# Challenge 67: An Almost Symmetric Functional Equation

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- Published on Nov 15, 2018
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Zicar10 months agoI 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 Sanghavi10 months agoJust 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 AGARWAL10 months agoThe answer os 2019

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 Zhao10 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

Tamerlane10 months agoAnswer is 2020

Tamerlane10 months agoLet'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 Costa10 months agoAns: 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 Louchart10 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 Louchart10 months agoThe 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 Louchart10 months ago@Numer in that case. You're right. The function isn't continuous in 0.

Numer10 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 Louchart10 months ago@Numer the function f est defined over R*. You can thus take its limit anywhere on that domain

02511332210 months agoThe answer is 2019

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 Caeiro10 months agoLet 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.

Brianchon10 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 X10 months agoDividing 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 ASHOK10 months agoThe answer is 2019.

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 Schöpf10 months agorearrange 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 kumar10 months agoLet a,b,c,d are roots of

(X^2+7X+6)^2+7(X^2+7X+6)+6=X , given that a

Ashish kumar10 months ago@attyfarbuckle thank you , one question is f(f(x))=x ,but you take f(r)=r (why?)

attyfarbuckle10 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 kumar10 months ago@Jackson Singh f(f(x))=x then next step ?

Thanks Jackson

Ashish kumar10 months agoAfter f(f(x))=x , then ?

Next step I can't understand about taking inverse

Jackson Singh10 months agoLets the function be f(x) =x^2 +7x +6

Will Bishop10 months agoGeez. 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 g10 months agoAnswer-2019

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.

NoName10 months agoRearrange 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 Datta10 months agoAnswer=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 Jones10 months ago2019 is the answer

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 Pledge10 months agof(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 Un10 months agoSolution:

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 p10 months agoThe 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 Gupta10 months agoAns is 2019... F(x)=2020-(2018/x)

adandap10 months agoThe 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 Bhat10 months agoThe answer is 2019.

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 L10 months agoThe 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 Agarwal10 months agoThe 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 Mogollon10 months agothe answer is 2019.

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 Gamucero10 months agof(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 Singh10 months agoThe answer is 2019

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 MANCILLA10 months agoI 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 :)

Mathias ICARTE MANCILLA10 months agoYes, is 2019 :( my mistake

Pedro Cardoso10 months ago^{+1}great solution, but you forgot to subtract that 1 in the end, the answer is 2019

el tapa10 months agoIt's seems like you have yo use the CFT. Calculus fundamental theorem

Elias Caeiro10 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 Caeiro10 months ago@LetsSolveMathProblems Okay, thank you for responding.

LetsSolveMathProblems10 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.

GreenMeansGO10 months agoWhat does f’ even mean in this context? How can we take the derivative if the function has inputs x AND y?

Cobalt31410 months ago@GreenMeansGO You can implicitly differentiate by equating the differentials of both sides.

GreenMeansGO10 months agoI understand but again, I am unable to find an expression for f’.

Jonty De Pledge10 months agoThe 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.

GreenMeansGO10 months agoOk but I still don’t see how to derive an expression involving f’ due to there being two variables.

Cobalt31410 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 Pawar10 months agoAnswer is 2019.

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 Caeiro10 months agoWhen 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 Cardoso10 months agoApply 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.

Cobalt31410 months agoThe answer is 2019.

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 Caeiro10 months agoMy 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 Caeiro10 months agoSorry 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 Nguyen10 months agoThe answer is 2019

Let f(-1)= c = constant, substituting y = -1 we have f(x)=(2x+2+xc-c)/(2x) then f'(x)= (c-2)/(2x^2)

f'(1)=(c-2)/2=2018 which means that c=4038. Finally we have f(2018)=2019.

Minh Cong Nguyen10 months ago^{+1}The answer is 2019

Minh Cong Nguyen10 months ago^{+2}f(x)=2020-2018/x

JHawk2410 months ago^{+1}first