If `f((x+y)/3)=(2+f(x)+f(y))/3` for all real `xa n dy` and `f^(prime)(2)=2,` then determine `y=f(x)dot`

Explanation

`f((x+y)/(3))=(2+f(x)+f(y))/(3)`
Replacing x by 3x and y by 0, we get
`f(x)=(2+f(3x)+f(0))/(3)`
`"or "f(3x)-3f(x)+2=-f(0)`
In (1) putting x = 0 and y = 0, we get
`f(0)=2`
`"Now, "f'(x)=underset(hrarr0)lim(f(x_+h)-f(x))/(h)`
`=underset(hrarr0)lim(f((3x+3h)/(3))-f(x))/(h)`
`=underset(hrarr0)lim((2+f(3x)+f(3h))/(3)-f(x))/(h)`
`underset (hrarr0)lim(f(3x)-3f(x)+f(3h)+2)/(3h)`
`=underset(hrarr0)lim(f(3h)-f(0))/(3h)" [from (2)]"`
`=f'(x)=c`
`therefore" "f'(x)=c`
`"At "x=2,f'(2)=c=2" (Given )"`
`therefore" "f'(x)=2`
Integrating bout sides, we get
`f(x)=2x+d" [From (2)]"`
`therefore" "d=2`
Hence, f(x)=2x+2.
Alternative method:
`f((x+y)/(3))=(2+f(x)+f(y))/(3)" (1)"`
Differentiating w.r.t. x keeping as constant, we get
`f'((x+y)/(3))(1)/(3)=(f'(x))/(3)`
Put x = 2 and y = 3x -2. Then
f'(x)=2
Integrating, we get `f(x)=2x+c.`
Now, put x=y=0 in (1). Then 3 f (0)= 2 + 2 f(0)`
`"or " f(0)=2`
`"Hence, "f(x)=2x+2`