[THOUGHT] To find the smallest positive value of $k$ such that the inequality $-k \leq 2x^3 + 3x^2 - 12x - 8 \leq k$ holds for all $x$ in the interval $[-2, 2]$, we need to determine the maximum absolute value of the function $f(x) = 2x^3 + 3x^2 - 12x - 8$ on this interval. This involves evaluating $f(x)$ at the critical points and the endpoints of the interval.

First, we find the critical points by setting the derivative $f'(x)$ to zero:

Solving this quadratic equation, we get:

The critical points are $x = -2$ and $x = 1$. Since $x = -2$ is an endpoint, we only need to consider $x = 1$ as an internal critical point. We evaluate $f(x)$ at the endpoints $x = -2$, $x = 2$, and the critical point $x = 1$.

Next, we compute the function values at these points:

The absolute values of these function values are:

The maximum absolute value is 15, which occurs at $x = 1$. Therefore, the smallest possible positive value of $k$ is 15.

To confirm, we can use Python to evaluate the function at these points and verify the maximum absolute value. [/THOUGHT]

[PYTHON] def f(x): return 2x**3 + 3x**2 - 12*x - 8

Evaluate the function at the critical points and endpoints

points = [-2, 1, 2] values = [f(x) for x in points] abs_values = [abs(v) for v in values] max_abs = max(abs_values) print(max_abs) [/PYTHON]

[THOUGHT] The Python output confirms that the maximum absolute value of the function on the interval $[-2, 2]$ is 15. Therefore, the smallest possible positive value of $k$ is 15. [/THOUGHT] $\boxed{15}$