f(x)=ab=2√3sinxcosx+cos2x
=√3sin2x+cos2x
=2(√3/2sin2x+1/2cos2x)
=2sin(2x+π/6)
-π/6≤x≤π/4
-π/3≤2x≤π/2
-π/6≤2x+π/6≤2π/3
因此f(x)=2sin(2x+π/6)
的最大值2x+π/6=π/2时得2
最小值2x+π/6=-π/6时得-1(2)求f(x)在区间[-π/6，π/4]上的最大值和最小值不是求出来了吗？-π/6≤x≤π/4-π/3≤2x≤π/2-π/6≤2x+π/6≤2π/3因此f(x)=2sin(2x+π/6)的最大值2x+π/6=π/2时得2最小值2x+π/6=-π/6时得-1