f(x+y+z) =f(x).f(y).f(z) for all x , y , z and f(2)=4 , f(0) =3 then f(2) =ans is 12plzz explain..

f(x+y+z)=f(x)f(y)f(z) for all x,y,z

f(0)=f(0)f(0)f(0) {Putting x=y=z=0}

f(0)= f(0)[1-(f(0))^2]=0

=> f(0)=1

putting z=0 and y=2

we get f(x+2)=f(x)f(2)f(0)

f(x+2)=4f(x) {by differentiating}

f''(2)=4f''(x)

put x=0

f''(2)=4f''(0)

f''(2)= 4*3=12