Let f be a differentiable function such that f(0) = 0 , f(1) = 3 , and f^ (0) = 4 . A new function g is defined as g(x) = f(2x+1) e^ 2f(x) . If g^ (0) = 40 , then the value of f^ (1) is equal to:

Step-by-step solution

Given g(x) = f(2x+1) e^ 2f(x) . Differentiating g(x) with respect to x using the product rule and chain rule, we get: g^ (x) = [ f^ (2x+1) 2 ] e^ 2f(x) + f(2x+1) [ e^ 2f(x) 2f^ (x) ] Substitute x = 0 into the derivative: g^ (0) = 2 f^ (1) e^ 2f(0) + 2 f(1) e^ 2f(0) f^ (0) We are given f(0) = 0 , f(1) = 3 , f^ (0) = 4 , and g^ (0) = 40 . Since f(0) = 0 , we have e^ 2f(0) = e^0 = 1 . Substituting these values into the equation for g^ (0) : 40 = 2 f^ (1) (1) + 2 (3) (1) (4) 40 = 2 f^ (1) + 24 2 f^ (1) = 16 f^ (1) = 8