- 一道高一的函数f(x)=ax^2+bx+c且过(
- f(x)=ax^2+bx+c且过(-1,0)要使x
- ok,finally find it...
已知二次函数f(x)=ax^2+bx+c
(1)f(-1)=0
(2)x<=f(x)<=(x^2+1)/2
对于任意实数x恒成立,求a,b,c
解: a-b+c=0 b=a+c
f(x)-x≥0
ax^2+(b-1)x+c≥0
a>0
△=(b-1)^-4ac=(a+c-1)^-4ac=(a-c)^-2(a+c)+1≤ 0
f(x)-(x^2+1)/2≥0
(a-1/2)x ^+bx+c-1/2≤0 (a-1/2)<0
△=b^-4(a-1/2)(c-1/2)=(a+c)^-4(a-1/2)(c-1/2)
=(a-c)^+2(a+c)-1≤0
两式相加: (a-c)^≤0 ∵(a-c)^≥0
∴(a-c)^=0 a=c
带回(a-c)^+2(a+c)-1≤0 a≤1/4
带回(a-c)^-2(a+c)+1≤ a≥1/4
∴a=1/4=c b=1/2