Given the x-intercepts J(-1; 0), K((1)/(2); 0), and L(3; 0), the factors of f(x) are (x+1), (x-(1)/(2)), and (x-3).

Given the x-intercepts $J(-1; 0)$, $K(\frac{1}{2}; 0)$, and $L(3; 0)$, the factors of $f(x)$ are $(x+1)$, $(x-\frac{1}{2})$, and $(x-3)$.
The general form of a cubic function is $f(x) = a(x-x_1)(x-x_2)(x-x_3)$.
From the given equation $f(x) = 2x^3 + bx^2 + cx + d$, the leading coefficient $a=2$.
So, we can write $f(x)$ as:

$$f(x) = 2(x+1)\left(x-\frac{1}{2}\right)(x-3)$$

Multiply $(x+1)$ by $(x-\frac{1}{2})$:

$$(x+1)\left(x-\frac{1}{2}\right) = x^2 - \frac{1}{2}x + x - \frac{1}{2} = x^2 + \frac{1}{2}x - \frac{1}{2}$$

$$\left(x^2 + \frac{1}{2}x - \frac{1}{2}\right)(x-3) = x^3 - 3x^2 + \frac{1}{2}x^2 - \frac{3}{2}x - \frac{1}{2}x + \frac{3}{2}$$

Combine like terms:

$$= x^3 + \left(-3 + \frac{1}{2}\right)x^2 + \left(-\frac{3}{2} - \frac{1}{2}\right)x + \frac{3}{2}$$ $$= x^3 - \frac{5}{2}x^2 - 2x + \frac{3}{2}$$

$$f(x) = 2\left(x^3 - \frac{5}{2}x^2 - 2x + \frac{3}{2}\right)$$ $$f(x) = 2x^3 - 5x^2 - 4x + 3$$

Comparing this with $f(x) = 2x^3 + bx^2 + cx + d$, we find the values of $b$, $c$, and $d$:

$$b = -5$$ $$c = -4$$ $$d = 3$$

The values are $b = -5, c = -4, d = 3$.