Question

a function f:R-R satisfies the equation f(x+y)=f(x).f(y) for all x belongs to R f(x) not equal to 0.suppose that function is differentiable at x=0 and f'(0)=2. prove that f'(x)=2f(x)

Answer 1

Answer: it is given that f(x+y)=f(x).f(y)...... 1

now taking x=0 and y=0 and putting it in 1

f(0+0)=f(0).f(0)

f(0)=f(0).f(0)

1=f(0)

also it is given that f'(0)=2

solving it using the first principle

f'(0)=limit f(0+h)-f(0)/h

         h-0

2=limit f(h)-1/h.....2

    h-0

also f'(x)=limit  f(x+h)-f(x)/h

                 h-0

and it is given that f(x+y)=f(x).f(y) therefore f(x+h)=f(x).f(h)

thus f'(x)= f(x) limit f(h)-1/h

                      h-0

now using 2

f'(x)=f(x) * 2 hence proved