- #1
How to Solve a Cubic Equation Involving Derivatives and Substitutions?
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Homework Help Overview
The discussion revolves around solving a cubic equation that involves derivatives and substitutions. Participants are examining critical points derived from the first derivative of a function, with specific focus on the conditions under which these points occur.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants are discussing the values of x for which the first derivative f' equals zero, noting discrepancies in their findings. There is speculation about potential errors in the problem statement and the implications of the discriminant on the number of solutions. Some participants suggest that substituting values should be approached carefully and at a later stage in the problem-solving process.
Discussion Status
The discussion is active, with participants providing insights and questioning the original problem's integrity. Some have offered guidance on the implications of the discriminant and the substitution of variables, while others are still exploring the critical points and their conditions.
Contextual Notes
There is mention of a potential typing error in the problem statement, which could affect the interpretation of the cubic equation. Additionally, participants express concerns about the appropriateness of posting images instead of text for the problem and solutions.
- #2
Buzz Bloom
Gold Member
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Hi Ameer:
I also found two values for x for which f'=0, but my values are different than yours. Might you have copied the problem incorrectly?
Regards,
Buzz
I also found two values for x for which f'=0, but my values are different than yours. Might you have copied the problem incorrectly?
Regards,
Buzz
- #3
Mark44
Mentor
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- 10,644
Since the problem is shown as an image, I don't see how it could have been copied incorrectly. I'm wondering if there is a typo in the problem itself. I too get two values for critical points. The only way that there will be only one solution for x in solving for f'(x) = 0, is the the discrimant has to be zero. IOW, ##4c^2 - 24m = 0##. If ##c^2 = 8m##, the discriminant is ##32m - 24m = 8m##.Buzz Bloom said:
One comment about the OP's work: if ##c^2 = 8m##, then it's not necessarily true that ##c = 2\sqrt{2m}##. Corrected, it would be ##c = \pm 2\sqrt{2m}##.
- #4
Mark44
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@Ameer Bux, it is frowned on here to post only images of the problem and your work. All of the work you showed can be written directly in the text window using LaTeX. We have a tutorial here: https://www.physicsforums.com/help/latexhelp/
- #5
Ameer Bux
- 19
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Mark44 said:
I'm sorry. I'll check your link and won't post it like I did the next time. Thank you
- #6
Ameer Bux
- 19
- 0
Hi people. I emailed my teacher and he said that : C² = 6M
There was a typing error on the page.
There was a typing error on the page.
- #7
Ameer Bux
- 19
- 0
- #8
Ameer Bux
- 19
- 0
Got it, thanks guys
- #9
Mark44
Mentor
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It's a lot simpler to NOT substitute for c^2 until later.Ameer Bux said:
If f'(x) = 0, then ##x = \frac{2c \pm \sqrt{4c^2 - 24m}}{12}##. Now, replace ##c^2## with 6m.
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