Question

已知函数f(x)=1 +2sin(2x-3分之兀)当x∈[4分之兀,2分之兀]求f(x)的最大值

最小值

Answer 1

x∈[π/4，π/2]
→2x-π/3∈[π/6，2π/3].
∴x=π/2时，
f(x)|max=1+2sin(2π/3)=1+√3;
当x=π/4时，
f(x)|min=1+2sin(π/6)=2。