Sketching Antiderivatives: Tips and Links for Quick Refresh

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Homework Help Overview

The original poster is tasked with sketching the antiderivative of a line based on a graph that does not specify the function. The discussion revolves around understanding the relationship between derivatives and antiderivatives in the context of graphical representation.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the concept of derivatives as slopes and how to reverse that reasoning to find the antiderivative. There are questions about the nature of the function that results from integrating a linear function and the implications of critical points on the shape of the antiderivative.

Discussion Status

The discussion is ongoing, with participants providing insights and links to resources. Some are exploring the implications of critical points and concavity, while others are questioning their interpretations of the graph and the resulting function.

Contextual Notes

There is mention of specific values and critical points derived from the graph, but there is uncertainty regarding the correct identification of these points and their implications for the antiderivative. Participants are also navigating the challenge of visualizing the antiderivative from the given derivative information.

swears
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Wow, I totally forgot how to do these.

I have a graph of a line, but it does not say the function. I have to sketch the antiderivative.

Does anyone have any links or advice to start me out.

TIA
 
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Remember that the derivative dy/dx is the change in the y value per change in the x value. So the derivitave at a point is basically the slope of the line.

Now turn that reasoning around to figure out the antiderivative. If you have a plot of the slope of the line at all points, then you can figure out what the original function's line looked like. Quiz question -- what other thing do you have to add into the final antiderivative answer to make it correct?
 
Yeah, I can do the derivative if I see the graph. I just can't visualize that backwards in my head.

These are the values I'm dealing with: f'(x)| 0 | 1 | 0 | -1 | 0
............x |-1 | 0 | 1 | 2 | 3
 
What kind of function is the integral of a line? A parabola right. Well, were is the vertex of that parabola? You should think about what a critical point is. Maybe the first derivative test. Does any of this help?
 
It is going up until x=0, then goes down. It is concave down there.

The at x = 1 it change to concave up and at x = 2 it starts increasing.

So, I guess the critical points are x = 0, 1, 2
 
A critical point is where f'(x) = 0. If I am reading your notation right, then the critical points are x = -1, 1, 3. So if there are 3 critical points, then the [EDIT]antiderivative should be a function of [tex]x^4[/tex]. What does this kind of function look like?
 
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Well actually, f'(x) is only 0 at x= 0 and 2, but it switches concavity at 1, so I figured that would be one too. (guess not)

x^4, yeah I believe so, I don't have a calculator to confirm(working on it), but it has two parabolas in it.(1 upside down).
 
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Oh, Sorry, I had some trouble reading your points. Sounds like you are on the right track though.
 

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