Question

求函数f(x)=tan(x+派/4)的单调增区间？

Answer 1

tanx单增区间是(kπ-π/2,kπ+π/2)
所以kπ-π/2<x+π/4<kπ+π/2
得kπ-3π/4<x<kπ+π/4
所以f(x)=tan(x+π/4)的单调增区间(kπ-3π/4,kπ+π/4)

Answer 2

kπ-π/2<x+π/4<kπ+π/2(k属于Z)
kπ-3π/4<x+π/4<kπ+π/4
(kπ-3π/4,kπ+π/4)(k属于Z)
在所有连续区间内均为单调递增.
请采纳！

Answer 3

单增区间是：K派－3/4派＜X＜K派＋1/4派 应该是这个

Answer 4

kπ-π/2<=x+π/4<=kπ+π/2
kπ-3π/4<=x<=kπ+π/4
区间为[kπ-3π/4,kπ+π/4](k∈z)

Answer 5

kπ-π/2<x+π/4<kπ+π/2(k属于Z)
kπ-3π/4<x+π/4<kπ+π/4
(kπ-3π/4,kπ+π/4)(k属于Z)