- #1
How to Solve a Cubic Equation Involving Derivatives and Substitutions?
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SUMMARY
The discussion centers on solving a cubic equation involving derivatives and substitutions, specifically addressing the critical points where the derivative f' equals zero. Participants identified a typing error in the problem statement, clarifying that C² should equal 6M instead of the previously assumed values. The correct approach involves substituting c² with 6m after determining the critical points using the formula x = (2c ± √(4c² - 24m)) / 12. The conversation emphasizes the importance of accurate problem representation and the use of LaTeX for clarity.
PREREQUISITES- Understanding of cubic equations and their derivatives
- Familiarity with critical points and the discriminant in calculus
- Knowledge of substitution methods in algebra
- Proficiency in LaTeX for mathematical formatting
- Learn how to apply the discriminant to determine the nature of roots in cubic equations
- Study the process of finding critical points in calculus
- Explore substitution techniques in solving polynomial equations
- Practice using LaTeX for formatting mathematical expressions in online forums
Students studying calculus, particularly those tackling cubic equations and derivatives, as well as educators and tutors looking for effective problem-solving strategies in mathematics.
- #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
- 19
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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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