Relative extrema problem for Essentials of Calculus

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Homework Help Overview

The discussion revolves around finding the relative extrema of the function f(x) = 1 - x^(2/3). Participants are tasked with identifying extremum points and sketching the graph of the function.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the derivative of the function and its implications for finding extrema. There is an exploration of points where the derivative is zero and consideration of other potential extrema points. Questions arise about the identification of extreme points and the reasoning behind them.

Discussion Status

Some participants have expressed understanding of the extremum at (0, 1), while others question the reasoning behind this identification. There is a mix of attempts to clarify the conditions under which extrema can occur, with some guidance provided on the nature of critical points.

Contextual Notes

Participants note that the original problem requires listing extrema and sketching the graph, but there is some confusion regarding the conditions for identifying these points, particularly concerning the derivative being zero.

Nawz
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Homework Statement



Find the relative extrema of each function, if they exist. List each extremum along with the x-value at which it occurs. Then sketch a graph of the function.

25. f(x)=1-x2/3

Homework Equations





The Attempt at a Solution



f prime of (x)= -2/3x-1/3

-2/3x-1/3= 0


??
 
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Nawz said:

Homework Statement



Find the relative extrema of each function, if they exist. List each extremum along with the x-value at which it occurs. Then sketch a graph of the function.

25. f(x)=1-x2/3

Homework Equations





The Attempt at a Solution



f prime of (x)= -2/3x-1/3

-2/3x-1/3= 0
??
There are no solutions to the equation f'(x) = 0. Extreme values can occur at points other than where the deriviative is zero. What are these other points?
 
Thanks for the help, I figured it out.

It was (x)=0 for f'(x) , idk why I couldn't see that. so when i put that in the original equation i got 1. So the x point was (0,1) . And then I found the remaning points on the graph and it was correct with the back of the book answer.

Thank you


If i have another questions, can i just ask here or should i make a new topic? It's the same type of questions just another problem?
 
Last edited:
Nawz said:
Thanks for the help, I figured it out.

It was (x)=0 for f'(x) , idk why I couldn't see that. so when i put that in the original equation i got 1. So the x point was (0,1) . And then I found the remaning points on the graph and it was correct with the back of the book answer.
But x = 0 is not a solution of f'(x) = 0, and that's what you were looking for. Do you understand why x = 0 is an extreme point?
Nawz said:
Thank you


If i have another questions, can i just ask here or should i make a new topic? It's the same type of questions just another problem?
Please start a new thread.
 
Mark44 said:
But x = 0 is not a solution of f'(x) = 0, and that's what you were looking for. Do you understand why x = 0 is an extreme point?

Please start a new thread.

No i don't know what you mean. I came back from class but I remember this problem still. (0,1) was the realtive maximum at least I am fairly sure. I don't need a solution, i just needed the relative extrema which is (0,1) and then sketch... that's right, isn't it?
 
Yes, (0, 1) is the maximum point. An extreme point of a function f can occur at any of the following:
1. Points at which f'(x) = 0
2. Points in the domain of f at which f' is undefined
3. Endpoints of the domain of f

The work you showed in your first post was toward the first point above, but for your function, there are no values x for which f'(x) = 0. It wasn't clear to me that you understood where to look for extreme values.
 

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