函数f(x)=In(4+3x-x²)的单调递减区间是 A.(负无穷,3/2] B.[3/2,正无穷) C.(-1,3/2] D.[3/2,4)
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![函数f(x)=In(4+3x-x²)的单调递减区间是 A.(负无穷,3/2] B.[3/2,正无穷) C.(-1,3/2] D.[3/2,4)](/uploads/image/z/14774295-39-5.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3DIn%284%2B3x-x%26%23178%3B%EF%BC%89%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F%E5%8C%BA%E9%97%B4%E6%98%AF+A.%EF%BC%88%E8%B4%9F%E6%97%A0%E7%A9%B7%2C3%2F2%5D+B.%5B3%2F2%2C%E6%AD%A3%E6%97%A0%E7%A9%B7%29+C.%28-1%2C3%2F2%5D+D.%5B3%2F2%2C4%29)
函数f(x)=In(4+3x-x²)的单调递减区间是 A.(负无穷,3/2] B.[3/2,正无穷) C.(-1,3/2] D.[3/2,4)
函数f(x)=In(4+3x-x²)的单调递减区间是 A.(负无穷,3/2] B.[3/2,正无穷) C.(-1,3/2] D.[3/2,4)
函数f(x)=In(4+3x-x²)的单调递减区间是 A.(负无穷,3/2] B.[3/2,正无穷) C.(-1,3/2] D.[3/2,4)
考查的是复合函数的单调性
把复合函数分成二次函数和对数函数
函数f(x)=log1/2 (6-x-x²)的定义域:
6-x-x²>0
x²+x-6<0
(x+3)(x-2)<0
-3<x<2
故定义域:(-3,2)
令t=-x²-x+6=-(x+1/2)²+25/4
则函数t在(-3,-1/2)上递增,在[-1/2,2)上递减
又函数y=log1/2(x)在定义域上单调递减
故函数f(x)=log1/2 (6-x-x²)的单调增区间是:[-1/2,2)
选B.