Graph of f' using graphical methods

  • Thread starter member 731016
  • Start date
  • Tags
    Graph
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.
  • #1
member 731016
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!
 

Attachments

  • 1682903312357.png
    1682903312357.png
    3.8 KB · Views: 85
Last edited by a moderator:
Physics news on Phys.org
  • #2
ChiralSuperfields said:
Dose
 
  • Like
Likes member 731016
  • #3
Thank you @Mark44, I have fixed the description now.
 
  • #4
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.
 
  • Like
Likes member 731016
  • #5
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!
 
  • #6
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.
 
  • Like
Likes member 731016
  • #7
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.
 
  • Like
Likes member 731016
  • #8
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.
 
  • Like
Likes member 731016
  • #9
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.
 
  • Like
Likes member 731016
  • #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.
 
  • Like
Likes member 731016

FAQ: Graph of f' using graphical methods

What is the significance of the graph of f'?

The graph of f' (the derivative of a function f) represents the rate of change or the slope of the original function f at any given point. It provides insights into the behavior of the function, such as where it is increasing or decreasing, and the locations of any local maxima or minima.

How can you identify critical points using the graph of f'?

Critical points of the original function f occur where its derivative f' is zero or undefined. On the graph of f', these points are where the curve crosses the x-axis (f'(x) = 0) or where there are discontinuities. These critical points indicate potential locations of local maxima, minima, or points of inflection.

What does it mean when the graph of f' is above or below the x-axis?

When the graph of f' is above the x-axis (f'(x) > 0), the original function f is increasing at those intervals. Conversely, when the graph of f' is below the x-axis (f'(x) < 0), the original function f is decreasing. The sign of f' indicates the direction of change of the function f.

How do you determine concavity using the graph of f'?

The concavity of the original function f can be determined by looking at the slope of the graph of f'. If the graph of f' is increasing (positive slope), then the original function f is concave up. If the graph of f' is decreasing (negative slope), then the original function f is concave down. Points where the slope of f' changes sign are inflection points of f.

How can you use the graph of f' to sketch the graph of the original function f?

To sketch the graph of the original function f from the graph of f', start by identifying critical points where f' is zero or undefined. Determine the intervals where f' is positive or negative to find where f is increasing or decreasing. Analyze the concavity by looking at the slope of f'. Use this information to piece together a rough sketch of f, ensuring continuity and smooth transitions based on the behavior indicated by f'.

Similar threads

Back
Top