若-π/2≤x≤π/2,求f(x)=√3sinx+cosx的最大值和最小值,并求出此时的x的值

题目

若-π/2≤x≤π/2,求f(x)=√3sinx+cosx的最大值和最小值,并求出此时的x的值

答案

f(x)=2(√3/2sinx+1/2cosx)
=2(sinxcosπ/6+cosxsinπ/6)
=2sin(x+π/6)
-π/2<=x<=π/2
-π/3<=x+π/6<=2π/3
所以x+π/6=-π/3,sin(x+π/6)最小=-√3/2
x+π/6=π/2,sin(x+π/6)最大=1
所以
x=-π/2,f(x)最小=-√3
x=π/3,f(x)最大=2