Recovering a function from its derivative
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Question

Recovering a function from its derivative

b. Repeat part (a), assuming that the graph starts at (−2, 0) instead of (−2, 3).

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says for a linear function F of X, defined on the closed interval from -3 to 7, sketch the function given the following conditions. One, the graph of F of X is composed of connected line segments. 2, the initial point of the graph is -3.0. And 3, the derivative of FX is represented by a step function, and this function is plotted in the graph below the problem. And so on this graph, F vex begins at an open point of minus 3.1, and the first segment ends at the open point at 0.1. The second segment begins at a close point at 0. 2.5 and goes to the open point at 2. 2.5. The next segment. Begins at the close of 2.0 and goes to the open point at 5.0. And the final segment on this graph begins with the close point at 5.2.5 and ends at the open point of 7.2.5. We're also given an empty graph on which to plotter function. So we need to create a sketch over a graph of F of X. Given our three conditions. So, we're given the initial point of our graph, which is -3.0. So, we'll go ahead and plot that point on our graph. And to get the next point, we're going to need to look at our graph of FXx. And for FXx, we see that over the interval, going from X equal to -3 to X equal to 1, the FXx has a value of 1. So that means that our graph F of X is going to have a slope of 1 between X equal to -3 and X equal to 0. So to get a slope of 1 on our graph, we'll have to move to the right 3 units, but since we have a slope 1, we'll also need to move up by 3 units. So the next point on our graph of FX is going to occur at 0.3. So we're going to connect those two points with the lines, so going from -3.0 to the point of 0.3, and that's because our graph is composed of connected line segments. So the next segment, is going to begin at 0.3, and to get the next point, we'll need to look at our graph of FX again. And here over the ittal going from 0 to 2, we see that FX has a value of -2.5. So that means we're going to have a slope of -2.5 for the next line segment. And so since we're going 2 units to the right, that means we'll need to go down 5 units. So, the next point is going to be located at 2.2. And so we're going to connect the point of 0.3. To the point of 2-minus 2 with a straight line. Now for the next point that we need to find, we'll again look at FX since the next segment is going to begin at 2 minus 2, and over the interval from X equal to 2 to X equal to 5, we see that FXx has a value of 0. So that means that our slope is 0, so we should have a horizontal line on our graph of F of X. So, we're going to draw a horizontal line starting at 2 minus 2 and going to the point. Of 5 minus 2, since our slope of our graph here is 0 over this interval. And now, for the final segment for a graph of F of X. We'll again look at F of X, and over the interval going from X equal to 5 to X equals to 7, we see that FX has a value of 2.5. So this means that our next segment should have a slope of 2.5. And since we're going to the right by 2 units, we'll need to go up. By 5 units. So, the end point for the next segment is going to be at 7.3. So we're gonna draw for a final line segment, beginning at the point of 5.2, going to the point of 7.3. And so the graph that we have drawn on the screen here is going to be the answer to this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Recovering a function from its derivative

b. Repeat part (a), assuming that the graph starts at (−2, 0) instead of (−2, 3).

Key Concepts

Antiderivative

Initial Condition

Constant of Integration

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