已知函数fx=ax*2+2ax+4(a.>0)若x1

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已知函数fx=ax*2+2ax+4(a.>0)若x1已知函数fx=ax*2+2ax+4(a.>0)若x1已知函数fx=ax*2+2ax+4(a.>0)若x1f(x)=ax^2+2ax+4=a(x+1)

已知函数fx=ax*2+2ax+4(a.>0)若x1
已知函数fx=ax*2+2ax+4(a.>0)若x1

已知函数fx=ax*2+2ax+4(a.>0)若x1
f(x)=ax^2+2ax+4=a(x+1)^2-a+4
∵x1∴x1<0,x2>0,且x1=-x2
又∵a>0
∴f(x1)-f(x2)
=[a(-x2+1)^2-a+4]-[a(x2+1)^2-a+4]
=a(-x2+1+x2+1)(-x2+1-x2-1)
=-4ax2<0所以
f(x1)