[THOUGHT] I need to find the smallest positive value of $k$ such that $-k \leq 2x^3 + 3x^2 - 12x - 8 \leq k$ for all $x \in [-2, 2]$.

This means I need to find the maximum absolute value of $f(x) = 2x^3 + 3x^2 - 12x - 8$ on the interval $[-2, 2]$.

So $k = \max_{x \in [-2, 2]} |f(x)|$.

To find this, I need to:

1. Find the critical points of $f(x)$ by setting $f'(x) = 0$
2. Evaluate $f(x)$ at the critical points and endpoints
3. Find the maximum and minimum values
4. The smallest $k$ is $\max(|f_{max}|, |f_{min}|)$

Let me compute this using Python. [/THOUGHT] [PYTHON] import numpy as np from sympy import symbols, diff, solve, Rational

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

Find critical points

f_prime = diff(f, x) print("f'(x) =", f_prime)

critical_points = solve(f_prime, x) print("Critical points:", critical_points)

Evaluate f at critical points and endpoints

points_to_check = [-2, 2] + [cp for cp in critical_points if -2 <= cp <= 2] print("Points to check:", points_to_check)

values = [] for p in points_to_check: val = f.subs(x, p) values.append((p, val)) print(f"f({p}) = {val}")

Find max and min

max_val = max(v[1] for v in values) min_val = min(v[1] for v in values) print(f"\nMaximum value: {max_val}") print(f"Minimum value: {min_val}")

k must satisfy both -k <= min_val and max_val <= k

So k >= max(|max_val|, |min_val|)

k = max(abs(max_val), abs(min_val)) print(f"\nSmallest k = {k}") [/PYTHON]

[THOUGHT] The function $f(x) = 2x^3 + 3x^2 - 12x - 8$ has critical points at $x = -2$ and $x = 1$ within the interval $[-2, 2]$.

Evaluating at the critical points and endpoints:

• $f(-2) = 12$
• $f(1) = -15$
• $f(2) = -4$

The maximum value is 12 and the minimum value is -15.

For the inequality $-k \leq f(x) \leq k$ to hold for all $x \in [-2, 2]$, we need $k \geq \max(|12|, |-15|) = 15$.

Therefore, the smallest possible positive value of $k$ is 15. [/THOUGHT]

\boxed{15}