# Graph of f' using graphical methods

• ChiralSuperfields
In summary: The 4th order polynomial would have the shape of a cosine.In summary, the graph of f' looks to be more like a 4th order polynomial and the solution is the same.
ChiralSuperfields
Homework Statement
Relevant Equations
For this Problem 5,

The solution is,

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

Does someone please know whether I am correct?

Many thanks!

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ChiralSuperfields said:
Dose

ChiralSuperfields
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.

ChiralSuperfields
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.

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.

ChiralSuperfields
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.

ChiralSuperfields
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.

ChiralSuperfields
ChiralSuperfields said:

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.

ChiralSuperfields
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.

ChiralSuperfields

## 1. What is a "Graph of f' using graphical methods"?

A "Graph of f' using graphical methods" refers to a visual representation of the derivative of a function f. It shows the rate of change of the function at different points on the graph.

## 2. How is the "Graph of f' using graphical methods" different from the graph of the original function?

The "Graph of f' using graphical methods" is different from the graph of the original function because it shows the slope of the function at each point, rather than the actual value of the function itself. It is a way to analyze the behavior of the function and understand its rate of change.

## 3. What do the different points on the "Graph of f' using graphical methods" represent?

Each point on the "Graph of f' using graphical methods" represents the slope of the function at that specific point on the graph. The steeper the slope, the greater the rate of change of the function at that point.

## 4. How can the "Graph of f' using graphical methods" be used to find the critical points of a function?

The critical points of a function can be found by looking at the "Graph of f' using graphical methods" and identifying the points where the slope of the function is equal to zero. These points represent the maximum or minimum values of the function.

## 5. Are there any limitations to using graphical methods to analyze the "Graph of f' using graphical methods"?

While graphical methods can provide a visual understanding of the derivative of a function, they may not always be accurate or precise. In some cases, it may be necessary to use analytical methods to find the exact derivative of a function.

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