函数f(x)=(√2)sin[2x-(π/4)],x∈R.
【1】
T=π.
【2】
∵π/8≤x≤3π/4
∴π/4≤2x≤6π/4
∴0≤2x-(π/4)≤π+(π/4)
∴-√2/2≤sin[2x-(π/4)]≤1
∴-1≤(√2)sin[2x-(π/4)]≤√2.
即-1≤f(x)≤√2.
∴f(x)min=-1,f(x)max=√2.
【1】由题设可知:x∈[π/8,3π/4].即有:π/8≤x≤3π/4.∴0≤2x-(π/4)≤5π/4.即此时(2x-π/4)∈[0,5π/4]【2】数形结合,你画出正弦曲线图,可知:在区间[0,5π/4]上,sinx的值域是-√2/2≤sinx≤1.即-√2/2≤sin[2x-(π/4)]≤1.∴-1≤(√2)sin[2x-(π/4)]≤√2.即-1≤y≤√2.∴ymin=-1,ymax=√2