Question

二次函数f(x)满足f(x+1)-f(x)=2x,且f(0)=1,做出f(x)的图像,并求f(x)在区间[-1,1]的值

Answer 1

设f(x)==ax2+bx+c.f(0)=c=1

f(x+1)-f(x)=a(x+1)2+b(x+1)+c-(ax2+bx+c)=2ax+a+b=2x

2a=2,a+b=0...a=1,b=-1

f(x)==x2-x+1=(x-1/2)2+3/4 ...开口向上,x=1/2时f(x)最小值为3/4

x＞1/2,f(x)随x增加而增加,x=1时f(x)为1

x＜1/2,f(x)随x减少而增加,x=-1时f(x)为3

f(x)在区间[-1,1]的值域【3/4,3】