f(x)=sinx+cosx
=√2(√2/2sinx+√2/2cosx)
=√2sin(x+π/4)
∵-π/2抱歉，我需要运用导数的方法解决早说嘛，f(x)=sinx+cosxf'(x)=cosx-sinx=0 (令它=0，找极值点)√2/2cosx-√2/2sinx=0cos(x+π/4)=0∵-π/2<=x<=π/2∴-π/4<=x+π/4<=3π/4∴当x+π/4=π/2为极值点x=π/4当 -π/4<=x+π/4<=π/2时cos为增区间当π/2<=x+π/4<=3π/4时cos为减区间∴最大值是f(π/4)=√2/2+√2/2=√2∵f(-π/2)=sin(-π/2)+cos(-π/2)=-1f(π/2)=sinπ/2+cosπ/2=1∴最小值是f(-π/2)=-1为什么当x+π/4=π/2为极值点我晕，你不是学过导数吗，这个忘了f'(x)=cosx-sinx=√2(√2/2cosx-√2/2sinx)=√2cos(x+π/4)=0∵-π/2<=x<=π/2∴-π/4<=x+π/4<=3π/4∴当x+π/4=π/2时 f'(x)=0∴x+π/4=π/2为极值点