二次函数y=f(x)满足：（1）f(0)=1;(2)f(x+1)-f(x)=2x,求f(x)的解析式,求f(x)在区间【-1,1】最小值

设 y = ax² + bx + c
因为f(0) = 1
所以 c = 1
因为f(x+1)-f(x)=2x,
所以 ax² + 2ax + a + bx + b + 1 - ax² - bx - 1 = 2x
化简得：2ax + a + b = 2x
所以 a = 1 , b = -1
所以f(x) = x² - x + 1
f(x) = x² - x + 1
=(x - 1/2)² + 3/4
当 x = 1/2 时
fmin = f(1/2) = 3/4