Sketch a continuous function f on some interval that has the p...

Question

Sketch a continuous function f on some interval that has the properties described. Answers will vary.

The function f satisfies f'(-2) = 2, f'(0) = 0, f'(1) = -3 and f'(4) = 1.

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the values of the following using the graph of F of X. F of -2, F-1, F 0, F1 and F of 3, where F represents the derivative of F of X. And here we have our graph showing F of X. Now how are we going to determine all of these derivatives? What do we know about the derivative on a graph? Well, recall that the derivative of FFX really just represents the slope of FFX at particular points, OK? So even if we don't know what the function is, just by looking at the graph, we can determine the slopes at each of these points. So for example, let's start at The point where X is -2, OK. Now if we go to X being negative 2 here, what do we notice? Well, we notice that the line, OK, there's a line segment here, OK, where FFX on -3, 1 is a line segment, and thus, and that goes up all the way to where FFX or sorry, where X is -1 and Y is 1. So in this case here, that means the value of F of -2 will be the slope of this line. No, what's its slope? Well, notice here, the slope is the change in the Y direction divided by the change in the X direction. OK? So in this case, notice that it goes up by how many units 123456, OK. And then it goes across by 12 units, OK. So, the slope is 6 divided by 2, which is 3 units. And that makes sense because what it's telling us is that for every 1 unit increase in X, Y increases by 3 units and we can see that here. Now let's apply the same idea to the rest of the derivatives. How about the derivative of. F of -1. Well, here, when we go to the 0.-1 OK, what do you notice now? Well, there exists a sharp point. The left hand slope and the right hand slope are not equal. So this tells us that this slope, this derivative rather, is undefined. And if we go to F of 0, OK. F of 0. Here, notice that we have a slope. So no, I think I can put my line here. We have a slope, OK, where X equals 0. And no, notice our line is horizontal. So that means the slope equals 0, OK? Next, when we go to F1. This is in a similar scenario to F of -1. There exists a sharp turn here where the left hand slope is not equal to the right hand slope. Therefore, this function, or is this derivative is also undefined. Finally, let's go to F3. So this is where X is 3, so that point will be right here. Now, notice that the point at X equals 3 is a local minimum, and thus the tangent line to the slope is going to be horizontal, which means that its derivative will be equal to 0. Thus, F-2 is 3 units, F-1 and 1 are undefined, F 0 is 0, and F 3 is also 0. Thanks a lot for watching everyone. I hope this video helped.

Sketch a continuous function f on some interval that has the properties described. Answers will vary.

The function f satisfies f'(-2) = 2, f'(0) = 0, f'(1) = -3 and f'(4) = 1.

Key Concepts

Derivative

Critical Points

Continuity

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