hao87fei
春芽
共回答了21个问题采纳率:95.2% 举报
f(x)=A sin(Bx+C),A改变函数幅度,B改变周期,C使函数偏移,周期距离为π/2,说明2ω=2π/(π/2)=4,解出ω=2,
f(x)关于(-π/4,0)对称,说明
将该点代入,ψ+-π/4*2ω=正负π/2,解出ψ-π=正负π/2,根据题目条件,解出ψ=π/2
f(x)=2√3sin(2ωx+ψ)=2√3sin(4x+π/2)
周期为函数在区间(-π/8,0)与(π/4,3π/8)上单调递增,(0,π/4)上单调递减
ax小于π/4时满足条件,a最大为3/4
1年前
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