已知二次函数y=f(x)满足条件f(0)=1/2m和f(x+1-f(x-1)=4x-2m,求f(x)的解析式及其最小值.
问题描述:
已知二次函数y=f(x)满足条件f(0)=1/2m和f(x+1-f(x-1)=4x-2m,求f(x)的解析式及其最小值.
答
二次函数y=f(x)满足条件f(0)=(1/2)m,∴设f(x)=ax^2+bx+m/2,由f(x+1)-f(x-1)=4x-2m,得a(x+1)^2+b(x+1)+m/2-[a(x-1)^2+b(x-1)+m/2]=4x-2m,∴4ax+2b=4x-2m,比较得a=1,b=-m.∴f(x)=x^2-mx+m/2=(x-m/2)^2+m/2-m^2/4,f(x)...