Challenge 67: An Almost Symmetric Functional Equation

  • Published on Nov 15, 2018
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Comments • 67

  • Zicar
    Zicar 7 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
    Yaman Sanghavi 7 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 7 months ago

    The 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 Zhao
    miner Zhao 7 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 well we derivate this to X,we get f(x)+x.f'(x)=y.f(y)-2/ 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
    Tamerlane 7 months ago

    Answer is 2020

    • Tamerlane
      Tamerlane 7 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

  • Leonardo Costa
    Leonardo Costa 7 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
    Sébastien Louchart 7 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 :/

    • Brian talks about
      Brian talks about 7 months ago

      Come solve my problem... it's my latest video.

    • Sébastien Louchart
      Sébastien Louchart 7 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
      Sébastien Louchart 7 months ago

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

    • Numer
      Numer 7 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
      Sébastien Louchart 7 months ago

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

  • 025113322
    025113322 7 months ago

    The 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 Caeiro
    Elias Caeiro 7 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
    Brianchon 7 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
    iQuickdraw X 7 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 7 months ago

    The 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 Schöpf
    Harald Schöpf 7 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
    Ashish kumar 7 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
      Ashish kumar 7 months ago

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

    • attyfarbuckle
      attyfarbuckle 7 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
      Ashish kumar 7 months ago

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

    • Ashish kumar
      Ashish kumar 7 months ago

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

    • Jackson Singh
      Jackson Singh 7 months ago

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

  • Will Bishop
    Will Bishop 7 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
    A g 7 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
    NoName 7 months ago

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

  • Haradhan Datta
    Haradhan Datta 7 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), f(X)=2020_(2018/X). f(2018)=2019.

  • Jack Jones
    Jack Jones 7 months ago

    2019 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
    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
    Jonty De Pledge 7 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
    Kim Jong Un 7 months ago

    Take x=1
    Thus, for xyf(x)-xyf(y)+2x=xf(x)-yf(y)+2y we get
    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:
    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
    aby p 7 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
    Ankit Gupta 7 months ago

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

  • adandap
    adandap 7 months ago

    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
    Shakti Bhat 7 months ago

    The 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 L
    Mr L 7 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
    Rohit Agarwal 7 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
    Ricardo Mogollon 7 months ago

    the 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 Gamucero
    Juan Pablo Gamucero 7 months ago

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

  • Jackson Singh
    Jackson Singh 7 months ago

    The 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


    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
    el tapa 7 months ago

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

  • Elias Caeiro
    Elias Caeiro 7 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
      Elias Caeiro 7 months ago

      @LetsSolveMathProblems Okay, thank you for responding.

    • LetsSolveMathProblems
      LetsSolveMathProblems  7 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
    GreenMeansGO 7 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
      Cobalt314 7 months ago

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

    • GreenMeansGO
      GreenMeansGO 7 months ago

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

    • Jonty De Pledge
      Jonty De Pledge 7 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
      GreenMeansGO 7 months ago

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

    • Cobalt314
      Cobalt314 7 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
    Parth Pawar 7 months ago

    Answer is 2019.
    First perform partial differentiation on both sides with respect to x, you get:
    Now put y=x, you get:
    Simplify and you get a variable separable differential equation:
    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
    Finally put everything, our function is:
    f(x)=-2018(1-x)/x +2
    Therefore f(2018)=2019

    • Elias Caeiro
      Elias Caeiro 7 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
    Pedro Cardoso 7 months ago

    Apply g(x)=xf(x) to get to
    Then, apply h(x)=g(x)-2 to get to
    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
    Cobalt314 7 months ago

    The 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 Caeiro
    Elias Caeiro 7 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
      Elias Caeiro 7 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
    Minh Cong Nguyen 7 months ago

    The 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 Nguyen
    Minh Cong Nguyen 7 months ago +1

    The answer is 2019

  • JHawk24
    JHawk24 7 months ago +1