- 函数问题函数fx=ax2
- fx=ax2-(3a-1)x+a2在-1到正无穷大上是增函数,求实数a 的范围
- f(x)=ax^2-(3a-1)x+a^2在-1到正无穷大上是增函数:
(1)a=0时,f(x)=x在-1到正无穷大上是增函数;
(2)a≠0时,
f(x)=ax^2-(3a-1)x+a^2
=a[x-(3a-1)/2a]^2+[a-(3a-1)^2/(4a^2)]
(i)a>0时,f(x)的图形开口向上,对称轴是x=(3a-1)/2a,当且仅当(3a-1)/2a≦-1时在-1到正无穷大上是增函数,此时0