 Hello, and welcome to this screencast for calculus 2. Today we're going to try to construct the graph of a function from the graph of its derivative. Actually, we will need a little bit more information than just the derivative in order to construct the graph of the function. The primary tools we'll be using within this are ideas we learn from calculus 1 about where a function is increasing, where it's decreasing, where a function is concave up, and where a function is concave down. We should be able to use the graph of the derivative in order to do this. The last piece we will need in order to actually construct the graph will be the total change theorem. So, here's the problem we're going to be looking at. Pause the screen, screencast for a minute, and make sure you understand what the problem is asking for. And before proceeding, you should actually try to answer the questions at the bottom of the screen by being able to use the graph of the derivative to help answer these questions. Okay, so here's what we can do. If we know where the derivative of a function is positive, we can say, okay, the function is increasing on that interval. So, if we look at this graph, and we can notice, kind of as we go around that graph, that the graph, the derivative is positive on the interval from 0 to 60. So, what that tells me then is that f of x is increasing. And we can now do kind of the same thing. You can look at the graph and see where the derivative is negative. And we can see on this, for this particular derivative, that's on the interval from 60 to 100. And so, that's where the function f of x is decreasing. Now, one thing we're going to want to do with this is also to help determine where the graph of a function would be concave up. The condition for that is where the derivative is increasing. So, what we want to look at is, where is the derivative increasing? And again, we can look at the graph and see, okay, it's increasing there and there. So, those are the intervals on which the graph of the function will be concave up. And we get two intervals for that. We get from 0 to 40, and from 80 to 100. So, on those intervals, the graph of f of x is concave up. And then, of course, we can also use the graph to see where the derivative is decreasing. And we can see that that occurs when x is between 40 and 80. And that is where the graph of f of x is concave down. Got that? Okay. Here's then a summary of what we have just done. The other thing that we should know that's not on this list here is that at 60, the derivative is equal to 0. And prior to 60, the graph is increasing. After 60, the graph is decreasing. So, what we can say is that the function will have a relative maximum at x equal to 60. And so we'll keep that in mind as we go through this also. Now, with this information that we have so far, we cannot actually sketch the graph of the function because we don't know where the graph is located vertically on the coordinate axes. We need one further piece of information. And generally, what that's going to be is the value of the function at one particular point. In this case, what we're going to do is assume that the function has this value down here, that f of 0 is equal to 100. So, pause the screen again. Think about the total change theorem and how that might be helpful in determining the graph of the function. Actually, here is a statement of the total change theorem and what we're going to be using for that. If f is continuously differentiable on an interval with derivative f prime, then we can get the total change in the function over that interval by taking the integral of the derivative. So, what we're going to use here is area interpretation of the derivative. We set this up so that we could easily calculate areas by calculating areas of triangles and circles. So, try this one. See if you can determine the value of the function at x equal to 40. So, cutting it there, remember we are assuming that f of 0 equals 100. And we'll start that in just a minute. So, here we are. We're going to try to calculate f of 40. And so, what we're going to do is form this triangle and use the total change theorem. And what we can conclude then is that f of 40 minus f of 0 is the integral from 0 to 40 of f prime of x dx. And so, f of 40 will be equal to 100 plus the area of this triangle because that is what will be equal to the integral of the derivative. And so, we use the formula one-half base times altitude. So, this triangle has a base length of 40 and an altitude of 20. So, if we do that computation, we can see we get, for that second term there, 400. So, this turns out to be 500. So, f of 40 is equal to 500. And now we can proceed to get some other values for the function as well. In particular, if we do the same thing using f of 60, for example. Again, that will be equal to f of 0, which is 100, plus the integral from 0 to 60 of the derivative. Okay, so that's going to be 100 plus the area of the triangle as we go from 0 to 40. We've already calculated that as 400 plus the area of that quarter of a circle. So, it'll be one fourth pi. And here's our circle we're thinking about. The radius of that circle is 20. So, we get 20 squared. And again, if we carry out that computation, we will see we get 500 plus 100 pi, which is approximately 814.16. So, we've got two values for that. We're going to try to do two more. In particular, we're going to try to do f of 80 and f of 100. Well, look at what happens with f of 80. Again, we're going to get 100 as f of 0, plus the integral from 0 to 80. Well, the integral from 0 to 80, what we will do is we'll take care of this triangle here, of course. We've got that area as 400. But look what happens next. What we're going to use, of course, is the net signed area of the rest of it. And we have a quarter of a circle that's above the axis right there, and a quarter of a circle that's below the axis right there. So, those two areas cancel out. We add the area above the axis, and we subtract the area below the axis. So, f of 80 is also equal to 500. The last one we're going to calculate is f of 100. And we've got a little tight space here on our screen. But you can see that if we try to determine the area, the net signed area for this graph, again, we're going to get this, the first triangle, the two quarter circles will cancel out. And then the last part will be the area of this triangle, which we subtract, because it's below the x-axis. So, what we will get is f of 100 will again be 100 plus 400, the first triangle. The two quarter circles cancel out. And then we get minus the area of the second triangle right here. And that has a length of 20 and a height of 20. And so, we end up subtracting 200 from that. And so, we get f of 100 is equal to 300. So, now we've got points we can plot along with the qualitative information that we got about increasing, decreasing, and concavity. Here's a summary again of what we have done. We have information about where the function's increasing, where it's decreasing, where it's concave up, where it's concave down. And in addition, we know some function values. In particular, remember, although it's not written here, we were starting with f of 0 equal to 100. So, if we now plot those points and then combine that with this information here, we can get a reasonably accurate picture of the graph of the function. Pause the screen and see if you can do that. The next screen, of course, will show the graph of the function. So, here we are. And as you can see, if we look at these two graphs together, we can see where the derivative is increasing from 0 to 60. I'm sorry, where the derivative is positive from 0 to 60. The function down here is increasing from 0 to 60 and decreasing from 60 to 100 exactly where the derivative is negative. We've also got the concavity measured, and we've plotted the five points that we have. And again, notice that the relative maximum now is at the point 60, and it's about 814.16. If we remember, it was actually, I think, 500 plus 100 pi. As a last note, remember that we needed the value of one point. In this case, we had f of 0 equal to 100. If we have changed the value for f of 0, we would change the location of the graph, but not the shape. So, in other words, if we tried f of 0 equal to 200, the only thing we would do with this graph is start it here at 200 and proceed to get a graph something like that. Of course, that is not nearly as well drawn as what the computer drew for us. That's it. See you next time.