Solving A Homemade Functional Equation

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[AlexandreRibeiroXRV]

There's another solution: f(x) = -1. Not as cool as the given in the video, but it also works.
It arises as the case when k equals 0, but then we have no domain restrictions for it unlike the log based solution, so I reckon it's good to list as a separate case. Or, we could write h(x) as k*log|x| to be more thorough. (Which still doesn't work for f(0), but f(x) = -1 does satisfy it).

[crisant.]

the problem is incomplete the domains of the function aren't given. Assuming its the real numbers by plugging y=0 you get f(0)²(f(x)+1)=0 which means either f(0)=0 or f(x)=-1
if f(0)=0, now plug x=y=0 you get 0=0/0 which isn't consistent, so the only correct function is f(x)=-1

[allanmarder]

If f(x) is defined on the real numbers then f(0) as defined in the video would have ln(0) in the denominator and would be undefined.
But letting x=0 and y=0 by definition we get f(0) = f(0)^2/[2*f(0) +f(0)^2]] Cross multiply gives f(0)^2 =2*(f(0)^2) +f(0)^3 or 0=f(0)^2 +f(0)^3
so that 0=(f(0)^2)*[f(0)+1]. This gives f(0)=0 or f(0)= -1. Now f(0) cannot equal 0 because the denominator of f(0) =f(0*0) as defined would be 0.
So we must have f(0)=-1. Let x=0 y=any real number. Then f(0*y) = [f(0)*f(y)]/[f(0)+f(y) + f(0)*f(y)].or f0)= [f(0)*f(y)]/[f(0)+f(y) + f(0)(f(y)]
Cross multiplying and simplifying gives 0=[f(0)^2]*[1+f(y)] since f(0)=-1 we must have f(y)= -1 for all real y. If k=0 in the video we do get f(x)= -1.
If k is not equal to 0 then f(x) is not defined for x=0 and log(x) =1/k which in turn depends on the base for the log. To get a specific f(x)
you probably need to specify the domain for f(x) along with some initial conditions.

[Anmol_Sinha]

This functional eqn question is a masterpiece! Defining new functions to simplify the equation into a recognizable form is quite clever

[Blaqjaqshellaq]

I'd go a bit further: take 1/f(x*y)=1/f(x) + 1/f(y) +1, and suppose that y=1.
Therefore 1/f(x)=1/f(x*1)=1/f(x) + 1/f(1) +1,
therefore 1/f(1) +1=0, therefore f(1)=-1.
Now suppose that y=0.
Then 1/f(0)=1/f(x*0)=1/f(x) + 1/f(0) + 1,
therefore 1/f(x) +1=0, therefore f(x)=-1 for all values of x!
This fits f(x)=1/[k*ln(x)-1] if k=0.

[macelomendes]

Pretty question!! Hugs from Teresina-PI, Brazil!!

[ACheateryearsago]

As a person who has just studied the basics of functions, this looks like a nightmare

[shacharh]

f = -1 (which you would get for the case k=0) also works. The difference is, it works even if you allow f to be defined for x=0, unlike a function with logarithms as you found

[leickrobinson]

That’s so funny! I solved it exactly the same way you did, and I mean *exactly*. I used exactly the same intermediate functions and called them g(x) and h(x), just like you did. 😄

[someperson]

If k <> 0, then the solution given isn't defined for x = e^(1/k). So, the only differentiable solution which works for all real x is when k = 0; i.e., f(x) = -1.

[Anmol_Sinha]

I got f(x) = -1 assuming that it lied in the functions domain. It only turns out to be true for k=0. I am not sure what happend because it does seem that 0 lies in the domain for a lot of values of k(?)

[dasmith]

I calculated f(0•y). When you simplify you have to take note that f(0) cannot be 0. Bottom line, f(y) has to be -1 for all y. Done.

[erikroberts]

f(x) = -1 also works for this problem.

[paultoutounji]

Great ! Really ! Thank you very much Sybermath ! 😀

[HoSza]

dividing with functions and hoping it doesn'g have a zero somewhere, because otherwise GAME OVER