题目

二次函数f(x)满足f(x+1)-f(x)=2x,且f(0)=1 ,当在区间【-1,2】上求y=f(x)的值域

提问时间：2020-10-09

答案

f(x) = ax^2 + bx + c
f(0)=1:c = 1
f(x) = ax^2 + bx + 1
f(x+1) = a(x+1)^2 + b(x+1) + 1 = ax^2 + 2ax + a + bx + b + 1
f(x+1)-f(x) = 2ax + a + b = 2x
a = 1
a+b = 0,b = -1
f(x) = x^2 - x + 1 = (x - 1/2)^2 + 3/4
f(x)对称轴为x = 1/2,顶点(1/2,3/4)
x^2系数>0,f(x)开口向上,最小值3/4
区间【-1,2】以x = 1/2为对称轴,最大值=f(-1) = f(2) = 3
在区间【-1,2】上的值域:[3/4,3]