设函数f(x)=ax+bxcot2x+c×cos(πx/2),若f(1)=10,则f(-1)=?
问题描述:
设函数f(x)=ax+bxcot2x+c×cos(πx/2),若f(1)=10,则f(-1)=?
答
f(1) = a + bcot2 + ccos(π/2)
=a + bcot2 = 10
f(-1) = -a - bcot2 + ccos(-π/2)
= -a - bcot2 = -10