Question

求函数y=3sin(2x+π/4),x∈[0,π]的单调递减区间和单调递增区间及最大值与最小值.

Answer 1

x∈[0,π]
那么π/4<=2x+π/4<=9π/4
所以最大值为3*1=3,最小宽乎值=3*(-1)=-3
设t=2x+π/4
当π/4<=t<=9π/4
y=3sint增区间为[π/4,π/2]与[3π/2,9π/4]
对应x的递增毕掘区间为[0,π/8]与[5π/8,π]
x的递减区间为[π/手巧核8,5π/8]