[THOUGHT] To find the smallest possible positive value of $k$, I need to determine the maximum absolute value of the function $f(x) = 2x^3 + 3x^2 - 12x - 8$ on the interval $[-2, 2]$. The condition $-k \leq f(x) \leq k$ is equivalent to $|f(x)| \leq k$. Thus, the smallest such $k$ is $k = \max_{x \in [-2, 2]} |f(x)|$.

I will define the function $f(x)$ in Python and find its critical points by taking the derivative and setting it to zero. Then I will evaluate $f(x)$ at the critical points that lie within the interval $[-2, 2]$ and at the endpoints $x = -2$ and $x = 2$. Finally, I will find the maximum of the absolute values of these results.

Steps:

1. Define $f(x) = 2x^3 + 3x^2 - 12x - 8$.
2. Find $f'(x)$.
3. Solve $f'(x) = 0$ for $x$.
4. Filter roots to keep those in $[-2, 2]$.
5. Evaluate $f(x)$ at these roots and at the endpoints $-2$ and $2$.
6. Compute the absolute values of these function values.
7. The maximum of these absolute values is the answer $k$. [/THOUGHT] [PYTHON] import sympy as sp

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

Find derivative

f_prime = sp.diff(f, x)

Find critical points

critical_points = sp.solve(f_prime, x)

Filter points within [-2, 2]

points_to_check = [-2, 2] for cp in critical_points: if -2 <= cp <= 2: points_to_check.append(cp)

Evaluate f at these points

values = [f.subs(x, p) for p in points_to_check] abs_values = [abs(val) for val in values]

print(f"Critical points: {critical_points}") print(f"Points to check: {points_to_check}") print(f"Values at points: {values}") print(f"Absolute values: {abs_values}") print(f"Max absolute value: {max(abs_values)}") [/PYTHON]

The function is $f(x) = 2x^3 + 3x^2 - 12x - 8$. We want to find $k$ such that $|f(x)| \leq k$ for all $x \in [-2, 2]$. This means $k$ must be at least the maximum of $|f(x)|$ on this interval.

The critical points are found by setting $f'(x) = 0$: $f'(x) = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x+2)(x-1)$. The critical points are $x = -2$ and $x = 1$. Both lie in the interval $[-2, 2]$.

We evaluate $f(x)$ at the critical points and the endpoints:

• At $x = -2$: $f(-2) = 2(-8) + 3(4) - 12(-2) - 8 = -16 + 12 + 24 - 8 = 12$.
• At $x = 1$: $f(1) = 2(1) + 3(1) - 12(1) - 8 = 2 + 3 - 12 - 8 = -15$.
• At $x = 2$: $f(2) = 2(8) + 3(4) - 12(2) - 8 = 16 + 12 - 24 - 8 = -4$.

The values are $12, -15, -4$. The absolute values are $|12|=12$, $|-15|=15$, $|-4|=4$. The maximum absolute value is $15$. Thus, the smallest positive value of $k$ is $15$.

\boxed{15}