求y=4sinx*cosx在0到2的区间 上的最小值?

2025-07-26 16:03:22
推荐回答(1个)
回答1:

y=4sinxcosx=2sin2x
x=0,y=2sin0=0
x=π/4,y=2sinπ/2=2
x=π/2,y=2sinπ=0
x=3π/4,y=2sin3π/2=-2
y=2sin2x在[π/4,3π/4]的区间内递减,在[0,π/4]的区间内递增
π/2,6,