f(x)=2+√2sin(2x+π/4) 1.求减区间 2.x∈[-π/2,π/2],求f(x)值域 3.f(x)≥1/2,求X取值范围

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f(x)=2+√2sin(2x+π/4) 1.求减区间 2.x∈[-π/2,π/2],求f(x)值域 3.f(x)≥1/2,求X取值范围
f(x)=2+√2sin(2x+π/4) 1.求减区间 2.x∈[-π/2,π/2],求f(x)值域 3.f(x)≥1/2,求X取值范围

f(x)=2+√2sin(2x+π/4) 1.求减区间 2.x∈[-π/2,π/2],求f(x)值域 3.f(x)≥1/2,求X取值范围
1,令2kπ+π/2≤2x+π/4≤2kπ+3π/2
那么kπ+π/8≤x≤kπ+5π/8
所以减区间为[kπ+π/8,kπ+5π/8] (k∈Z)
2,当-π/2≤x≤π/2时,
-3π/4≤2x+π/4≤5π/4
-1≤sin(2x+π/4)≤1
那么2-√2≤2+√2sin(2x+π/4)≤2+√2
即值域为:[2-√2,2+√2]
3,f(x)=2+√2sin(2x+π/4)≥1/2
sin(2x+π/4)≥-3√2/4
那么2kπ+arcsin(-3√2/4)≤2x+π/4≤2kπ-arcsin(-3√2/4)
解得:kπ-π/8+1/2*arcsin(-3√2/4)≤x≤kπ-π/8-1/2*arcsin(-3√2/4)

1,减区间只给你求法,其余你自己算吧 2x+π/4大于π/2+2kπ小于3/2π+2kπ然后求出x值就是减区间