函数f(x)=sinx+x/2(x∈【0,2π】)的最大值是最小值是

题目

函数f(x)=sinx+x/2(x∈【0,2π】)的最大值是最小值是

答案

f(x)′=cosx+1/2
画单位圆可知
当x∈[0,2π/3]∪x∈[2π/3,2π]时,f(x)′＞0 f(x)单调递增
当x∈[2π/3,4π/3]时,f(x)′＜0 f(x)单调递减
列增减性表可知
∴当x=2π/3,f(x)有最大值f(2π/3)=sin2π/3+π/3=√3/2+π/3
当x=4π/3,f(x)有最小值f(4π/3)=sin4π/3+π/3=-√3/2+2π/3