很急设函数fx=ax²+bx+c,且f（1）=－a/2,3a＞2c＞2b,求证：
问题描述：
很急设函数fx=ax²+bx+c,且f（1）=－a/2,3a＞2c＞2b,求证：
（1）a＞0且－3＜b/a＜－3/4
（2）函数fx在区间（0,2）内至少有一个零点
很急!谢谢
还有第三问
设x1，x2是函数fx两零点，则
根号2≤|x1-x2|＜根号57 /4
答
f(1) = a + b + c = -a/2 => 3a + 2b + 2c = 0
if 3a 3a > 2b > 2C => 3a + 2b + 2c if 3a = 0, then 0= 3a > 2c > 2b => 2c + 2b hence 3a > 0 => a > 0;
OR we have 3a + 2b + 2c = 0 a > 0;
then because c > b, 0= 3a + 2b + 2c > 3a + 2b + 2b => 3a + 4b Similarly, 3a > 2c => 0= 3a + 2b + 2c b > -3a
(2). Because a > 0, then f(1) = -a/2 if c a - c > 0 ,
if c = 0 => a - c = a > 0,
if c > 0 => f(0) = c > 0.
So no matter what value c takes, f(0)f(1) at least we have one root in (0,2).
The End