已知函数f(x)=cos(2x+π/6)求f(x)在区间【-π/6,π/2】上的最大值和最小值,并求出相应的x的值.

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已知函数f(x)=cos(2x+π/6)求f(x)在区间【-π/6,π/2】上的最大值和最小值,并求出相应的x的值.

已知函数f(x)=cos(2x+π/6)求f(x)在区间【-π/6,π/2】上的最大值和最小值,并求出相应的x的值.
已知函数f(x)=cos(2x+π/6)
求f(x)在区间【-π/6,π/2】上的最大值和最小值,并求出相应的x的值.

已知函数f(x)=cos(2x+π/6)求f(x)在区间【-π/6,π/2】上的最大值和最小值,并求出相应的x的值.
-π/6<=x<=π/2
-π/3<=2x<=π
-π/6<=2x+π/6<=7π/6
所以2x+π/6=0时最大,2x+π/6=π时最小
所以
x=-π/12,最大值=1
x=5π/12,最小值=-1

-π/3<=2x<=π
-π/6<=2x+π/6<=7π/6
f(x)=cos(2x+π/6)
2x+π/6=0
x=-π/12
f(max)=1
2x+π/6=π
2x=5π/6
x=5π/12
f(min)=-1