One last functional equation

[BMAD]

(f(x+y)-f(xy))/(3x) = f(y/(3x))-11-y

Find f(x) where f(x) is a polynomial.

[EventHorizon]

Spoiler

1,0
(f(1)-f(0))/3 = f(0)-11
f(1) = 4*f(0)-33

-1,0
(f(-1)-f(0))/(-3) = f(0)-11-y
f(-1) = -2*f(0)+33

-1,1
(f(0)-f(-1))/(-3) = f(-1/3)-12
f(0)-f(-1) = 36 - 3*f(-1/3)

1,-1
(f(0)-f(-1))/(3) = f(-1/3)-10
f(0)-f(-1) = 3*f(-1/3) - 30

3*f(-1/3) - 30 = 36 - 3*f(-1/3)
6*f(-1/3) = 66
f(-1/3) = 11
f(0)-f(-1) = 3

(f(0)-3) = -2*f(0)+33
3*f(0) = 36
f(0) = 12
f(-1) = 9

f(1) = 4*(12)-33
f(1) = 48-33 = 15

f(x) = 3x+12??

(f(x+y)-f(xy))/(3x) = f(y/(3x))-11-y
(3(x+y)+12 - (3xy+12))/3x = (y/x +12)-11-y
(3x+3y+12-3xy-12)/3x = y/x +1 -y
(3x+3y-3xy)/3x = y/x +1 -y
1+y/x-y = y/x +1 -y
yup.

f(x) = 3x+12

 

[EventHorizon]

Alternatively,

Spoiler

x,0
(f(x)-f(0))/(3x) = f(0)-11
f(x)-f(0) = 3x * (f(0)-11) = ax, for some constant a.
clearly f(x) is linear.

f(x) = ax+b

(f(x+y)-f(xy))/(3x) = f(y/(3x))-11-y
(a(x+y)+b-axy-b)/3x = ay/3x+b-11-y
(ax+ay-axy)/3x = ay/3x +b-11-y
a/3 + ay/3x - ay/3 = ay/3x +b-11-y
a/3 - ay/3 = b-11-y
ay/3 = y since the variable terms must be equal
a = 3
a/3 = b-11
1 = b-11
b=12

f(x) = 3x+12