若函数f(x)=(1+tanx)cosx,0≤x
问题描述：
若函数f(x)=(1+tanx)cosx,0≤x
答
f(x)=(1+tanx)cosx
=cosx+sinx
=√2((√2/2)cosx+(√2/2)sinx)
=√2sin(x+π/4)
f(x)在[0,π/4]内增,在[π/4,π/2]内减
因此最大值为：当x=π/4时,f(π/4)=√2
最小值为：当x=0时,f(0)=√2/2