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

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