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Your approach is so elegant! I went through the process of finding all the inflection points of f(x) by setting f'(x) = 0, which came up with cos2(x) = 0, cos2(x) = 1, or cos2(x) = 2/3, and then I had to plug them all back in to find which one was the minimum.
Hi, the intended method is the mark scheme is even simpler. As you said this is a hidden quadratic, let y=cos²x, you end up with 9y² - 12y +7, the complete square, 9(y-2/3)²+3. Minimum is 3, when y=2/3 like you had.
Yes I mentioned that this can be done more directly, but I find completing the square to be more work than taking a derivative. Which means more room for mistakes!
I would love to do a format like that too! It’s getting really hard to find the time to record even these short problems though. I’ve done some longer format things in the past (contest papers rather than entrance exams) but that was back when I wasn’t so busy with real life stuff.
The function I’m taking a derivative of is a parabola with a positive coefficient, so it’s definitely a minimum and not a maximum. Maybe I could have mentioned this while going through the work.
