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Definite Integrals As Net Change - Problem 3
If given a derivative graph f'(x), it is possible to sketch the graph of f(x) by using properties of the graph of f'(x).
The local minima and maxima of f(x) occur at points where the graph of f'(x) crosses the x-axis. The point is a minimum if f'(x) changes from negative (below the x-axis) to positive (above the x-axis) at that point, and a maximum if f'(x) changes from positive to negative at that point.
If given the areas under certain regions of the graph, you know that the net change of f(x) between from the beginning to the end of that region is the same value as the area. For example, if the area under a curve between x=a and x=b is 28, then f(b) - f(a) = 28.
You can use other facts about the derivative's graph to determine end behavior (the value of the derivative is the slope of the original function, etc).
Let's take a look at a really nice problem that involves the net change theorem. First of all, recall that the net change theorem allows us to calculate a net change using the integral of the function's derivative.
Well here, in this problem says the graph of the derivative F'(x) of a function F(x) is shown below. It's actually shown up to the right, and values of some areas are given. If F(0) is 25, sketch a graph of F(x), and label coordinates of local max or min.
So here is the graph that the problem refers to. The graph of F'. What we need to try to do is graph F(x) given this graph. We're also told what the areas of each of these little regions is. So this area here is 56. This area is 81. This one is 49. So this is a perfect net change theorem problem.
We're also told that F(0) is 25. So in order to find the values of the coordinates at the local max, and min, we should first figure out where the local max, and min are.
Well, this is the derivative of the function we're going to draw. Local max, and min happen where the derivative is 0 where it changes from positive to negative or negative to positive. So it's precisely at x equals 2, and x equals 5 that we're going to get local max and min, and which is which? Well, remember when the derivative goes from positive to negative, the function whose derivative this is will go from increasing to decreasing. So we're going to have a local max right here at 2.
At 5, the derivative goes from negative to positive, so the function is going to go from decreasing to increasing, so we'll have a local min. This will be really clear once we see the values of the function, but that's what we have to figure out. We have to figure out what the value of the function is at 2. What the value is at 5, and we might as well find out what the value is at 6. So we can do that.
First let's find what the value is at 2. So find F(2). First of all what net change will give me F(2)? F(2) minus F(0) will do it. I know F(0) is 25, so I can relate this net change to the integral of the derivative of this function. F'(x) from 0 to 2.
Now that integral is precisely equal to this area. The integral from 0 to 2 of this function is going to be this area, because the function is non-negative on that interval. So it's exactly 56. F(0) is 25, and F(2) is what we're trying to find. So if you add 25 to both sides, you'll get F(2). You'll see that F(2) is 81. So we'll plot that in a minute.
Now that we know F(2), let's find F(5). F(2) is what we think is the local max. F(5) is what we think is going to be the local min. We can integrate from 2 to 5 to get that. So that would use this net change, F(5) minus F(2). Of course that's going to equal the integral from 2 to 5 of F'.
Now this integral is related to the area here, but it's not equals to it. Remember when the function is below the x axis, the definite integral doesn't equal the area, it equals the opposite of the area. The area is 81, so the definite integral will be -81. This definite integral is -81, so F(5) minus F(2) which is 81. It looks like F(5) needs to be 0, so that's F(5), our local minimum.
Finally, let's find F(6). Again we use the net change F(6) minus F(5) that's going to equal the integral from 5 to 6 of F'. Now that integral according to this graph is 49. It's exactly equal to the area under the curve from 5 to 6, because the function F' is non-negative on that interval. So we have 49 for the integral. F(5) remember was 0, so F(6) is exactly 49.
Let's plot these points. First F(0). F(0) was 25. This is 20, this is 30 so 25 is right about here. F(2) is our next point. F(2) is 81. Here is 80, a little more than that is 81. F(5) is our next point. F(5) is 0, that's easy to plot right down here. Then F(6) is 49. Here is 40, this is 60, 49 is just a hair short of 50, so let's say about there.
When we draw this graph, remember that we can use the derivative to help us out. The function's increasing from 0 to 2. Then it reaches a local max where it levels off, you can see the derivative is 0 here. So that will look like this. It levels off, and it decreases, because F' is negative between 2, and 5. It's going to have a horizontal slope at 2 and 5. So starts up here decreases. I've got to make sure I'm on the right trajectory here and then it comes back to a horizontal slope again, that'll do.
Then from 5 to 6, it's going to increase again. Starting with the slope of 0, and then increasing from there. So let me just label these points; (6, 49), (5, 0), (2, 81) and (0, 25. So this is my graph of y equals F(x); the function whose derivative I was originally given. I just used the net change theorem a bunch of times to find these values.