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|
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![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|](/uploads/image/z/6074754-42-4.jpg?t=%E5%BE%88%E6%80%A5++++%E8%AE%BE%E5%87%BD%E6%95%B0fx%3Dax%26sup2%3B%2Bbx%2Bc%2C%E4%B8%94f%EF%BC%881%EF%BC%89%3D%EF%BC%8Da%2F2%2C3a%EF%BC%9E2c%EF%BC%9E2b%2C%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BC%881%EF%BC%89a%EF%BC%9E0%E4%B8%94%EF%BC%8D3%EF%BC%9Cb%2Fa%EF%BC%9C%EF%BC%8D3%2F4%EF%BC%882%EF%BC%89%E5%87%BD%E6%95%B0fx%E5%9C%A8%E5%8C%BA%E9%97%B4%EF%BC%880%2C2%EF%BC%89%E5%86%85%E8%87%B3%E5%B0%91%E6%9C%89%E4%B8%80%E4%B8%AA%E9%9B%B6%E7%82%B9%E5%BE%88%E6%80%A5%21%E8%B0%A2%E8%B0%A2%E8%BF%98%E6%9C%89%E7%AC%AC%E4%B8%89%E9%97%AE%E8%AE%BEx1%EF%BC%8Cx2%E6%98%AF%E5%87%BD%E6%95%B0fx%E4%B8%A4%E9%9B%B6%E7%82%B9%EF%BC%8C%E5%88%99%E6%A0%B9%E5%8F%B72%E2%89%A4%7Cx1-x2%7C)
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|
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
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|
f(1) = a + b + c = -a/2 => 3a + 2b + 2c = 0
if 3a < 0, then 0 > 3a > 2b > 2C => 3a + 2b + 2c < 0, so this condition is false;
if 3a = 0, then 0= 3a > 2c > 2b => 2c + 2b < 0, this condition is also false.
hence 3a > 0 => a > 0;
OR we have 3a + 2b + 2c = 0 < 9a => a > 0;
then because c > b, 0= 3a + 2b + 2c > 3a + 2b + 2b => 3a + 4b < 0;
Similarly, 3a > 2c => 0= 3a + 2b + 2c < 6a + 2b => b > -3a
(2). Because a > 0, then f(1) = -a/2 < 0, then f(2) = 4a + 2b + c = 3a + 2b + 2c + a - c = a - c
if c < 0 => 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) < 0 OR f(2)f(1) < 0, => at least we have one root in (0,2).
The End