[THOUGHT] We need to find the smallest positive $k$ such that for all $x \in [-2, 2]$, the cubic polynomial $f(x) = 2x^3 + 3x^2 - 12x - 8$ satisfies $-k \le f(x) \le k$.

That is equivalent to $|f(x)| \le k$ for all $x \in [-2, 2]$. So $k$ must be at least the maximum absolute value of $f(x)$ on $[-2, 2]$. The smallest possible $k$ is $\max_{x \in [-2, 2]} |f(x)|$.

We can find this maximum by:

1. Finding critical points inside $(-2, 2)$ by solving $f'(x) = 0$.
2. Evaluating $|f(x)|$ at critical points and endpoints.
3. Taking the maximum.

Let's compute $f'(x) = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x+2)(x-1)$. So critical points: $x = -2$ and $x = 1$. Note $x = -2$ is an endpoint, $x = 1$ is interior.

Thus interior critical point: $x = 1$. Endpoints: $x = -2, 2$.

We'll evaluate $f(x)$ at $x = -2, 1, 2$ and also check if there are other critical points? Wait, derivative zero at $x = -2$ and $x = 1$. But $x = -2$ is endpoint. So only interior critical point: $x = 1$.

But we must also consider that the maximum of $|f(x)|$ could occur at endpoints or at critical points. Let's compute $f$ at these points.

Let's do it in Python. [/THOUGHT] [PYTHON] import sympy as sp

x = sp.symbols('x') f = 2x**3 + 3x**2 - 12*x - 8

Evaluate at endpoints and critical point

points = [-2, 1, 2] values = [abs(f.subs(x, pt)) for pt in points] print("Values of |f(x)| at points -2, 1, 2:", values)

Also check if any other critical points? derivative = 6x^2 + 6x -12

Solve f'(x)=0: 6x^2+6x-12=0 => x^2+x-2=0 => (x+2)(x-1)=0 => x=-2,1.

So no other interior critical points.

But maybe maximum of |f| occurs at points where f' doesn't exist? f is polynomial, so differentiable everywhere.

So candidates: endpoints -2, 2, and critical point 1.

Let's also check if there are points where f(x)=0? Not needed for max |f|.

max_abs = max(values) print("Maximum |f(x)| among these points:", max_abs)

To be thorough, we can check the function's behavior numerically over the interval.

import numpy as np xs = np.linspace(-2, 2, 10001) ys = 2xs**3 + 3xs**2 - 12*xs - 8 abs_ys = np.abs(ys) max_abs_num = np.max(abs_ys) print("Maximum |f(x)| over fine grid:", max_abs_num) print("Corresponding x:", xs[np.argmax(abs_ys)])

Also, we can find exact maximum of |f| by considering f(x) positive and negative ranges.

But the numerical check should be fine.

Let's also verify that the inequality holds for this k.

k = max_abs print("k =", k)

Check inequality at a few random points

import random for _ in range(10): pt = random.uniform(-2, 2) val = 2pt**3 + 3pt**2 - 12*pt - 8 if not (-k <= val <= k): print("Violation at x =", pt, "val =", val) break else: print("Inequality holds for random points with k =", k)

Output exact k if possible.

The values at -2, 1, 2:

f_at_m2 = f.subs(x, -2) f_at_1 = f.subs(x, 1) f_at_2 = f.subs(x, 2) print("f(-2) =", f_at_m2, "|f(-2)| =", abs(f_at_m2)) print("f(1) =", f_at_1, "|f(1)| =", abs(f_at_1)) print("f(2) =", f_at_2, "|f(2)| =", abs(f_at_2))

The maximum absolute value is max(|f(-2)|, |f(1)|, |f(2)|).

max_exact = max(abs(f_at_m2), abs(f_at_1), abs(f_at_2)) print("Exact max |f|:", max_exact) print("Exact max |f| as integer?", max_exact.evalf()) [/PYTHON]

[THOUGHT] The maximum absolute value of $f(x) = 2x^3 + 3x^2 - 12x - 8$ on the interval $[-2, 2]$ is 15, occurring at $x = 1$. Therefore, the smallest possible positive $k$ is 15.

$\boxed{15}$