Câu trả lời hay nhất:
This has been a VERY interesting problem. Thanks for posting it! :)
"I found y⸍ = 3mx² – 6mx + (2m+1) .... ✔ agree
For f(x) to have both max and min, f ⸍(x) should have two (REAL) solutions, so ▲> 0 .... ✔ agree
For▲> 0, m(12m + 12) > 0" .... ✘ disagree
There is very small, but important, error in the last line. It should say:
For▲> 0, m(12m – 12) > 0
That small sign error changes a lot!
So either both m and (12m – 12) are (BOTH) positive, or (BOTH) negative.
Case ➊: (BOTH) positive
m > 0, AND 12m – 12 > 0 ➞ m > 1
➞ m > 1
Case ➋: (BOTH) negative
m < 0, AND 12m – 12 < 0 ➞ m < 1
➞ m < 0
Perhaps this small correction will be sufficient for you to come to your own solution....?
However, I have been playing with software that dynamically presents the graphs of f(x), f⸍(x), and the lines formed by the local max, local min, and the origin, based on sliding values of "m" that I can control (to whatever # of sig. figs I like.)
From this messing around, I have a solution for "m", (m is exactly = 10), and an I am now going to to try to solve the puzzle algebraically, and get my solution posted here as an addendum before the deadline expires. (Perhaps an extension of 2 days may give a chance for others to respond as well?)
Hope this is helpful! Cheers! (for now!) :)