Question

f is a function such that f(x+y)=f(x)xxf(y), f(0)!=0. If f(x) is differentiable at x=0 then for all x in R:

• f is differentiable
• f' is linear function of f
• Can't comment

Solution

We are talking about differentiablty on R which is only possible if f is continuous on R.

We will first check if Lt_(h->0^+) f(x+h)=Lt_(h->0^-) f(x+h)=f(x)

Since it is differentiable at 0 so it is continuous at 0.

=>Lt_(h->0^+) f(0+h)=Lt_(h->0^-) f(0+h)=f(0)

More over f(0)=f(0)xxf(0). Since f(0)!=0 we get f(0)=1

Now coming back to Lt_(h->0^+) f(x+h)

=Lt_(h->0^+) f(x)xxf(h)=f(x)Lt_(h->0^+) f(h)=f(x)

We can similarly prove Lt_(h->0^-) f(x-h)=f(x)

Now we look at the differentiablty of f(x) at any x

Which is Lt_(h->0)(f(x+h)-f(x))/h

=Lt_(h->0)(f(x)xxf(h)-f(x))/h

=f(x)Lt_(h->0)(f(h)-1)/h

=f(x)xxf'(0)

So f(X) is differentiable for all x in R

Moreover f' is a linear function of f!!