With t = f(x) + x + y we get f(t) = 3t - 3f(x) + 9x + 8 Now we set t = x: f(x) = 3x - 3f(x) + 9x + 8 4f(x) = 12x + 8 f(x) = 3x + 2 After watching the video this is similar to your 3rd method...
Looks linear, f(x)=ax+b, apply twice, then you get something like (.....)x + ay + (........) = 12x+3y+8, so a=3 and (........) = 8 which gives the b being 2. f(x)=3x+2.
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without watching:
set y = -f(x)
f(x) = 12x - 3f(x) + 8
f(x) = 3x + 2
check
f(f(x) + x + y)
= f(4x + y + 2)
= 3(4x + y + 2) + 2
= 12x + 3y + 8
nice, didnt realize i can replace y=-f(x)
Might be the shortest solution
Cool!
i replace x=0 ---> f(f(0)+y)=3y+8 ---> f(y+c)=3y+8=3(y+c)+8-3c ---> f(x)=3x+8-3c.
x=0 ---> c=f(0)=3*0+8-3c ---> c=2 ---> f(x)=3x+2.
y = - f(x)
👍✌️☮️😀😀☮️✌️👍
f(f(x)+x+y)=12x+3y+8
f(0)=c → f(c+y)=3y+8=3(c+y)+8-3c
c+y=t → f(t)=3t+8-3c
f(0)=c=8-3c → c=2
∴f(t)=3t+2 → f(x)=3x+2
put x=f(0) and y=-8 can give you f(0)=2
With t = f(x) + x + y we get f(t) = 3t - 3f(x) + 9x + 8
Now we set t = x: f(x) = 3x - 3f(x) + 9x + 8
4f(x) = 12x + 8
f(x) = 3x + 2
After watching the video this is similar to your 3rd method...
f(x)=3x+2
Looks linear, f(x)=ax+b, apply twice, then you get something like (.....)x + ay + (........) = 12x+3y+8, so a=3 and (........) = 8 which gives the b being 2. f(x)=3x+2.