Question

求函数y=—tan(2x-3π/4)的单调区间,定义域,对称中心

Answer 1

y=-tan2(x-3π/8)y=tanx是增函数,定义域是x≠(2k+1)π/2 k是整数 对称中心是(kπ,0) ,k是整数所以y=-tan2(x-3π/8)在定义域上是减函数单调区间是((2k+1)π/4,(2k+3)π/4),定义域是2(x-3π/8)≠(2k+1)π/2 x-3π/8≠(2k+1)π/4x≠(2k+1)π/4+3π/8 x≠(4k+5)π/8 对称中心是2(x-3π/8)=kπ x=kπ/2+3π/8=(4k+3)π/8 ((4k+3)π/8 ,0)