Question

函数f(x)=cosx-sinx(x属于【-π，0】)的单调递增区间为

Answer 1

f(x) = 根号(2) * cos(x+ π/4)
∵ x∈【-π，0】
所以 x+π/4 ∈【-3/4π，π/4】
由cos函数单调性可知(0,π）为减区间
f(x)单调减区间为 【-1/4π，0】

Answer 2

f(x)=cosx-sinx=√2sin（x+3π/4）
令2kπ-π/2<=x+3π/4<=2kπ+π/2(k是整数)
得2kπ-5π/4<=x<=2kπ-π/4
再令k=0，得到-5π/4<=x<=-π/4
再与【-π，0】求交得到[-π,-π/4]
所以函数f(x)=cosx-sinx(x属于【-π，0】)的单调递增区间为[-π,-π/4]