I solved it this way: The function is of the form f(u) = u - 4u^2 where u = 1/(x*3^x). f(u) is an inverted parabola with intercepts at u = 0 and u=1/4. It‘s maximum therefore lies at u = 1/8. f(1/8) = 1/8 - 4/64 = 1/8 - 1/16 = 1/16. So that‘s the maximum value. One still has to check whether u = 1/8 can be attained. This is equivalent to x*3^x = 8. The solution lies between x=1 and x=2. So the maximum can be attained.
You’re doing great! Keep going! I’d suggest zooming the camera in a bit when filming these so that it’s only you+the whiteboard in frame. Otherwise there’s a lot of wasted wall space and the writing is therefore smaller.
Oooohhhh the AM≥GM inequality is one of my favourite methods to solve function-range questions. But then there are those cases where it fails, and i hate those. Could we perhaps get a video on that as well? For example, when we need to find maximums/minimums of expressions of the form asec(x)+bcosec(x)+c Amazing video!
I simplified by factoring. Substituted y = x^-1 * 3^-x to get y(1-4y^2). Solved for vertex so where max occurs at y =1/8. Solved for x using lambert w function. Max occurs at x = W(8ln3)/ln3. Plug into original expression to get 1/16. Not as practical!
I have been subscribed to you for a few weeks now! You are doing an excellent job and I am rooting for you!
I admire both of you🙏🏻
the goat is here damn
thanks for sharing this video with us!
Thanks for sharing video !
Only because you resent whites.
I solved it this way: The function is of the form f(u) = u - 4u^2 where u = 1/(x*3^x). f(u) is an inverted parabola with intercepts at u = 0 and u=1/4. It‘s maximum therefore lies at u = 1/8. f(1/8) = 1/8 - 4/64 = 1/8 - 1/16 = 1/16. So that‘s the maximum value.
One still has to check whether u = 1/8 can be attained. This is equivalent to x*3^x = 8. The solution lies between x=1 and x=2. So the maximum can be attained.
Very nice!
You’re doing great! Keep going! I’d suggest zooming the camera in a bit when filming these so that it’s only you+the whiteboard in frame. Otherwise there’s a lot of wasted wall space and the writing is therefore smaller.
Thank you! Will do.
Yup
Clear explanation with a very satisfying result
that was so cool!
Great video, intuitive and instructive!
Very nice way of explaining 👍
Oooohhhh the AM≥GM inequality is one of my favourite methods to solve function-range questions.
But then there are those cases where it fails, and i hate those. Could we perhaps get a video on that as well? For example, when we need to find maximums/minimums of expressions of the form asec(x)+bcosec(x)+c
Amazing video!
For cases where it fails, perhaps try another inequality. Cauchy often deals with trig expressions nicely :)
This is pretty epic
This is a really cool concept and I think you explained it very clearly :)
Cool vid
Outstanding!
Makes perfect sense!
The max value is when x = W(8*ln3)/ln3
My teacher taught this method as well. I forgot about it.
Man i solved the eqn at the end but it was too smart to use the fact that x3^x is continuous
Nice
I simplified by factoring. Substituted y = x^-1 * 3^-x to get y(1-4y^2). Solved for vertex so where max occurs at y =1/8. Solved for x using lambert w function. Max occurs at x = W(8ln3)/ln3. Plug into original expression to get 1/16. Not as practical!
Bro are you on discord, I had some great math problems for discussion😅
you can write some here :)
every master has its cheap copy!