From inspection it can be seen to accept x=1 as a root. At that point, calculus can be used to show that this is a minimum, hence the only positive root.
A very very close approximation to the negative x solution of this equation is x = -1/sqrt(1+log_2(3-cbrt(1/4))). This can be found by re-writing the equation as 2^x + 2^(x^(-2)-1) = 3 and then approximating 2^x as a constant near x=-2/3
The thing with AM-GM was really cool. Never ever would I arrive at it myself :) I just analyzed functions f1(x) = 2^{x+1} and f2(x)=2^{1/x^2} (so that f(x) = f1(x) + f2(x)) separately for x>0. For x near 0, f approaches plus infinity like f2 (f1 simply approaches just a constant 2). So, just arround zero we have a monotonically decreasing function with negative derivative. For somewhat large positive x (x->+oo), f2 is negligible (appoaches just a constant 1), and the global behaviour is fully determined by the f1. So, somewhat remote from zero, f is monotonically increasing with positive derivative. So, there must be a point (say, x*) within (0,+oo), where derivative is zero, and because of these "local" monotonicities, this point can be only one. Then, the values of f(x) within that range will be from f(x*) up to +oo. There can be possibly three scenarios. If f(x*) > 6, we cannot have any real roots in that range. So, 0 positive roots. Imagine... Upon moving from x=0 to large x's, we are descending from infinity to f(x*) and then bouncing back up to the infinity never reaching 6. If f(x*)
I solved it for both solutions by using calculus to graph it and end behavior along with domain restriction aka x cannot equal 0. And then showed there’s only two solutions.
いつもいい問題をありがとう😊
From inspection it can be seen to accept x=1 as a root. At that point, calculus can be used to show that this is a minimum, hence the only positive root.
A very very close approximation to the negative x solution of this equation is x = -1/sqrt(1+log_2(3-cbrt(1/4))). This can be found by re-writing the equation as 2^x + 2^(x^(-2)-1) = 3 and then approximating 2^x as a constant near x=-2/3
I saw the positive answer immediately just looking at the problem.
The thing with AM-GM was really cool. Never ever would I arrive at it myself :) I just analyzed functions f1(x) = 2^{x+1} and f2(x)=2^{1/x^2} (so that f(x) = f1(x) + f2(x)) separately for x>0. For x near 0, f approaches plus infinity like f2 (f1 simply approaches just a constant 2). So, just arround zero we have a monotonically decreasing function with negative derivative. For somewhat large positive x (x->+oo), f2 is negligible (appoaches just a constant 1), and the global behaviour is fully determined by the f1. So, somewhat remote from zero, f is monotonically increasing with positive derivative. So, there must be a point (say, x*) within (0,+oo), where derivative is zero, and because of these "local" monotonicities, this point can be only one. Then, the values of f(x) within that range will be from f(x*) up to +oo. There can be possibly three scenarios. If f(x*) > 6, we cannot have any real roots in that range. So, 0 positive roots. Imagine... Upon moving from x=0 to large x's, we are descending from infinity to f(x*) and then bouncing back up to the infinity never reaching 6. If f(x*)
Thanks so much!
You bet!
@@SyberMath 🙏🙏
Very Graceful And Informative Video .Thank you Syber Math for your original and interesting ways of solving problems.Bravoo👏👏👏
Np! Thank you! 😍
thank you .very nice
That's a great insight to the problem!
You didn't draw the smiley face, though... I miss that.
Ooops!
I solved it for both solutions by using calculus to graph it and end behavior along with domain restriction aka x cannot equal 0. And then showed there’s only two solutions.
Gide how to solve problems like this:
1. Try to do it with a cases 0, 1, 2
2. With probability 99% you are found an answer
:)
The "x=1" was by inspection, but I couldn't figure out easily if there might be other solutions.
Shouldn't it be 2^x.2 + 2^(1/x^2) in the first line?
2 times 2^x he wrote it as 2^x + 2^x
Genial la manera original de resoldrr aixœ
I could see it from the start.
such equations do have a tendency to have a very guessable root
For the second am gm x has to be positive
x has to be positive already
It is
Nice one
Thanks 🔥
x = 1 is an obvious solution. Take the RHS to the LHS and differentiate, it’s always positive, ie unique solution.
Wow
x = 1
Of course x=0 duh
Ahahaha!
This expression is undefined!
Fantastic Maths
Give me heart and pin please
x=1