已知二次函数f(x)满足f(x+1)-f(x)=2x且f(0)=1,设g(t)=f(2t+a),t∈[-1,1],求g(t)的求g(t)的最大值
问题描述：
已知二次函数f(x)满足f(x+1)-f(x)=2x且f(0)=1,设g(t)=f(2t+a),t∈[-1,1],求g(t)的
求g(t)的最大值
答
二次函数f(x)满足 f(0)=1 ，
设 f(x) = ax^2 +bx +1
f(x+1)-f(x)= [a(x+1)^2 +b(x+1) +1]-[ax^2 +bx +1] = 2ax + a+b = 2x
a = 1 , b= -1
∴ f(x) = x^2 -x +1
g(t)=f(2t+1) = (2t+1)^2 -(2t+1) +1 = 4t^2 +2t +1 = (2t+1/2)^2 + 3/4
最小值：g(-1/4) = 3/4
g(-1) = 3 , g(1) = 7
∴ g(t)的最大值为：g（1）= 7
答
f(x)>2x+mx^2-x+1>2x+mx^2-3x+1-m>0令g(x)=x^2-3x+1-m可以得出当x0m=1/...