g(x+y)=g(x)+g(y)+3xy(x+y)∀x,y∈R and g′(0)=−4

Solution:

Given the functional equation g(x+y) = g(x) + g(y) + 3xy(x+y) for all x, y ∈ ℝ and g′(0) = −4, we find the function g(x).

1. Substitute y = 0 in the given equation:

   g(x+0) = g(x) + g(0) + 3x ⋅ 0 ⋅ (x+0)

   Simplifying, we get:

   g(x) = g(x) + g(0)

   This implies:

   g(0) = 0

2. Differentiate both sides of the given functional equation with respect to x:

   ∂/∂x [g(x+y)] = ∂/∂x [g(x) + g(y) + 3xy(x+y)]

   Since y is treated as a constant:

   g'(x+y) = g'(x) + 3y(x+y) + 3xy

   Substitute x = 0:

   g'(y) = g'(0) + 3y^2

   Given g'(0) = -4:

   g'(y) = -4 + 3y^2

3. Integrate g'(y) to find g(y):

   g(y) = ∫(-4 + 3y^2) dy = -4y + y^3 + C

   Given g(0) = 0, we find C:

   g(0) = -4(0) + (0)^3 + C = 0 ⟹ C = 0

   Therefore:

   g(y) = y^3 - 4y

Answer: The function g(x) is g(x) = x^3 - 4x.