f（x）=2sin（兀/4x+兀/4）当x属于[-6,-2/3]时求f（x）+f（x+2）最大值与最小值及相应x的值.

Question

f（x）=2sin（兀/4x+兀/4）当x属于[-6,-2/3]时求f（x）+f（x+2）最大值与最小值及相应x的值.

Answer 1

f(x)=2sin(π/4*(x+1))
f(x+2)=2sin(π/4*((x+2)+1))=2sin(π/4*((x+1)+2))
=2sin(π/2+π/4*(x+1))
=2cos(π/4*(x+1))
f(x)+f(x+2)=2sin(π/4*(x+1))+2cos(π/4*(x+1))
=2√2*(sin(π/4*(x+1))*1/√2+cos(π/4*(x+1))*1/√2)
=2√2*(sin(π/4*(x+1))*cosπ/4+cos(π/4*(x+1))*sinπ/4)
=2√2*sin(π/4*(x+1)+π/4)
=2√2*sin(π/2+π/4*x)
=2√2*cos(π/4*x)
当 x∈[-6,-2/3] 时
-π3/2≤π/4*x≤-π/6
f(x)+f(x+2)=2√2*cos(π/4*x)
在 π/4*x=-π 时 x=-4
有最小值
f(x)+f(x+2)=2√2*cos(-π)
=-2√2
当 x=-2/3 时有最大值
f(x)+f(x+2)=2√2*cos(π/4*x)
=2√2*cos(π/4*(-2/3))
=2√2*cos(-π/6)
=2√2*√3/2
=+√6