欧阳淼
幼苗
共回答了23个问题采纳率:95.7% 举报
∵f(x)=√2sin(ωx+π/4)+b 的最小正周期T=π,
T=2π/ω=π,
∴ω=2.
∵f(x)的最大值为2√2.只有当sin(2x+π/4)=1时,f(x)取得最大值.
即,f(x)max=√2*1+b=2√2.
∴b=√2.
1.∴解得ω=2,b=√2.f(x)=√2sin(2x+π/4)+√2.
2.x∈[0,π],存在满足f(x)=2√2的x值.
令f(x)=√2sin(2x+π/4)+√2中的sin(2x+π/4)=1,
则,2x+π/2=π/2,
2x=π/2-π/4.
2x=π/4.
x=π/8,∵π/8∈[0,π].
∴当x=π/8时,f(x)=√2*1+√2=2√2.
3.F(x)=f(x)-f(x-π/4).
=√2sin(2x+π/4)+√2-{√2sin[2(x-π/4)+π/4]+√2}.
=√2[sin(2x+π/4)-sin(2x-π/4)].
=√2*2cos(2x)*sin(π/4)
=2cos2x.
当cos2x=1.x=2kπ时,F(x)取得最大值,F(x)max=2;
当cos2x=-1,x=2kπ+π时,F(x)取得最小值,F(x)min=-2.
1年前
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