This made me tear up. I first started taking math seriously a year ago. Now i feel like i sometimes get to peak a little under the surface of the ocean, and begin to see the most beautiful patterns of logic in the world.
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.
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.
Love this, thanks for also diving into the expressions in the numerator needing to both be negative or both positive and showing possible ways that equation can sometimes be confusing! Great job!!
Great job. Thanks for sharing your method. By the way, the derivative is an excellent way to go, and you'll get two local extrema, and one of these local extrema turns out to be a global maximum at x3^x = 8. Solving for x may be difficult, but finding the maximum using x3^x=8 will yield a maximum of 1/16.
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!
alternate: you can factor then cancel a factor of x^2, then letting y = x3^x it is (y - 4)/y^2, say, equals 1/k (makes it easier later) for some k then y^2 - ky + 4k must have real solutions so discriminant k^2 - 16k >= 0 getting k = 16
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.
This made me tear up. I first started taking math seriously a year ago. Now i feel like i sometimes get to peak a little under the surface of the ocean, and begin to see the most beautiful patterns of logic in the world.
Math is everywhere, even in your name 😅
@@felipedutra5276 😂😂
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
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!
Love this, thanks for also diving into the expressions in the numerator needing to both be negative or both positive and showing possible ways that equation can sometimes be confusing! Great job!!
Great job. Thanks for sharing your method. By the way, the derivative is an excellent way to go, and you'll get two local extrema, and one of these local extrema turns out to be a global maximum at x3^x = 8. Solving for x may be difficult, but finding the maximum using x3^x=8 will yield a maximum of 1/16.
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 :)
alternate: you can factor then cancel a factor of x^2, then letting y = x3^x it is (y - 4)/y^2, say, equals 1/k (makes it easier later) for some k
then y^2 - ky + 4k must have real solutions so discriminant k^2 - 16k >= 0 getting k = 16
been subsribed for the past month amazing videos
If you let y=x3^x then you can simplify to finding the max of f(y):=(y-4)/y^2. Let M be a value such that f(y)
The max value is when x = W(8*ln3)/ln3
Clear explanation with a very satisfying result
This is a really cool concept and I think you explained it very clearly :)
Man i solved the eqn at the end but it was too smart to use the fact that x3^x is continuous
Great video, intuitive and instructive!
Very nice way of explaining 👍
Great video!
My teacher taught this method as well. I forgot about it.
Outstanding!
helloo I have some math discoveries but I don't know where to start. Can I mail you?
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!
that was so cool!
Makes perfect sense!
Cool vid
This is pretty epic
4/3
Nice
Bro are you on discord, I had some great math problems for discussion😅
you can write some here :)
every master has its cheap copy!
Too hot topic just lacking some clarity in body language I hope u do it better next time