Question

已知函数f(x)=2sin(x-4/ π)的最小值

已知函数 f(x)=2sin(π/4x+π/4)当X∈【-2,2】时,求函数y=f(x-1)+f(x+1)的最小值及取最小值时相应的x的值

Answer 1

解由y=f(x-1)+f(x+1)
=2sin(π/4(x-1)+π/4)+2sin(π/4(x+1)+π/4)
=2sinπ/4x+2sin(π/4x+π/2)
=2sinπ/4x+2cos(π/4x)
=2√2(√2/2sinπ/4x+√2/2cos(π/4x))
=2√2sin(π/4x+π/4)
由X∈【-2,2】
得πX/4∈【-π/2,π/2】
即πX/4+π/4∈【-π/4,3π/4】
故当πX/4+π/4=-π/4时，
即x=-2时，y=f(x)有最小值y=2√2×(-√2/2)=-2