Question

函数f(x)=sinx+cosx在[-π/2,π/2]上的最大,最小值分别是

运用导数的方法解决

Answer 1

f(x)=sinx+cosx

=√2(√2/2sinx+√2/2cosx)
=√2sin(x+π/4)
∵-π/2<=x<=π/2
∴-π/4<=x+π/4<=3π/4
∴当x+π/4=π/2时有最大值√2
当x+π/4=-π/4时有最小值√2*(-√2/2)=-1

追问

抱歉，我需要运用导数的方法解决

追答

早说嘛，
f(x)=sinx+cosx
f'(x)=cosx-sinx=0 (令它=0，找极值点)
√2/2cosx-√2/2sinx=0
cos(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)=-1
f(π/2)=sinπ/2+cosπ/2=1
∴最小值是f(-π/2)=-1

追问

为什么当x+π/4=π/2为极值点

追答

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