[THOUGHT] We need the smallest k such that -k ≤ f(x) ≤ k for all x in [-2,2], where f(x)=2x^3+3x^2-12x-8. This k is the maximum of |f(x)| over the interval. I will find critical points from f'(x)=0, evaluate f at endpoints and critical points, then take the maximum absolute value. [/THOUGHT] [PYTHON] import sympy as sp

x=sp.Symbol('x', real=True) f=2x**3+3x**2-12*x-8 fp=sp.diff(f,x) crit=sp.solve(sp.Eq(fp,0), x) # critical points crit_simplified=[sp.nsimplify(c) for c in crit] crit, crit_simplified, fp print("f:", f) print("f':", fp) print("critical points:", crit)

filter those in [-2,2]

crit_in=[] for c in crit: if sp.N(c) >= -2-1e-12 and sp.N(c) <= 2+1e-12: crit_in.append(c) crit_in_sorted=sorted(crit_in, key=lambda t: float(sp.N(t))) print("critical points in [-2,2]:", crit_in_sorted)

evaluate f at endpoints and these points

points=[sp.Rational(-2), sp.Rational(2)] + crit_in_sorted vals=[] for p in points: v=sp.simplify(f.subs(x,p)) vals.append((p, v, sp.N(v))) print("values:") for p,v,vn in vals: print(p, v, vn)

compute max absolute value exactly (using radicals) and numerically

abs_vals=[sp.Abs(v) for ,v, in vals] max_abs=sp.simplify(sp.Max(*abs_vals)) print("abs values:", abs_vals) print("max_abs symbolic:", max_abs) print("max_abs numeric:", sp.N(max_abs))

also compute with calculus and verify by direct maximization numeric sampling

import numpy as np, math def f_num(t): return 2t**3+3t**2-12*t-8 grid=np.linspace(-2,2,2000001) m=max(abs(f_num(t)) for t in grid) tmax=grid[np.argmax([abs(f_num(t)) for t in grid])] print("grid max abs approx:", m, "at", tmax, "f=", f_num(tmax)) [/PYTHON][THOUGHT] We need the smallest k such that -k ≤ f(x) ≤ k for all x in [-2,2], where f(x)=2x^3+3x^2-12x-8. This k is the maximum of |f(x)| over the interval. I will find critical points from f'(x)=0, evaluate f at endpoints and critical points, then take the maximum absolute value. [/THOUGHT] [PYTHON] import sympy as sp

x=sp.Symbol('x', real=True) f=2x**3+3x**2-12*x-8 fp=sp.diff(f,x) crit=sp.solve(sp.Eq(fp,0), x)

print("f:", f) print("f':", fp) print("critical points:", crit)

crit_in=[] for c in crit: if sp.N(c) >= -2-1e-12 and sp.N(c) <= 2+1e-12: crit_in.append(c) crit_in_sorted=sorted(crit_in, key=lambda t: float(sp.N(t))) print("critical points in [-2,2]:", crit_in_sorted)

points=[sp.Rational(-2), sp.Rational(2)] + crit_in_sorted vals=[] for p in points: v=sp.simplify(f.subs(x,p)) vals.append((p, v, sp.N(v)))

print("values:") for p,v,vn in vals: print(p, v, vn)

abs_vals=[sp.Abs(v) for ,v, in vals] max_abs_val = max([sp.N(av) for av in abs_vals]) print("abs values:", abs_vals) print("max_abs numeric from candidates:", max_abs_val)

high-resolution numeric verification

import numpy as np def f_num(t): return 2t**3+3t2-12t-8 grid=np.linspace(-2,2,2000001) fv = 2grid3 + 3grid**2 - 12grid - 8 absfv = np.abs(fv) idx = int(absfv.argmax()) print("grid max abs approx:", float(absfv[idx])) print("at x approx:", float(grid[idx])) print("f(x) approx:", float(fv[idx])) [/PYTHON]

[THOUGHT] The smallest such (k) is (\max_{x\in[-2,2]} |2x^3+3x^2-12x-8|). From the evaluated critical points and endpoints, the maximum absolute value is (15) (at (x=1)). Therefore (k=15). [ \boxed{15} ]