln goes to infinity.
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The function actually goes towards infinity, as we get close to 0.
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That makes 0 a vertical asymptote.
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In other words, from your perspective 0 is here, e² is here.
00:26:44.700 --> 00:26:50.700
As we get close here, the function is going to fly up vertically towards infinity.
00:26:50.700 --> 00:26:52.600
I will draw that in just a minute.
00:26:52.600 --> 00:26:54.600
We are just working out the analytics.
00:26:54.600 --> 00:27:11.800
As x approaches e², ln(x) is going to approach 2.
00:27:11.800 --> 00:27:19.100
2 - ln(x) is going to approach 0.
00:27:19.100 --> 00:27:29.000
Ln of 2 - ln x is going to approach negative infinity.
00:27:29.000 --> 00:27:37.700
e² is also a vertical asymptote, when the function goes down.
00:27:37.700 --> 00:27:51.200
From your perspective, 0, e², as I get close to e², my function is going to go down to negative infinity.
00:27:51.200 --> 00:27:58.900
What I have is this, I will just go ahead and put e² there.
00:27:58.900 --> 00:28:01.100
That is a vertical asymptote.
00:28:01.100 --> 00:28:06.700
I know that as I get close to 0, my function is going to fly up.
00:28:06.700 --> 00:28:14.000
As I get close to this, my function is going to fly down.
00:28:14.000 --> 00:28:16.800
It is going to cross somewhere there, we will deal with that later.
00:28:16.800 --> 00:28:20.500
At least I have my end behavior of my graph.
00:28:20.500 --> 00:28:25.900
Nothing happens over here, nothing happens over here, nothing happens over here.
00:28:25.900 --> 00:28:32.400
Because again, this is our domain in the center.
00:28:32.400 --> 00:28:36.600
Let us see what is next.
00:28:36.600 --> 00:28:52.700
It crosses the x axis when f(x) is equal to 0, the root.
00:28:52.700 --> 00:28:58.900
F(x) is ln of 2 – ln x is equal to 0.
00:28:58.900 --> 00:29:01.700
I’m going to exponentiate both sides.
00:29:01.700 --> 00:29:10.700
When I do that, I'm left with 2 – ln x e⁰ is equal to 1.
00:29:10.700 --> 00:29:13.500
I’m going to move the ln x over to this side, bring the 1 over this side.
00:29:13.500 --> 00:29:18.000
I end up with ln x is equal to 1.
00:29:18.000 --> 00:29:24.500
I’m going to exponentiate both sides, I get x is equal to e.
00:29:24.500 --> 00:29:33.500
It crosses the axis at x = e.
00:29:33.500 --> 00:29:44.200
The graph crosses at x is equal to e.
00:29:44.200 --> 00:29:51.800
Our graph is starting to come together nicely here.
00:29:51.800 --> 00:30:01.100
This is x = e² which is 7.4.
00:30:01.100 --> 00:30:04.100
E is somewhere around there.
00:30:04.100 --> 00:30:13.400
I know the graph looks something like, this is e, 2.718.
00:30:13.400 --> 00:30:17.400
The graph is going to look something like this.
00:30:17.400 --> 00:30:22.400
Let us go ahead and see what we can do.
00:30:22.400 --> 00:30:24.600
Clearly, there is no local max or min here.
00:30:24.600 --> 00:30:30.000
We already know what the graph looks like, just from this basic analysis which did not involve any derivatives.
00:30:30.000 --> 00:30:34.800
We already know that there is no local max, local min.
00:30:34.800 --> 00:30:38.800
But there is a concavity and there is a point of inflection.
00:30:38.800 --> 00:30:42.600
We can go ahead and find that.
00:30:42.600 --> 00:31:01.700
Clearly, there is no, I should say local max, min.
00:31:01.700 --> 00:31:05.100
Let us confirm this analytically.
00:31:05.100 --> 00:31:22.300
Let us confirm this analytically.
00:31:22.300 --> 00:31:25.100
It is exhausting writing all this out.
00:31:25.100 --> 00:31:33.200
We have f(x) is equal to ln(2) - ln(x).
00:31:33.200 --> 00:31:44.500
Our f’(x) is going to equal 1/ the argument which is 2 – ln x × the derivative of what is inside.
00:31:44.500 --> 00:31:52.000
The derivative is going to be -1/ x.
00:31:52.000 --> 00:32:04.700
That gives us -1/ x × 2 - ln(x).
00:32:04.700 --> 00:32:08.900
We set that first derivative = 0.
00:32:08.900 --> 00:32:15.700
This means -1 is equal to 0 because the only way this is equal to 0, is if the numerator is 0.
00:32:15.700 --> 00:32:19.400
We get -1 is equal to 0, there is no solution.
00:32:19.400 --> 00:32:21.900
This confirms the fact that there is no critical point.
00:32:21.900 --> 00:32:30.600
There is no possibility of a maximum or a minimum, local max or min, high point or low point.
00:32:30.600 --> 00:32:41.600
No solution, no local max or min.
00:32:41.600 --> 00:32:51.200
We have f’(x), we said it is equal to -1/ x × 2 - ln(x).
00:32:51.200 --> 00:32:56.500
Let us go ahead and find the second derivative.
00:32:56.500 --> 00:33:27.400
F”(x) equals this × the derivative of that is 0 - that × the derivative of what is down below, which is going to be,
00:33:27.400 --> 00:33:29.100
This is going to be product rule.
00:33:29.100 --> 00:33:48.400
It is going to be this × the derivative of this, which is going to be -1/ x × the derivative of that
00:33:48.400 --> 00:33:58.300
+ 2 - ln x × the derivative of that which is 1.
00:33:58.300 --> 00:34:10.100
It is going to be all of that /2x – x ln x².
00:34:10.100 --> 00:34:12.800
I’m going to let you work out.
00:34:12.800 --> 00:34:34.400
I write this on this paper, as opposed to here, what we end up with is 2 - 1 - ln x/ 2x – x ln x².
00:34:34.400 --> 00:34:48.200
We end up with f”(x) is equal to 1 - ln x/ 2x – x ln x².
00:34:48.200 --> 00:34:50.500
We set that equal to 0.
00:34:50.500 --> 00:35:01.500
We have 1 - ln x is equal to 0, we have 1 = ln x.
00:35:01.500 --> 00:35:06.800
We exponentiate both side, we get x is equal to e.
00:35:06.800 --> 00:35:30.200
Therefore, at x = e which happened to be also be the root, there is a point of inflection.
00:35:30.200 --> 00:35:38.600
Now we have our final graph.
00:35:38.600 --> 00:35:48.600
Let us go ahead and, 1, 2, 3, 4, 5, 6, 7, 8, 7.4, this is that.
00:35:48.600 --> 00:35:52.000
1 to 2.718, this is there.
00:35:52.000 --> 00:35:59.300
Our graph is going to look something like this.
00:35:59.300 --> 00:36:05.200
We are going to be concave up from 0 to e.
00:36:05.200 --> 00:36:14.800
We are going to be concave down from e to e².
00:36:14.800 --> 00:36:17.900
I think that is it, we have everything that we need.
00:36:17.900 --> 00:36:19.300
Final graph looks something like this.
00:36:19.300 --> 00:36:27.200
There you go, 0, notice it is steep rise, this is our e.
00:36:27.200 --> 00:36:31.400
It also happens to be our inflection point, in addition to being our root.
00:36:31.400 --> 00:36:48.500
Over here 7.4 that is our vertical asymptote, that is a graph.
00:36:48.500 --> 00:36:55.500
Make sure you examine the graph, examine the function.
00:36:55.500 --> 00:37:00.600
Take a good look at the function before you just jump right on in and start taking derivatives.
00:37:00.600 --> 00:37:18.000
Make sure you examine the function before you just jump in and start differentiating.
00:37:18.000 --> 00:37:27.500
It is very tempting to want to just run through the process, and hopefully that algorithmic process will give you what you need.
00:37:27.500 --> 00:37:33.100
It is not always true, it was true in simple math but this is reasonably sophisticated math.
00:37:33.100 --> 00:37:37.400
We have to bring all of all our resources to bear.
00:37:37.400 --> 00:37:41.900
Examine the function, check the domain, check the asymptotic behavior.
00:37:41.900 --> 00:37:44.800
If you take the first derivative, is it going to give you all the critical points.
00:37:44.800 --> 00:37:47.400
We have to places where the function is not differentiable.
00:37:47.400 --> 00:37:57.300
There is a lot going on and that is the nature of more complex material, more complex functions.
00:37:57.300 --> 00:38:02.800
Let us take a look at this one, find the cubic equation ax³ + bx² + cx + d,
00:38:02.800 --> 00:38:11.500
that achieves a local max at -3, 4 and a local min at 4, -2.
00:38:11.500 --> 00:38:16.800
They are telling me that, basically, I need to find a, I need to find b,
00:38:16.800 --> 00:38:23.600
I need to find c, and I need to find d, that satisfies these conditions.
00:38:23.600 --> 00:38:26.700
I already know what f(-3) is, it is 4.
00:38:26.700 --> 00:38:29.200
I already know what f(4) is, it is -2.
00:38:29.200 --> 00:38:35.400
I can plug that in here to get two equations.
00:38:35.400 --> 00:38:50.400
F(-3) = 8 × -3³ + b × -3² + c × - 3 + d.
00:38:50.400 --> 00:38:55.100
I know that equals 4.
00:38:55.100 --> 00:39:04.700
When I solve this, I get -27a + 9b, I hope to God that I have done my arithmetic correct.
00:39:04.700 --> 00:39:12.000
-3c + d, that is equal to 4, this is the first of my four equations.
00:39:12.000 --> 00:39:17.400
I have 4 variables, I’m going to need 4 equations and 4 unknowns, it is that simple.
00:39:17.400 --> 00:39:23.100
That is the only tool that we have at our disposal, at this point.
00:39:23.100 --> 00:39:27.000
We have dealt with linear systems, four equations and four unknowns.
00:39:27.000 --> 00:39:30.900
This is a linear system now because everything is to the first power.
00:39:30.900 --> 00:39:34.000
This is our first equation.
00:39:34.000 --> 00:39:38.000
I have a second equation, I know what f(4) is, it is equal to -2.
00:39:38.000 --> 00:39:52.100
F(4) which = a × 4³ + b × 4² + c × 4 + d.
00:39:52.100 --> 00:39:54.000
I know that = -2.
00:39:54.000 --> 00:40:05.600
When I work this out, I get 64a + 16b + 4c + d = -2.
00:40:05.600 --> 00:40:10.100
This is my second equation.
00:40:10.100 --> 00:40:11.500
We also know something else.
00:40:11.500 --> 00:40:22.800
It tells me that at these points, it achieves a local max and a local min which means the derivative of this, at these points, is equal to 0.
00:40:22.800 --> 00:40:28.100
Local max/local min, the derivative there at those points is equal to 0.
00:40:28.100 --> 00:40:40.800
I find the derivative f’(x) is equal to 3ax² + 2bx + c.
00:40:40.800 --> 00:40:56.500
I know that f’ at -3 which is going to equal 3a × -3² + 2b × -3 + c, I know that it equal 0.
00:40:56.500 --> 00:41:02.400
F’ at that point = 0 because it is a local max and f’ at this point is a local min.
00:41:02.400 --> 00:41:04.700
Its derivative is also equal to 0.
00:41:04.700 --> 00:41:17.900
I get 27 a - 6 b + c = 0, this is the third of my four equations.
00:41:17.900 --> 00:41:31.400
I do the same thing for f’ at 4, that is going to equal 3a × 4² + 2b × 4 + c, I know that equal 0.
00:41:31.400 --> 00:41:42.100
Here I get 48a + 8b + c, that is equal to 0.
00:41:42.100 --> 00:41:45.500
I have my 4 equations and 4 unknowns.
00:41:45.500 --> 00:41:54.500
This is my 1st, this is my 2nd, this is my 3rd, this is my 4th.
00:41:54.500 --> 00:42:04.100
I’m going to solve this by putting it in matrix form and I’m going to convert that matrix to something called reduced row-echelon form.
00:42:04.100 --> 00:42:06.400
Some of you have actually seen reduced row-echelon form.
00:42:06.400 --> 00:42:10.900
Do you remember back in pre-calculus, when you guys were doing row reduction
00:42:10.900 --> 00:42:18.400
and solving simultaneous systems, 3 equations, 4 equations, row reduction is what you guys did.
00:42:18.400 --> 00:42:30.200
Reduced row reduction is taking it a step further and making it so all of the coefficients in that matrix end up just equaling 1 along the diagonal.
00:42:30.200 --> 00:42:31.300
I will show you in just a minute.
00:42:31.300 --> 00:42:36.700
I’m not going to run through the process, there is plenty of mathematical software available online.
00:42:36.700 --> 00:42:43.200
Just do a Google search for reduced row-echelon calculator and a whole number of things will come up.
00:42:43.200 --> 00:42:47.700
You basically plug these numbers in and it will give you the answers.
00:42:47.700 --> 00:42:50.200
Here is what it actually looks like.
00:42:50.200 --> 00:43:24.200
Solve by converting the augmented matrix of equations to reduced row-echelon form.
00:43:24.200 --> 00:43:26.500
Our equations, we just put the coefficients in.
00:43:26.500 --> 00:43:33.800
We end up with -27, 9, -3, 1, is equal to 4.
00:43:33.800 --> 00:43:42.500
We have 64a, 16b, 4c, + d, = -2.
00:43:42.500 --> 00:43:51.800
27a - 6b + c, 0, 0, there is no d.
00:43:51.800 --> 00:43:57.500
And of course we have 48 and 8 and 1 and 0 and 0.
00:43:57.500 --> 00:44:00.800
This matrix, the first column is the a.
00:44:00.800 --> 00:44:07.000
The 2nd column is the variable b, 3rd column is the variable c, 4th column is the variable d.
00:44:07.000 --> 00:44:16.900
When I subject this to reduced row-echelon form, a reduced row-echelon form of a matrix is unique.
00:44:16.900 --> 00:44:21.100
Row-echelon form, Gaussian elimination is not a unique matrix.
00:44:21.100 --> 00:44:23.100
You get the answer but is not unique.
00:44:23.100 --> 00:44:24.800
Reduced row-echelon is always unique.
00:44:24.800 --> 00:44:27.000
There is only one place it always ends up.
00:44:27.000 --> 00:44:32.700
In this particular case, what you end up with is the following, when you have your computer do it.
00:44:32.700 --> 00:44:59.300
1, 0, 0, 12/343, 0, 1, 0, 0, -18/343, 0, 0, 1, 0, -432/343, and 0, 0, 0, 1, 562/343.
00:44:59.300 --> 00:45:07.400
There you go, a is equal to this, b is equal to this, c is equal to this, d is equal to this.
00:45:07.400 --> 00:45:12.500
That is it, that simple, that is what beautiful about reduced row-echelon form.
00:45:12.500 --> 00:45:25.400
This is a, this is b, this is c, and this is d.
00:45:25.400 --> 00:45:55.800
We get f(x) is equal to 12/ 343 x³ - 18/ 343 x² – 432/ 343 x + 562/343.
00:45:55.800 --> 00:45:58.400
Again, you can use whatever method you want to solve the system.
00:45:58.400 --> 00:46:03.400
I think reduced row-echelon is just about fastest and best because it just gives you your answer, your final matrix.
00:46:03.400 --> 00:46:10.800
You just read it off, it is the last column, a, b, c, and d.
00:46:10.800 --> 00:46:20.600
Let us do one more function here.
00:46:20.600 --> 00:46:29.900
For the function f(x) = 3x²/3 – x, find the intervals of increase/decrease, local maxes and mins, intervals of concavity, normal stuff.
00:46:29.900 --> 00:46:33.600
Use it to graph the function.
00:46:33.600 --> 00:46:45.400
The domain here is all real numbers so we can just start with f’(x).
00:46:45.400 --> 00:46:51.600
You know what, I will stick with blue, sorry about that.
00:46:51.600 --> 00:46:55.600
I will just use red if I absolutely need to.
00:46:55.600 --> 00:47:16.100
We have f’(x) is equal to 2x⁻¹/3 - 1 is equal to 0, which gives us 2/ x¹/3 - 1 is equal to 0,
00:47:16.100 --> 00:47:32.100
which gives 2/ x¹/3 = 1, which gives us x¹/3 is equal to 2.
00:47:32.100 --> 00:47:48.700
Therefore, when I cube both sides, I get x is equal to 8.
00:47:48.700 --> 00:47:53.500
I have 8, that is one of the critical points.
00:47:53.500 --> 00:47:56.600
That is where the derivative is equal to 0.
00:47:56.600 --> 00:48:06.500
Let us find other critical points, if they exist.
00:48:06.500 --> 00:48:18.000
In this particular case, places where f’(x) does not exist.
00:48:18.000 --> 00:48:23.100
We just have to be careful because when we see this function,
00:48:23.100 --> 00:48:28.200
we are not dealing with some ordinary polynomial where you have integral powers.
00:48:28.200 --> 00:48:36.300
This right here, leads me to believe that there might be places where it is not differentiable.
00:48:36.300 --> 00:48:39.400
Notice that f is defined at 0.
00:48:39.400 --> 00:48:49.300
Notice that f(x) is defined at 0, the domain is all real numbers including 0.
00:48:49.300 --> 00:48:52.700
When I put 0 in here, I get f = 0.
00:48:52.700 --> 00:49:00.600
It is defined at 0, in other words f(0) = 0.
00:49:00.600 --> 00:49:27.600
But f’(0) which = 2 × 0⁻¹/3 - 1 which = 2/ 0¹/3 – 1, does not exist.
00:49:27.600 --> 00:49:33.800
2/0 is undefined, so the derivative at 0 does not exist.
00:49:33.800 --> 00:49:36.000
The value of the function does.
00:49:36.000 --> 00:49:45.100
The graph passes through 0, hits 0, but it is not differentiable there.
00:49:45.100 --> 00:50:04.500
F’ at 0 is undefined, in other words, f, the original function is not differentiable there.
00:50:04.500 --> 00:50:13.700
0 is another critical point, not differentiable at x = 0, even though it is defined there.
00:50:13.700 --> 00:50:18.000
It is defined but it is not differentiable which means that it is another critical point.
00:50:18.000 --> 00:50:23.600
When we check our number line, we have to include 0 and 8.
00:50:23.600 --> 00:50:25.900
Let us go ahead and do that.
00:50:25.900 --> 00:50:39.300
0 and 8 are critical points.
00:50:39.300 --> 00:50:46.600
0, 8, I’m going to check here, I’m going to check here, and I’m going to check there.
00:50:46.600 --> 00:50:53.100
F’(x) is equal to 2/ x¹/3 – 1.
00:50:53.100 --> 00:50:57.900
When I check a point to the left here, I’m going to go ahead and check -1.
00:50:57.900 --> 00:51:11.700
When I check -1, I'm going to get 2/-1 -1 which is equal to -3.
00:51:11.700 --> 00:51:12.800
It is going to be decreasing.
00:51:12.800 --> 00:51:16.500
It is going to be decreasing there.
00:51:16.500 --> 00:51:19.800
I’m going to go ahead and check 1.
00:51:19.800 --> 00:51:24.600
When I check 1, I get 2/1 - 1 is equal to 1.
00:51:24.600 --> 00:51:29.100
It is increasing here.
00:51:29.100 --> 00:51:32.500
However, we just said it is actually not differentiable at 0.
00:51:32.500 --> 00:51:36.200
Even though it is decreasing and increasing, this is not a local min.
00:51:36.200 --> 00:51:56.500
It is actually it cusp but not differentiable at 0.
00:51:56.500 --> 00:52:07.200
This is not a local min, what you have is a cusp here.
00:52:07.200 --> 00:52:14.500
The function is decreasing to the left of 0.
00:52:14.500 --> 00:52:17.300
It is increasing but it is not differentiable there.
00:52:17.300 --> 00:52:21.800
The graph actually looks like this, there is a cusp there.
00:52:21.800 --> 00:52:29.100
That is what the graph looks like there.
00:52:29.100 --> 00:52:42.100
Let us go ahead and try, we have done these two, we still have try a region and number over there.
00:52:42.100 --> 00:52:45.800
I’m going to pick a number that is actually convenient to work with.
00:52:45.800 --> 00:52:47.500
I’m going to pick the number 27.
00:52:47.500 --> 00:52:52.800
When I try the number 27 because the 3√27 is really easy to take,
00:52:52.800 --> 00:53:01.200
it is going to give me 2/3 – 1, which is a negative number.
00:53:01.200 --> 00:53:05.200
Therefore, it is actually decreasing there.
00:53:05.200 --> 00:53:09.100
Increasing/decreasing, 8 was a normal critical point.
00:53:09.100 --> 00:53:12.800
It is what we found by setting the derivative, first derivative equal to 0.
00:53:12.800 --> 00:53:19.000
This is definitely a local max.
00:53:19.000 --> 00:53:29.500
We have a local max at x = 8.
00:53:29.500 --> 00:53:33.200
That is a pretty wacky looking 8.
00:53:33.200 --> 00:53:35.800
We have got 8 there.
00:53:35.800 --> 00:53:51.100
F(8) is equal to 4, the point 8, 4 is our local max.
00:53:51.100 --> 00:54:00.100
The function is increasing from 0 to 8.
00:54:00.100 --> 00:54:13.300
Our function is decreasing from negative infinity to 0 union 8 to +infinitive.
00:54:13.300 --> 00:54:17.300
Let us go ahead and deal with some other aspects, the second derivative.
00:54:17.300 --> 00:54:25.700
We said that f’(x) is equal to 2x⁻¹/3 – 1.
00:54:25.700 --> 00:54:28.600
We want to take the second derivative.
00:54:28.600 --> 00:54:41.100
F”(x) is equal to -2/3 x ^- 4/3 -1.
00:54:41.100 --> 00:54:44.900
We are going to set that equal to 0.
00:54:44.900 --> 00:54:55.300
We are going to get -2/ 3x⁴/3 - 1 is equal to 0.
00:54:55.300 --> 00:55:04.000
We are going to get -2/3 x⁻⁴/3 = 1.
00:55:04.000 --> 00:55:21.500
Rearranging, we are going to end up with -2 is equal to 3x⁴/3 - 2/3 = x⁴/3.
00:55:21.500 --> 00:55:29.400
This is -2/3 = x¹/3⁴.
00:55:29.400 --> 00:55:35.300
I separated this 4/3 that way, this is even.
00:55:35.300 --> 00:55:39.100
Even, something when you have an even exponent, you are never going to get a negative number.
00:55:39.100 --> 00:55:43.600
This is no solution.
00:55:43.600 --> 00:55:57.900
No solution so there are no points of inflection.
00:55:57.900 --> 00:56:01.900
Let us see what we have got here.
00:56:01.900 --> 00:56:14.700
F” at x is equal to -2/ 3x⁴/3 – 1.
00:56:14.700 --> 00:56:21.200
This is actually less than 0 for all x.
00:56:21.200 --> 00:56:25.700
The graph is concave down everywhere.
00:56:25.700 --> 00:56:40.000
F(x) is concave down everywhere.
00:56:40.000 --> 00:56:43.400
Let us see, where else f(x) actually equal 0.
00:56:43.400 --> 00:56:46.500
In other words, let us see if it actually hits the x axis, someplace else.
00:56:46.500 --> 00:56:49.100
We know it hits the x axis at 0, that cusp.
00:56:49.100 --> 00:56:55.500
Let us see if it actually hits someplace else.
00:56:55.500 --> 00:57:10.700
Let us see where else f(x) = 0, beside x = 0.
00:57:10.700 --> 00:57:23.700
F(x) = 3x²/3 - x is equal to 0.
00:57:23.700 --> 00:57:26.800
I’m going to go ahead and factor out the x²/3.
00:57:26.800 --> 00:57:35.800
I get x²/3 × 3 – x¹/3, that is equal to 0.
00:57:35.800 --> 00:57:40.800
I'm going to get x²/3 = 0 that gives me x = 0.
00:57:40.800 --> 00:57:43.100
I already know that one.
00:57:43.100 --> 00:57:51.000
I'm going to get 3 - x¹/3 = 0, x¹/3 = 3.
00:57:51.000 --> 00:57:57.200
Therefore, when I cube both sides, I get x = 27.
00:57:57.200 --> 00:58:03.400
27 is the other root, it is the other place where it actually hits the x axis.
00:58:03.400 --> 00:58:08.700
I have all of the information that I need.
00:58:08.700 --> 00:58:12.900
My graph is concave down everywhere.
00:58:12.900 --> 00:58:22.800
It hits at 0 and it hits at 27, there is a cusp here and there is a cusp here.
00:58:22.800 --> 00:58:28.400
At 8, it hits a local max of 4.
00:58:28.400 --> 00:58:35.500
When it comes back down, passes through 27 again, and goes down that way and goes up that way.
00:58:35.500 --> 00:58:37.400
That is my graph.
00:58:37.400 --> 00:58:39.100
Let us see what it looks like.
00:58:39.100 --> 00:58:40.800
Yes, that is exactly right.
00:58:40.800 --> 00:58:43.900
Here is my point where f is defined but it is not differentiable.
00:58:43.900 --> 00:58:45.600
There is a cusp there.
00:58:45.600 --> 00:58:49.800
Concave down, concave down, there is no concave up.
00:58:49.800 --> 00:58:57.200
Here is my max at 8, 4, and here is where I’m at 27, 0, it is my other root.
00:58:57.200 --> 00:59:00.600
There we go, thank you so much for joining us here at www.educator.com.
00:59:00.600 --> 00:59:01.000
We will see you next time, bye.