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로그인 회원가입 You are solving a Korean CSAT mathematics problem. Reason carefully, but the final response must contain exactly one final answer inside \boxed{}.

For multiple-choice problems, put only one of ①, ②, ③, ④, ⑤ inside \boxed{}. For short-answer problems, put only the final integer inside \boxed{}. Do not use external tools.

Problem 19 For all real numbersxwith−2≤x≤2, the inequality −k≤2x 3 + 3x2 −12x−8≤k holds. Find the smallest possible positive value ofk. [3 points] 더보기

The function 𝑓 ( 𝑥 )

2 𝑥 3 + 3 𝑥 2 − 12 𝑥 − 8 f(x)=2x 3 +3x 2 −12x−8 must satisfy − 𝑘 ≤ 𝑓 ( 𝑥 ) ≤ 𝑘 −k≤f(x)≤k for all 𝑥 ∈ [ − 2 , 2 ] x∈[−2,2], which is equivalent to ∣ 𝑓 ( 𝑥 ) ∣ ≤ 𝑘 ∣f(x)∣≤k. The smallest such 𝑘 k is max ⁡ 𝑥 ∈ [ − 2 , 2 ] ∣ 𝑓 ( 𝑥 ) ∣ max x∈[−2,2] ​

∣f(x)∣.

Find the critical points:

𝑓 ′ ( 𝑥 )

6 𝑥 2 + 6 𝑥 − 12

6 ( 𝑥 + 2 ) ( 𝑥 − 1 )

0    ⟹    𝑥

− 2 ,    𝑥

f ′ (x)=6x 2 +6x−12=6(x+2)(x−1)=0⟹x=−2,x=1.

Evaluate 𝑓 f at the critical points and endpoints:

𝑓 ( − 2 )

2 ( − 8 ) + 3 ( 4 ) − 12 ( − 2 ) − 8

− 16 + 12 + 24 − 8

12 , f(−2)=2(−8)+3(4)−12(−2)−8=−16+12+24−8=12, 𝑓 ( 1 )

2 ( 1 ) + 3 ( 1 ) − 12 ( 1 ) − 8

2 + 3 − 12 − 8

− 15 , f(1)=2(1)+3(1)−12(1)−8=2+3−12−8=−15, 𝑓 ( 2 )

2 ( 8 ) + 3 ( 4 ) − 12 ( 2 ) − 8

16 + 12 − 24 − 8

− 4. f(2)=2(8)+3(4)−12(2)−8=16+12−24−8=−4.

The absolute values are 12 , 15 , 4 12,15,4. The maximum is 15 15.

Thus the smallest possible positive 𝑘 k is 15 15.

\boxed{15}

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