If Rolle's theorem holds for the function f(x) = 2x³ + bx² + cx, x ∈ [-1, 1], at the point x = 1/2, then 2b + c equals:

f(x) = 2x³ + bx² + cx and x ∈ [-1, 1]
For Rolle's theorem to be applicable, f(1) = f(-1) and f'(1/2) = 0
f(1) = 2(1)³ + b(1)² + c(1) = 2 + b + c
f(-1) = 2(-1)³ + b(-1)² + c(-1) = -2 + b - c
Since f(1) = f(-1), 2 + b + c = -2 + b - c
2c = -4
c = -2
f'(x) = 6x² + 2bx + c
f'(1/2) = 6(1/2)² + 2b(1/2) + c = 0
6(1/4) + b + c = 0
3/2 + b + c = 0
Since c = -2, 3/2 + b - 2 = 0
b = 2 - 3/2 = 1/2
Therefore, 2b + c = 2(1/2) + (-2) = 1 - 2 = -1