Question

已知函数f(x)=(x^2-ax+a)/x,x∈[1.+∞) (1)当a=4时，求函数f(x)的最小值

若对任意x∈[1.+∞) ，f(x)>0恒成立，试求实数a的取值范围

Answer 1

(1) when a = 4
f(x) = (x-2)^2 / x
when x = 2, f(2) = 0, which is minimum

(2) f(x) = [(x-a/2)^2 + a - a^2/4]/x
if f(x) > 0
we must have (x-a/2)^2 + a - a^2/4 > 0
when x = a/2 its minimum is a-a^2/4 which must be > 0
so a(1-a/4) > 0
we may have a > 0 and 1-a/4 > 0 or a < 4, so a in (0,4) is the range
a <0 and 1-a/4 < 0 has no solution

Answer 2

我认为：先求出对称轴x=a/2，
当a/2>=1时 a>2 当x=1时，取最小值，f（1）=1-a+a>0；
当a/2<1时a<2，则x=a/2，代入-a^2+4a/2a>0，同乘2a，得-2a^3+8a^2>0 提a^2，得a^2（-2a+8)>0则a^2>0 a为一切实数，-2a+8>0 a<1/4，则a<1/4，在综上得……a=0也包括了
答案是对的，所以放心看吧因为我做过