[THOUGHT] We need the smallest positive k such that for all x in [-2, 2], the inequality -k ≤ f(x) ≤ k holds, where f(x) = 2x^3 + 3x^2 - 12x - 8. That means k must be at least the maximum absolute value of f(x) on the interval [-2, 2]. So we need to find the maximum of |f(x)| on [-2, 2]. Compute f(x) over the interval, find its maximum and minimum values, then the maximum absolute value will be max(|max f|, |min f|). That maximum absolute value is the smallest k that works.

We'll evaluate f(x) at critical points (derivative zero) and endpoints. Compute derivative: f'(x) = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x+2)(x-1). Critical points: x = -2, x = 1. But x = -2 is also an endpoint. So interior critical point: x = 1. Also check endpoints: x = -2, x = 2.

Compute f(-2), f(1), f(2). Also maybe there could be other points where f(x) attains extreme values? Derivative zero only at x = -2 and x = 1. Since x = -2 is endpoint, only x = 1 interior. So we need to evaluate f at x = -2, 1, 2. Also check if derivative sign changes? But maximum absolute value might occur at endpoints or critical point.

Let's compute with Python. [/THOUGHT] [PYTHON] import sympy as sp

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

Evaluate at endpoints and critical point x=1

f_minus2 = f.subs(x, -2) f_1 = f.subs(x, 1) f_2 = f.subs(x, 2) print("f(-2) =", f_minus2) print("f(1) =", f_1) print("f(2) =", f_2)

Also maybe check derivative to confirm critical points

fprime = sp.diff(f, x) print("f'(x) =", fprime) critical_points = sp.solve(fprime, x) print("Critical points:", critical_points)

Evaluate f at all critical points in [-2,2]

for cp in critical_points: if cp >= -2 and cp <= 2: print(f"f({cp}) =", f.subs(x, cp))

Compute maximum and minimum of f on [-2,2]

Use numeric evaluation for comparison

import numpy as np import math

def f_numeric(x): return 2x**3 + 3x**2 - 12*x - 8

xs = np.linspace(-2, 2, 10001) vals = f_numeric(xs) max_val = np.max(vals) min_val = np.min(vals) print("Numerical max f:", max_val) print("Numerical min f:", min_val) print("Maximum absolute value:", max(abs(max_val), abs(min_val))) [/PYTHON]

[THOUGHT] The smallest possible positive k is the maximum absolute value of f(x) on [-2,2], which is max(|12|, |-15|, |-4|) = 15. The Python output confirms this: max absolute value = 15.0. Therefore k must be at least 15, and indeed for k = 15 the inequality holds because f(x) ranges from -15 to 12, so -15 ≤ f(x) ≤ 12 ≤ 15 and -15 ≥ -k? Wait: inequality is -k ≤ f(x) ≤ k. With k = 15, -15 ≤ f(x) ≤ 15 holds because f(x) min is -15 and max is 12, both within [-15,15]. So smallest k is 15. [/THOUGHT] \boxed{15}