Let f: R (0, ) be a differentiable function satisfying f(x+y) = 2f(x)f(y) for all x, y R . It is given that f(x) = 1 2 + x g(x) , where _ x 0 g(x) = 5 2 . The value of _ k=1 ⁴ f(k) is equal to:

Step-by-step solution

Given the functional equation: f(x+y) = 2f(x)f(y) Substitute x = 0, y = 0 : f(0) = 2(f(0))^2 Since f(x) > 0 for all x , we get f(0) = 1 2 . We are given f(x) = 1 2 + x g(x) . Rearranging this: f(x) - 1 2 x = g(x) Taking the limit as x 0 : _ x 0 f(x) - f(0) x-0 = _ x 0 g(x) This is the definition of the derivative at x=0 , so f'(0) = 5 2 . Now, differentiate the functional equation f(x+y) = 2f(x)f(y) with respect to x (treating y as a constant): f'(x+y) = 2f'(x)f(y) Substitute x = 0 and replace y with x : f'(x) = 2f