f(x)=1+2sin(2x-π/3),x属于[π/4,π/2],求f(x)的最值并求出相应的x

1个回答

  • 解析:

    y = sin a 在a∈[π/4,π/2]上是增函数,在a∈[π/2,π]上是减函数

    所以令2x-π/3 = a

    则当a = π/2时有最大值,此时 2x -π/3 = π/2

    解得:x = 5π/12 < π/2,所以,可确定当 x = 5π/12时,函数有最大值,f(x)= 3

    当a在对称区间内取对称值时,函数值不变(即x=π/n 与x=π- π/n,a≠0)

    所以将x∈[π/4,π/2]带入sin(2x-π/3),比较两端的值:

    当x = π/4时,a = π/6

    当x = π/2时,a = 2π/3

    π/6较2π/3更靠近x轴,所以,sin(π/6)< sin(2π/3)

    所以sin(π/6)更小一些

    所以当x = π/4时,函数有最小值,f(x)= 2