While I'm not at all surprised that it wasn't mentioned, since it wasn't especially elegant, I did this problem a third way. When I saw f(x) - f(y) it immediately suggested (to me at least) a derivative in the limit that the two points are infinitesimally close. So I put x = y + epsilon and took the limit as epsilon --> 0, giving a DE with the solution of a family of curves passing through (1,2). The condition on f'(1) then lets you pick out the right curve.
@@SteveGottaGoFast he says its defined for,ALL real numbers so that means it is,continuous everywhere..so,you can,take a general,form of,the derivative,and plug,in the point you k ow exists for 1 which equals 2018. Even if,its,not,differentials,everywhere. That would,be correct,and thus,a valid approach.
10:01 You don't have to assume that they can't be 1. Having your denominator equal 0 just means that your numerator equals 0. This is particularly useful since you don't have to go back and plug in the 1 at the start.
@@BigDBrian In general, if you divide by 0, you don't get exactly 1 computational result - you may get a couple or infinitely many or none at all. This specific problem has a couple of reasons why we'll get only 1 result. Probably the simplest that you'll find most appealing is that the problem description also included a derivative at 1 - telling us that it's continuous at 1. That's probably my least favorite reason, but hey if it works.
@@BigDBrian Division is the inverse of multiplication. If you have ab=c, then you have c/b=a. So, if you have a0=0, then you have 0/0=a. Since most mathematical contexts have it so that a can be any number, it holds for any number a. The only way to narrow it down to a finite or singular amount of numbers a is to impose restrictions on a. This is often done in calculus with limits of expressions. While I'm here, I might as well proselytize you to how I do math with numbers. Basically, I work in a context where numbers aren't points, instead they are intervals bounded by addition of some set of numbers that is non archemedian smaller than the system we are working with. It provides a better framework for what would otherwise be the limit and does so with only basic algebra. Oh, and it makes the problems with 0 more visible when they appear and don't appear.
I also saw that this resembled a differential (and the f' in the question), so i said y = x+Δx. Since it has to be true for all x,Δx, and you get a differential equation with f'(x) and f(x), which I failed to solve, because it's still some weeks until we learn them in university, but I thought this was the solution.
There is something missing..okay we know that f'(1) exists but we dont know that f'(x) exists for all nonzero real values..So since we dont know it from the beggining how do we differentiate?
You are partially correct. That is the reason why you cant differentiate (in terms of the one variable) right from the beginning. But while solving the problem we conclude that the parametric solutions of f(x) without any constraints is f(x)=c+(2-c)/x RHS is differentiable thus f is differentiable for all x except 0
Or,another point how do we know if f of one that derivative is f of x or f of y..you assumed it was f of x without justifying it..it,could,as, easily be f of y...not enough info to solve the problem, I think
@@Azouzazu but my point: he says derivative of f at one is 2018..but is that f of x or y..he doesn't specify so not enough info to solve the problem! Am I the only one who caught that mistake?!
Both the solutions were amazing.
While I'm not at all surprised that it wasn't mentioned, since it wasn't especially elegant, I did this problem a third way. When I saw f(x) - f(y) it immediately suggested (to me at least) a derivative in the limit that the two points are infinitesimally close. So I put x = y + epsilon and took the limit as epsilon --> 0, giving a DE with the solution of a family of curves passing through (1,2). The condition on f'(1) then lets you pick out the right curve.
You can't do that since you don't know that the function is derivable or continuous everywhere, so it might not have a limit
@@SteveGottaGoFast he says its defined for,ALL real numbers so that means it is,continuous everywhere..so,you can,take a general,form of,the derivative,and plug,in the point you k ow exists for 1 which equals 2018. Even if,its,not,differentials,everywhere. That would,be correct,and thus,a valid approach.
@@leif1075 defined for all real numbers doesn't mean continuous. Take f(x)={x when x=0}. It is defined for all reals but discontinuous at x=0.
@@SteveGottaGoFast yea but zero is a number, so in your example it is NOT defined for all real numbers , thats a contradiction.
@@leif1075 my example IS defined for all real numbers
Ahhh I was so close, what a nice problem, it's seems Easy but kind of tricky
Nice approach...
why don't we differentiate the equation in terms of x, y will be constant and we know f(1)=2
The function isnt neccesarily derivable everywhere
Wait WAIT how do we know if f prime of one refers to f of x or,f of y..he doesn't specify..how can you solve without knowing this??
@@leif1075 its just the name of the variable
Try to differentiate it, you wil get 0=0 not useful information
10:01 You don't have to assume that they can't be 1. Having your denominator equal 0 just means that your numerator equals 0. This is particularly useful since you don't have to go back and plug in the 1 at the start.
the step to divide by (x-1)(y-1) is invalid if either of the terms are 0
@@BigDBrian
In general, if you divide by 0, you don't get exactly 1 computational result - you may get a couple or infinitely many or none at all.
This specific problem has a couple of reasons why we'll get only 1 result. Probably the simplest that you'll find most appealing is that the problem description also included a derivative at 1 - telling us that it's continuous at 1. That's probably my least favorite reason, but hey if it works.
I don't know what you mean by getting (various amounts of) computational results
@@BigDBrian
Division is the inverse of multiplication. If you have ab=c, then you have c/b=a. So, if you have a0=0, then you have 0/0=a. Since most mathematical contexts have it so that a can be any number, it holds for any number a. The only way to narrow it down to a finite or singular amount of numbers a is to impose restrictions on a. This is often done in calculus with limits of expressions.
While I'm here, I might as well proselytize you to how I do math with numbers. Basically, I work in a context where numbers aren't points, instead they are intervals bounded by addition of some set of numbers that is non archemedian smaller than the system we are working with. It provides a better framework for what would otherwise be the limit and does so with only basic algebra. Oh, and it makes the problems with 0 more visible when they appear and don't appear.
Thanks! It was really interesting to see the two different techniques and both were very good to learn from
That symmetry solution made a lot more sense than the substitution did to me. Really nice problem, I'm quite bummed I didn't try it.
Amazing, thanks for sharing this
Look at the description: Congratulations to Minh Cong Nguyen, ..., *Kim Jong Un*, ...
Wow.
Hi teacher I love this lesson but my English is bad at listening, Can teacher give some record’s pdf for me, please
Why f(-1) is a constant
I also saw that this resembled a differential (and the f' in the question), so i said y = x+Δx. Since it has to be true for all x,Δx, and you get a differential equation with f'(x) and f(x), which I failed to solve, because it's still some weeks until we learn them in university, but I thought this was the solution.
who used to exam IMO 2018? please show name
There is something missing..okay we know that f'(1) exists but we dont know that f'(x) exists for all nonzero real values..So since we dont know it from the beggining how do we differentiate?
You are partially correct. That is the reason why you cant differentiate (in terms of the one variable) right from the beginning.
But while solving the problem we conclude that the parametric solutions of f(x) without any constraints is f(x)=c+(2-c)/x
RHS is differentiable thus f is differentiable for all x except 0
Or,another point how do we know if f of one that derivative is f of x or f of y..you assumed it was f of x without justifying it..it,could,as, easily be f of y...not enough info to solve the problem, I think
@@Azouzazu but my point: he says derivative of f at one is 2018..but is that f of x or y..he doesn't specify so not enough info to solve the problem! Am I the only one who caught that mistake?!
Expensive