Question

求函数Y=-2sin(x+π/6)+3在下列区间的最大值和最小值及对应的x的值（1）R,(2)[0,π](3)[-π/2,π/2]

Answer 1

y=-2sin(x+π/6)+3
x属于R
所以 x+π/6属于R
最大值=5 x+π/6=2kπ-π/2 x=2kπ-2π/3
最小值=1 x+π/6=2kπ+π/2 x=2kπ+π/3
2 x属于[0,π]
x+π/6 属于[π/6,7π/6]
最大值=4 x+π/6=7π/6 x=π
最小值=1 x+π/6=π/2 x=π/3
3 x属于[-π/2,π/2]
x+π/6属于[-π/3,2π/3]
最大值=3+根号下3 x+π/6=-π/3 x=-π/2
最小值=1 x+π/6=π/2 x=π/3