T=2π/2=π
f（x）max=√2+2
此時sin（2x+π/4）=1
2x+π/4=π/2+2kπ k∈z
x=π/8+kπ k∈z
f（x）min=2-√2
此時sin（2x+π/4）=-1
2x+π/4=-π/2+2kπ k∈z
x=-3/8π+kπ k∈z
遞增區(qū)間
-π/2+2kπ＜2x+π/4＜π/2+2kπ k∈z
∴-3/8π+kπ＜x＜π/4+kπ k∈z
即：遞增區(qū)間是【-3/8π+kπ,π/4+kπ】k∈z
2.縱坐標乘以√2
向左平移π/4的單位
橫坐標乘以1/2
2.