For all real numbers with , the inequalityholds. Find the smallest possible positive value of . [3 points]\
[THOUGHT] The smallest k is the maximum of |p(x)| on [-2,2], where p(x) = 2x^3 + 3x^2 - 12x - 8. Find critical points by solving p'(x)=0. p'(x)=6x^2 + 6x -12=0 → x^2 + x -2=0 → x=-2, x=1. Evaluate p at x=-2,1,2. [/THOUGHT] [PYTHON] def p(x): return 2x**3 + 3x**2 -12*x -8
print(p(-2)) print(p(1)) print(p(2)) print(max([abs(p(-2)), abs(p(1)), abs(p(2))])) [/PYTHON]
[THOUGHT] The values are p(-2)=12, p(1)=-15, p(2)=-4. The maximum of the absolute values is 15, which is the smallest k. [/THOUGHT] \boxed{15}
Explanation
The reasoning identifies that the smallest k is the maximum absolute value of p(x) = 2x³ + 3x² - 12x - 8 on the interval [-2, 2]. To find this, locate the critical points by solving p'(x) = 0. The derivative is p'(x) = 6x² + 6x - 12, which simplifies to x² + x - 2 = 0. Factoring gives (x + 2)(x - 1) = 0, so critical points are x = -2 and x = 1. Evaluate p(x) at these points and the endpoint x = 2: p(-2) = 12, p(1) = -15, and p(2) = -4. The absolute values are |12| = 12, |-15| = 15, and |-4| = 4. The maximum is 15. As a cubic polynomial, p(x) has extrema at these critical points and endpoints, confirming 15 as the global maximum of |p(x)| on [-2, 2]. Thus, k must be at least 15 to satisfy -k ≤ p(x) ≤ k for all x in the interval.