Graph of f' using graphical methods

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Homework Help Overview

The discussion revolves around the graphical representation of the derivative function f' in relation to the original function f, specifically focusing on the end behavior and characteristics of these graphs. Participants are analyzing the implications of changes made to the graph of f' and how these relate to the properties of f.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the accuracy of altered graphs of f' and discussing the implications of steepness and asymptotic behavior. There are attempts to clarify the relationship between the order of the functions and their derivatives, as well as the effects of gradient changes on the graph.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning each other's interpretations. Some guidance has been offered regarding the estimation of derivatives through graphical methods, but no consensus has been reached on the correctness of the graphs presented.

Contextual Notes

Participants are navigating assumptions about the nature of the functions involved, including their orders and behaviors at extremes. There is mention of homework constraints that may limit the exploration of certain aspects of the problem.

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Homework Statement
Please see below
Relevant Equations
Please see below
For this Problem 5,
1682903343264.png

The solution is,
1682903366977.png

However, I though the graph of f' would have end behavior more like,
1682903523058.png

Does someone please know whether I am correct?

Many thanks!
 

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ChiralSuperfields said:
Dose
 
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Thank you @Mark44, I have fixed the description now.
 
ChiralSuperfields said:
Does someone please know whether I am correct?
I don't think you are correct. Why did you change the graph of f with the red dotted lines that are close to vertical?

In your altered graph of f', the parts you added look way too steep to me.
 
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Mark44 said:
I don't think you are correct. Why did you change the graph of f with the red dotted lines that are close to vertical?

In your altered graph of f', the parts you added look way too steep to me.
Thank you for your reply @Mark44!

Do you please agree with the solutions then?

Many thanks!
 
ChiralSuperfields said:
Do you please agree with the solutions then?
Yes, pretty much. You could confirm that their graph of f' looks reasonable by tracing the graph of f on some graph paper, and then using a straight-edge at a number of points on the graph to estimate the derivative, and then plotting each of these estimates.
 
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f looks more like 4th order equation and f' looks like a 3rd order equation by counting the inflection points.

Your dotted line adds to more orders to change the direction of the asymptote of the derivative.

4. is a negative cosine function so that f' is easy to find.
 
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You can expect the graph of the function to keep having more and more slopes at the extremes (negative on the left and positive on the right. Both the book answer and your answer have f' going more negative for negative x and more positive for positive x. So the next question is whether the slopes of f(x) change faster at the extremes. That is where I think the book answer is better. Notice that for x going more negative, the function has less curve and becomes straighter. That means that f' is not changing as fast (it is still becoming more negative, just slower). Likewise, on the right end, as x goes more positive, the function graph also becomes straighter. It is still sloped positively, and more positively as x gets larger, but the change is not as fast. That means that on the right side, f' keeps increasing positive, but not as fast as x gets larger.
That is why I prefer the book answer.
 
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ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For this Problem 5,
View attachment 325718
The solution is,
View attachment 325719
However, I though the graph of f' would have end behavior more like,
View attachment 325720
Does someone please know whether I am correct?

Many thanks!

One thing that may be helpful to ponder is how the gradient, ##m##, affects the line ##y=mx##. When ##m=1## the line will be on a ##45°## angle. However, when the gradient is doubled to ##m=2##, the angle does not double with it to ##90°##, rather the angle changes to ##63°##. When ##m=3##, the angle is ##72°##. Why does this matter? Well in the graph of ##f## we are examining the gradient of the tangents, and near the endpoints of ##f## these tangent lines are becoming closer and closer to being on the same angle. Because there is less and less increase of the gradient, the function value of ##f'## which represents the gradient, is also having less and less increase. Hence why ##f'## is flattening out.
 
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  • #10
TonyStewart said:
f looks more like 4th order equation and f' looks like a 3rd order equation by counting the inflection points.

Your dotted line adds to more orders to change the direction of the asymptote of the derivative.

4. is a negative cosine function so that f' is easy to find.
Furthermore in 5. (to explain why you are incorrect)

To extend f you cannot add another change in slope rather assume it is approaching a asymptotic slope so the derivative f' approaches a constant -y on the left and constant =Y on the right.

on 4. it could be a partial trig function for 1 cycle or it could also be a 4th order or more polynominal with the same shape.
 
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