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

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

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    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.

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