# Challenge 67: An Almost Symmetric Functional Equation

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• Published on Nov 15, 2018
• Congratulations to Jaleb, Kevin Tong, Hiren Bavaskar, reynolds45, Bigg Barbarian, Rohit Agarwal, Aswini Banerjee, Ricardo Mogollon, Albanovaphi7, and aby p for successfully solving the last week's math challenge question! Jaleb was the first person to solve the question.
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Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.
For more Weekly Math Challenges:

• Zicar 10 months ago

I divided the first formula by xy (x and y can't be 0) and arrived at f(x) - f(y) = f(x) / y - f(y) / x - 2/y +2/x.
then i plugged in y = 2018 and arrived at the formula f(x) - f(2018) = f(x) / 2018 - f(2018)/x - 2/2018 + 2/x.
i derived everything and arrived at f'(x) = f'(x) / 2018 + f(2018)/x^2 - 2/x^2.
then i plugged in x = 1 and arrived at 2018 = 1 + f(2018) - 2 f(2018)-1 = 2018 f(2018) = 2019

• Yaman Sanghavi 10 months ago

Just put y=2 and rearrange to get f(x)
And use f'(1) to find the unknown constant f(2) in the expression of f(x)

• PRAKHAR AGARWAL 10 months ago

Partial Differentiate the equation with respect to x to get a new functional equation. Put y=1 to get f(1)=2.
Again partial different the new functional equation with respect to x. After canceling out the factor of (y-1) from the new differential equation, we get a differential equation in x.
Which can be solved to get f(x) =2020-(2018/x)
Put x=2018 to get 2019

• miner Zhao 10 months ago +1

just give x=1 (or y=1)we can get f(1)=2.then changing this equation form,we can get x.f(x)-2 over x-1 equal to y.f(y)-2 over y-1.as well we derivate this to X,we get f(x)+x.f'(x)=y.f(y)-2/y-1.so let x=1, y.f(y)-2/y-1=f(1)+1.f'(1)=2020,after that let y=2018,so f(2018)=2019

• Tamerlane 10 months ago

• Tamerlane 10 months ago

Let's transform equation into the following:
xf(x) (y-1) + 2x = yf(y) (x-1) + 2y
Let's take y = 2:
xf(x) +2x = 2f(2)(x-1)+4
Therefore f(x) = (2f(2)-2) + (4-2f(2))/x
But then
f'(1)=2f(2)-4=2018 => f(2) = 1011.
Now we know that
f(x) = 2020-2018/x
And therefore
f(2018)=2020-2018/2018=2019

• Leonardo Costa 10 months ago

Ans: 2019.
By rewriting the equation you find that (yf(y)-2)/(y-1) is constant, which lets you write f(x) as a-b/x. Now by differentiating and plugging x=1 into the equation, you're able to solve for the constant and finally have f(x)=2020-2018/x. Meaning f(2018)=2019

• Sébastien Louchart 10 months ago +4

Hello. I proved the continuity of f over its domain (x=t, y=a than lim(t->a) f(t) = f(a)). This done I felt more comfy to try to play with the functional identity. I proved f(1) = 2 simply plugging x=1 and y=2. I then derived the expression with respect to x to find f'(x) = -f(x)/x + yf(y)/y-1 * 1/x - 2/(y-1) * 1/x. evaluating for f'(1) = 2018 leads to f(y)= 2020(y-1)/y + 2/y hence f(2018) = 2019. I'm still working to prove derivability of f over its domain, though :/

• Sébastien Louchart 10 months ago

The proposed function is defined over R* and it can be proved it is continuous over R*. For God's sake it's even differentiable over R* ;)

• Sébastien Louchart 10 months ago

@Numer in that case. You're right. The function isn't continuous in 0.

• Numer 10 months ago

@Sébastien Louchart I mean correct me, if i'm wrong, but just because the function is defined everywhere doesn't automatically imply, that the limit at every point exists, right? Take for example f: R->R defined by f(x) := sin(1/x) for x != 0 and f(0) := 0. Then lim_(x->0) f(x) doesn't exist, right?

• Sébastien Louchart 10 months ago

@Numer the function f est defined over R*. You can thus take its limit anywhere on that domain

• 025113322 10 months ago

First, differentiate with respect to x with replacement of y=1 we get f(1)=2 then we replace x=1 we get yf(y) = 2020y-2018 so we plug in y=2018 we get f(2018) = 2019 as desired.

• Elias Caeiro 10 months ago

Let g(x) = f(x)*x. Our relation becomes yg(x) + g(y) + 2x = xg(y) + g(x) + 2y for all non-zero x, y. Let k = g(2)-2 g(2) = k+2.
By plugging in y = 2, we get : 2g(x) + (k-2) + 4 = x(k-2) + g(x) + 2y g(x) = kx + 2 - k for all non-zero x.
We have g'(x) = xf'(x) + f(x) so g'(1) = f'(1) + f(1) = f'(1) + g(1) = 2020. However, we also have g'(x) = k so g'(1) = k. Therefore k = 2020==> g(x) = 2020x - 2018.
We get f(x) = (2020x-2018)/x = 2020 - 2018/x so f(2018) = 2019.

• Brianchon 10 months ago +3

We can subtract 2 from both sides and rearrange to get (y-1)(xf(x)-2) = (x-1)(yf(y)-2); for x and both not equal to 1, this further rearranges to (xf(x)-2)/(x-1) = (yf(y)-2)/(y-1), so (xf(x)-2)/(x-1) is some constant C for all x other than 0 and 1. It follows that f(x) = C + (2-C)/x for all x other than 0 and 1. Since f is differentiable at 1, it is also continuous there, so our equation for f also holds at 1, and we can calculate the derivative: 2018 = f'(1) = C-2/(1)^2, so C = 2020. Thus, f(x) = 2020 - 2018/x for all x other than 0, and in particular, f(2018) = 2019.

• iQuickdraw X 10 months ago

Dividing x-y on both sides and take limit as y->x gives x^2*f'(x)+2=x*f'(x)+f(x), solving the differential equation with initial value f'(1)=2018 gives f(x)=2020-2018/x, so f(2018)=2019

• AKASH ASHOK 10 months ago

Keeping y constant and differentiating wrt x,we get xyf'(x)+yf(x)-yf(y)+2=xf'(x)+f(x)
Putting y=1,we get f(1)=2.
Putting x=1,we get f(y)=2020-2018/y.
Therefore f(2018)=2019.

• Harald Schöpf 10 months ago

rearrange to x*f(x)(y-1)+2x=y*f(y)(x-1)+2y -> f(z) must be of the form f(z)=a(z-1)/z+2 -> f´(z)=a/z^2 -> a=2018 ->2018*2017/2018+2=2019

• Ashish kumar 10 months ago

Let a,b,c,d are roots of
(X^2+7X+6)^2+7(X^2+7X+6)+6=X , given that a

• Ashish kumar 10 months ago

@attyfarbuckle thank you , one question is f(f(x))=x ,but you take f(r)=r (why?)

• attyfarbuckle 10 months ago +1

Use f(x) =x^2 +7x +6.
The strategy is to find r with f(r) = r.
Any such r will also be a solution of the original equation.
So solving x^2 + 7x + 6 = x gives x = -3 + sqrt(3) & x = -3 - sqrt(3)
So now you've got 2 roots of the original equation, say r1 & r2.
This means (X - r1) & (X - r2) are factors of (X^2+7X+6)^2+7(X^2+7X+6)+6 - X
If you multiply out (X^2+7X+6)^2+7(X^2+7X+6)+6 - X = 0 then divide by (X^2 + 6X + 6) you get a quadratic which gives the other 2 roots:
Let p(x) = x^2 + 6x + 6.
Then (X^2+7X+6)^2+7(X^2+7X+6)+6 - X
= (p(X)+X)^2 + 7(p(X)+X) + 6 - X
= p(X)^2 + 2Xp(X) + X^2 + 7p(X) + 7X + 6 - X
= p(X)[p(X) + 2X +7] + p(X)
(it's just more algebra after this)

• Ashish kumar 10 months ago

@Jackson Singh f(f(x))=x then next step ?
Thanks Jackson

• Ashish kumar 10 months ago

After f(f(x))=x , then ?
Next step I can't understand about taking inverse

• Jackson Singh 10 months ago

Lets the function be f(x) =x^2 +7x +6

• Will Bishop 10 months ago

Geez. Lots of algebra mistakes before I got it right but here we go.
xy(f(x) - f(y)) + 2x = xf(x) - yf(y) + 2y
xyf(x) - xf(x) - xyf(y) + yf(y) + 2x - 2y = 0
(y-1)xf(x) - (x-1)yf(y) + 2x - 2y = 0
Differentiate with respect to x:
(y-1)(xf'(x) + f(x)) - yf(y) + 2 = 0
Plug in x = 2018, y = 1:
-f(1) + 2 = 0 --> f(1) = 2
Plug in x = 1, y = 2018:
2017(f'(1) + f(1)) - 2018f(2018) + 2 = 0
2017(2018 + 2) - 2018f(2018) + 2 = 0
2017*2020 + 2 = 2018f(2018)
f(2018) = (2017*2020 + 2)/2018
Let b = 2018 so we have f(b) = ((b-1)(b+2) + 2)/b = (b^2 + b - 2 + 2)/b = b + 1 = 2019
So f(2018) = 2019.

• A g 10 months ago

Differentiate wrt x:
y(f(x) -f(y)) + xyf'(x) +2 = f(x) + xf'(x)
Let x = y
(x^2-x)f'(x) = f(x)-2
Solving the differential equation
ln(f(x)-2) =ln(x-1/x) + c
f'(1)=2018 so differentiating the above equation and solving it , we get c = 2018
So f(x) -2 = 2018(x-1)/x
f(2018) = 2017+2=2019.

• NoName 10 months ago

Rearrange the equation:
x*(y-1)*f(x)+y*(1-x)*f(y)=2*(y-x)
We can get rid of on EOF the functions by x=1 or y=1.
Y=1
(1-x)*f(1)=2*(1-x)
f(1)=2
Due to f'(1)=2018, f(1+dz)->2+2018dz
Y=1+dz, X=2018
2018*dz*f(2018)+(1+dz)*(-2017)*(2+2018dz)=2*(-2017+dz)
2018dz*f(2018)=2017*2-2017*2+2*2017dz+2017*2018dz+2dz+2017*2018dz^2
We can drop the dz^2, as it is

• Haradhan Datta 10 months ago

Answer=2019 & f(X)=2020_(2018/X). By subtracting 2 from both sides and rearranging the functional equation,we have, (xf(X)_2)/(x_1)=(yf(y)_2)/(y_1) which is a constant for all values of X in domain of f. The given equation becomes an identity for X=y (in particular X=y=1).So we can take, (xf(X)_2)/(x_1)=k(constant) & hence f(X)=k_((k_2)/X) . Now f'(X)=(k_2)/x^2. But f'(1)=2018=(k_2).k=2020.so, f(X)=2020_(2018/X). f(2018)=2019.

• Jack Jones 10 months ago

put y=0 in the given equation we get x(f(x)-1)+2x=xf(x)-f(1)+2, from here we get f(1)=2
Now, differentiate the equation keeping y as a constant we get
yf(x)+xyf'(x)-yf(y)+2=f(x)+xf'(x)
Put x=1
2y+f'(1)-yf(y)+2=2+f'(1) put the value of f'(1)=2018 and put y = 2018 we get the required answer f(2018)=2019

• Jonty De Pledge 10 months ago

f(2018) = 2019
First we substitute x=2 and y=1
Rearranging gives us f(1)=2
Since the expression is true for all non zero x and y we can differentiate it with respect to x treating y as a constant.
Then substituting x=1, f(x)=2, f'(x)=2018 and rearranging gives f(y)=2020 - 2018/y.
Substituting y=2018 we get f(2018)=2019

• Kim Jong Un 10 months ago

Solution:
Take x=1
Thus, for xyf(x)-xyf(y)+2x=xf(x)-yf(y)+2y we get
yf(1)-yf(y)+2=f(1)-yf(y)+2y
isolating f(1) we get f(1)=2(y-1)/(y-1)=2
so f(1)=2
WLOG take the derivative of the expression w.r.t x and treating y as a constant you will get:
yf(x)+xf'(x)y-yf(y)+2x=f(x)+f'(x)x
plug in x=1 and substitute f'(1) for 2018
Then, you get yf(1)+2018y-yf(y)+2=f(1)+2018
Isolating f(y), you get yf(y)=2018y+yf(1)-f(1)-2016
This leaves you with f(y)=2018+f(1)-(f(1)+2016)=2020-2018/y
substitute y=2018 and you get f(2018)=2019

• aby p 10 months ago

The solution to this is a little lengthy but I managed to get the ans so if we put x=1 in the functional equation we get f(1)=2 then if we put y=2 in the equation and then differentiate w.r.t x we get f(x)+x.f1(x)-2f(2)+2=0 here if we put f1(x) as dy/dx and put f(x) as y and then when we solve we get as 2020-(2018)/x and then when we put x=1 we get 2018 and then to get the ans we put x=2018 and we get the ans as f(2018)=2020-1=2019

• Ankit Gupta 10 months ago

Ans is 2019... F(x)=2020-(2018/x)

The answer is 2019. Rearranging the given equation we find that y(x-1)(f(x)-f(y))=(x-y)f(x)+2(y-x). Divide by (x-y) and take the limit as x-->y and you get a DE (yay!) y(y-1)f'(y)-f(y)=-2 Using the usual tricks (integrating factor, partial fractions) we solve to get f(y) = 2/y + (y-1)C/y (note that f(1)=2 for any finite C) So f'(y) = (C-2)/y^2 + a term proportional to (y-1). The condition on f'(y) gives us C=2. So f(2018) = (2020x2017+2)/2018 = 2019

• Shakti Bhat 10 months ago

First, substitute y=1 and x=2 to get f(1)=2.
Now, put x=1 and solve for f(y) to get f(y) = (2020y-2018)/y.
Now, substitute y=2018 to get f(2018) = 2019.

• Mr L 10 months ago

The answer is 2019. Divide by x-y, let y->x to obtain x(x-1) f'(x) = f(x) -2 ; a set of solutions is f(x) = 2/x + c(x-1)/x , away from 0, for any real c . Compute c using the condition f'(1) = 2018; c turns out to be 2020 and f(x) = 2020 - 2018/x , f(2018) = 2019. It remains to show that this is the only possible solution of the original problem

• Rohit Agarwal 10 months ago

The answer is 2019. Take the partial derivative with respect to x on both sides. Plug in 1 for x and 2018 for y. You get that 2017*f'(1)=-2017f(1)+2018f(2018)-2. Next, plug in the same x and y for the original equation. The 2018*f(2018) cancels on both sides, and you get f(1)=2. Therefore, plugging this information in, you get into our partially solved one, you get f(2018)=2019.

• Ricardo Mogollon 10 months ago

Since the function in question is a single valued function, without loss of generality, we can take x as our variable and y as a positive real number, so the equation becomes more approachable, if we factorize the equation so that it becomes a more common notation of a function it becomes f(x)=(2(y-x)+yf(y)(x-1))/x(y-1), deriving the function and simplifying it becomes f'(x)=(y(f(y)-2)/((x^2)(b-1)) setting x as 1 and reordering we find out that y are inteconneted by the following relation. y=2018/(2020-f(y)), setting the previous relation on f(2018) we find out that f(2018)=(2019(f(b)-2))/(f(b)-2). Finally we can see that f(2018)=2019

• Juan Pablo Gamucero 10 months ago

f(2018)=2015
Fix y, and derive from both sides, we get:
xyf'(x)+y(f(x)-f(y))+2=xf'(x)-f(x)
then
xf'(x)(1-y)=2+f(x)+y(f(x)-f(y)).
From the second expression, if x=1
f'(1)(1-y)=2+f(1)+y(f(1)-f(y))
In particular if y=1. We conclude f(1)=-2.
So, our equation becomes
f'(1)(1-y)=y(f(1)-f(y))
And taking y=2018 we obtain:
f(2018)=(f'(1)(2018-1)/2018)+f(1)=2018-1+f(1)
=2015.

• Jackson Singh 10 months ago

Firstly we know that the identuty is true for all x,y >0 so put x=1 and y=2 then we get f(1)=2 now we differentiate the identity to get and then put x=1 to get a equation in y and f(y) then we put y=2018 to get the answer 2019

• Mathias ICARTE MANCILLA 10 months ago

I think that the answer is 2019. First take x=1 and y=2 for example, you will see that f(1)=2. Take the derivative with respect to x, which means that y is constant. Replace the x for 1 and solve for f(y) and that is 2020-2018/y. Thus f(2018) is 2019 :)

• el tapa 10 months ago

It's seems like you have yo use the CFT. Calculus fundamental theorem

• Elias Caeiro 10 months ago +1

Hi, I solved your problem but I had minor calculations mistakes… Is that a problem ? I replied to my first comment even though I shouldn't have…

• Elias Caeiro 10 months ago

@LetsSolveMathProblems Okay, thank you for responding.

• LetsSolveMathProblems  10 months ago +2

Could you post an entirely new comment by copy-and-pasting your full solution? I almost always do not accept reply comments to reduce confusion on the order in which correct answers are submitted--Although USclip can sort regular comments in chronological order, it cannot do the same for replies.

• GreenMeansGO 10 months ago

What does f’ even mean in this context? How can we take the derivative if the function has inputs x AND y?

• Cobalt314 10 months ago

@GreenMeansGO You can implicitly differentiate by equating the differentials of both sides.

• GreenMeansGO 10 months ago

I understand but again, I am unable to find an expression for f’.

• Jonty De Pledge 10 months ago

The function has only one input, this input could be either y or x.
For example consider the function f(x)=x^2. The function f(y)=y^2 is an identical function we are just substituting in y instead of x. We could then write f'(x)=2x by differentiating with respect to x, or f'(y)=2y by differentiating with respect to y. Either way, we substitute in for x or y we get f'(1) = 1*2 = 2, it doesn't matter what variable we use.

• GreenMeansGO 10 months ago

Ok but I still don’t see how to derive an expression involving f’ due to there being two variables.

• Cobalt314 10 months ago +1

It's a function of 1 variable; f' is well-defined. It's just saying that for any two numbers you plug in (individually), the above relationship holds.

• Parth Pawar 10 months ago

First perform partial differentiation on both sides with respect to x, you get:
xyf'(x)+yf(x)-yf(y)+2=xf'(x)+f(x)
Now put y=x, you get:
x^2f'(x)+2=xf'(x)+f(x)
Simplify and you get a variable separable differential equation:
dy/(y-2)=dx/(x^2-x)
Integrating and upon further simplification we get:
y=c(1-x)/x +2 where c is an arbitrary constant.
differentiate both side and plug in 1 in place of x and 2018 in place of y, we get
c=-2018
Finally put everything, our function is:
f(x)=-2018(1-x)/x +2
Therefore f(2018)=2019

• Elias Caeiro 10 months ago

When you differentiate whit respect to x, you don't know if f is differentiable so you cannot do that for all x and integrate. You can only do this for x=1, or you can prove that f is differentiable.

• Pedro Cardoso 10 months ago

Apply g(x)=xf(x) to get to
y(g(x))-x(g(y))+2x-2y=g(x)-g(y)
y(g(x)-2)-x(g(y)-2)=g(x)-g(y)
Then, apply h(x)=g(x)-2 to get to
y(h(x))-x(h(y))=h(x)-h(y)
(y-1)(h(x))=(x-1)(h(x))
h(x)/(x-1)=h(y)/(y-1)
Which means h(x)=m(x-1), where m is any real constant, and thus
f(x)=(m(x-1)+2)/x, by reversing the substitutions and
f'(x)=(m-2)/x^2, by differentiating.
Since f'(1)=2018, we have m-2=2018 and m=2020
Then f(2018) is (2020(2017)+2)/2018, which equals 2019.

• Cobalt314 10 months ago

First, setting y = 1, we have x(f(x) - f(1)) + 2x = xf(x) - f(1) + 2 -> x(f(1) - 2) = f(1) -2. As this must hold for all nonzero x, we require that f(1) = 2.
Next, if we fix y and differentiate with respect to x (implicity), we get y(f(x) - f(y)) + xyf'(x) + 2 = f(x) + xf'(x), or f'(x) = (f(x) + yf(y) - yf(x) - 2)/(x(y-1)). Plugging in x = 1, we have f'(1) = (f(1) + yf(y) - yf(1) - 2)/(y-1) = y(f(y) - 2)/(y-1), and as it holds for all y, we solve it to get f(y) = (2020y - 2018)/y. Note that y = 1 is a removable discontinuity, as 1 - 1 = f(1) - 2. Thus, having the formula for f(y), we plug in 2018 to get f(2018) = (2020*2018 - 2018)/2018 = 2019.

• Elias Caeiro 10 months ago

My first solution to a weekly math challenge : ) :
Let g(x) = f(x)/x. Our relation becomes yg(x) + g(y) + 2x = xg(y) + g(x) + 2y for all non-zero x, y. Let k = g(2)-2 g(2) = k+2.
By plugging in y = 2, we get : 2g(x) + (k-2) + 4 = x(k-2) + g(x) + 2y g(x) = kx + 2 - k for all non-zero x.
We have g'(x) = (f'(x)*x - f(x))/x^2 so g'(1) = f'(1)-f(1) = f'(1) - 1*g(1) = 2018 - (k*1 + 2 -k) = 2016. However, we also have g'(x) = k so g'(1) = k. Therefore k = 2016 ==> g(x) = 2016x - 2014 so f(x) = x(2016x - 2014).
So f(2018) = 2018(2016*2018 - 2014) and we get f(2018) = 8205740932 using a calculator.

• Elias Caeiro 10 months ago

Sorry if I reply to my comment, I had a few calculations mistakes/typos… Here's the last part corrected :
I meant of course g(x) = x*f(x) f(x) = g(x)/x.
We have g'(x) = xf'(x) + f(x) so g'(1) = f'(1) + f(1) = f'(1) + g(1) = 2020. However, we also have g'(x) = k so g'(1) = k. Therefore k = 2020==> g(x) = 2020x - 2018.
We get f(x) = (2020x-2018)/x = 2020 - 2018/x so f(2018) = 2019.

• Minh Cong Nguyen 10 months ago