若函数y=log1/2^[2x^2-3x+1]的递减区间?
问题描述:
若函数y=log1/2^[2x^2-3x+1]的递减区间?
答
定义域x1
y=log2(t)是减函数,复合函数y=log1/2^[2x^2-3x+1]是减函数
t=2(x-3/4)^2-1/8应该为增函数
x>=3/4,
y=log1/2^[2x^2-3x+1]的递减区间(1,正无穷)