Question

求函数y=-tan(x+π/6)+2的单调区间，周期，定义域

Answer 1

最小正周期T=2π 使函数y=-tan(x+π/6)+2有意义应有x+π/6 ≠π/2+kπ，即x≠π/3+kπ，k为任意整数。故 函数y=-tan(x+π/6)+2的定义域为{x| x≠π/3+kπ，k为任意整数} 由- π/2+kπ< x+π/6< π/2+kπ得-2 π/3+kπ