Put y=x
f(x + (x+1)f(x) ) = x + (x+1)f(x)
=> f(x)=x , x + (x+1)f(x) = 0
If and f is differentiable function satisfies :
then find f(x)
not a doubt....jus found it gud
Put y=x
f(x + (x+1)f(x) ) = x + (x+1)f(x)
=> f(x)=x , x + (x+1)f(x) = 0
@ dubey sir.....dat does not necessarily confirm f(x)=x is the only function satisfying it...
Also f(x)=-1 is another one satisfying this.
A small extension, division by k.f'(x) is valid only when f'(x) ≠0, f(x)=-1 is a valid soln in this case.