f(x)=sinx+cosx
=√2(√2/2sinx+√2/2cosx)
=√2sin(x+π/4)
∵-π/2抱歉，我需要運(yùn)用導(dǎo)數(shù)的方法解決早說(shuō)嘛，f(x)=sinx+cosxf'(x)=cosx-sinx=0 (令它=0，找極值點(diǎn))√2/2cosx-√2/2sinx=0cos(x+π/4)=0∵-π/2<=x<=π/2∴-π/4<=x+π/4<=3π/4∴當(dāng)x+π/4=π/2為極值點(diǎn)x=π/4當(dāng) -π/4<=x+π/4<=π/2時(shí)cos為增區(qū)間當(dāng)π/2<=x+π/4<=3π/4時(shí)cos為減區(qū)間∴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為什么當(dāng)x+π/4=π/2為極值點(diǎn)我暈，你不是學(xué)過(guò)導(dǎo)數(shù)嗎，這個(gè)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∴當(dāng)x+π/4=π/2時(shí) f'(x)=0∴x+π/4=π/2為極值點(diǎn)