How to Solve a Cubic Equation Involving Derivatives and Substitutions?

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Homework Statement


Refer to attached image. Please help. It's due tomorrow.

Homework Equations


Doesn't state

The Attempt at a Solution


Refer to attached image
 

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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
 
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Buzz Bloom said:
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?
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##.

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}##.
 
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Mark44 said:
@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/

I'm sorry. I'll check your link and won't post it like I did the next time. Thank you
 
Hi people. I emailed my teacher and he said that : C² = 6M

There was a typing error on the page.
 
IMG_0759.JPG
 
Got it, thanks guys
 
Ameer Bux said:
It's a lot simpler to NOT substitute for c^2 until later.
If f'(x) = 0, then ##x = \frac{2c \pm \sqrt{4c^2 - 24m}}{12}##. Now, replace ##c^2## with 6m.
 
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