在區(qū)間〔1,3〕上, f'(x) = x^2 + 2ax + 5 = (x - (-a))^2 + 5 - a^2 >= 0, f"(1)=2a+6, f'(3)=6a+14 ==>(1) -a = 0a < -3, 6a+14 >= 0a < -3, a >= -14/6 a的取值范圍 a >= -5^(1/2)如果這個(gè)函數(shù)在區(qū)間〔1，3〕上是單調(diào)函數(shù)，則A的取值范圍呢，謝謝了，看懂會(huì)加分。similar. 在區(qū)間〔1，3〕上單調(diào) decreasing<=>在區(qū)間〔1，3〕上, f'(x) <= 0 <=>f'(1) <= 0 and f'(3) <= 0 ( since f'(x) is open up )<=>2a + 6 <= 0 and 6a+14 <= 0<=>a <= -3anda <= -7/3 <=>a <= -3 如果這個(gè)函數(shù)在區(qū)間〔1，3〕上是單調(diào)函數(shù)，則A的取值范圍是 { a <= -3 } U { a >= -5^(1/2) }