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)
.
-
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
-
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
-
Integrate
g'(y)
to findg(y)
:g(y) = ∫(-4 + 3y^2) dy = -4y + y^3 + C
Given
g(0) = 0
, we findC
: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
.