二次函数f(x)满足f(x+1)-f(x)=2x+3,且f(0)=2

题目

二次函数f(x)满足f(x+1)-f(x)=2x+3,且f(0)=2
1.求f（x）的解析式
2.求f（x）在【-3,4】上的值域
3.若函数f（x+m）为偶函数,求f【f（m）】的值
4.求f（x）在【m,m+2】上的最小值

答案

f(x) = ax² + bx  + c,f(0) = c = 2
f(x+ 1) = a(x + 1)² + b(x+ 1) + c
f(x + 1) - f(x) = a[(x + 1) + x][(x + 1) - x]+b(x + 1 - x) = a(2x + 1) + b = 2ax + a + b = 2x + 3
2a = 2,a = 1,a + b = 3,b = 2
f(x) = x² + 2x + 2


f(x) = (x + 1)² + 1,对称轴x = -1在[-3,4]内,顶点(-1,1)
4离对称轴更远,f(4) = 26
[=3,4]上的值域:[1,26]

对称轴x = -1,f(x)向右平移一个单位时(x -> x -1),对称轴为y轴,新抛物线为偶函数,m = -1
f(m) = f(-1) = 1,f(f(-1)) = f(1) = 5


区间[m,m + 2]为[-1,1],f(x)为开口向上的抛物线,对称轴x = -1在[-1,1]内,最小值为顶点的纵坐标1