- #1
How to Solve a Cubic Equation Involving Derivatives and Substitutions?
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The discussion revolves around solving a cubic equation involving derivatives and substitutions, with a focus on finding critical points where f'(x) = 0. Participants identified discrepancies in their calculated values for x, suggesting a potential typo in the problem statement. It was clarified that the discriminant must equal zero for a single solution, leading to the equation 4c^2 - 24m = 0. A participant noted that substituting c^2 should be done later in the process for simplicity. Ultimately, the issue was resolved with confirmation of a typing error, leading to the correct substitution of c^2 with 6m.
- #2
Buzz Bloom
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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
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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
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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
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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
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- #8
Ameer Bux
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Got it, thanks guys
- #9
Mark44
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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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