[THOUGHT] To find the smallest possible 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 $[-2, 2]$, I need to determine the maximum absolute value of the function $f(x) = 2x^3 + 3x^2 - 12x - 8$ over this interval. The maximum value of $|f(x)|$ will give me the smallest $k$ that satisfies the given inequality.

First, I will find the critical points of $f(x)$ by taking its derivative and setting it to zero. Then, I will evaluate $f(x)$ at the critical points and at the endpoints of the interval $[-2, 2]$. Finally, I will determine the maximum absolute value of these evaluations. [/THOUGHT] [PYTHON] from sympy import symbols, diff, solve

Define the variable and the function

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

Calculate the derivative

f_prime = diff(f, x)

Find the critical points by solving f'(x) = 0

critical_points = solve(f_prime, x)

Evaluate the function at the critical points and the endpoints of the interval [-2, 2]

endpoints = [-2, 2] all_points = critical_points + endpoints all_points = [point.evalf() for point in all_points if -2 <= point <= 2] # Filter points in the interval

Evaluate f(x) at these points

values = [f.subs(x, point).evalf() for point in all_points]

Find the maximum absolute value of these evaluations

k = max(abs(value) for value in values) print(k)

[THOUGHT] Upon reviewing the Python code execution, I realize there was an error in handling the data types. I will re-evaluate the critical points and endpoints manually to ensure accuracy.

The derivative $f'(x) = 6x^2 + 6x - 12$. Solving $6x^2 + 6x - 12 = 0$ gives $x^2 + x - 2 = 0$, which factors to $(x+2)(x-1) = 0$, yielding critical points $x = -2$ and $x = 1$.

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

• $f(-2) = 2(-2)^3 + 3(-2)^2 - 12(-2) - 8 = -16 + 12 + 24 - 8 = 12$
• $f(1) = 2(1)^3 + 3(1)^2 - 12(1) - 8 = 2 + 3 - 12 - 8 = -15$
• $f(2) = 2(2)^3 + 3(2)^2 - 12(2) - 8 = 16 + 12 - 24 - 8 = -4$

The maximum absolute value among $12, -15, -4$ is $15$. Therefore, the smallest possible positive value of $k$ is $15$. [/THOUGHT] $\boxed{15}$