Question

设f(x)=(x^2+x+a)/(x+1),x属于[0,∞).当a=2时,求f(x) 的最小值. 当0

Answer 1

(1)a＝2时,

f(x)＝(x²+x+2)/(x+1)

→f(x)＝(x+1)+[2/(x+1)]-1

≥2√[(x+1)·2/(x+1)]-1

＝2√2-1.

故所求最小值为：

f(x)|min＝2√2-1,

此时,x+1＝2/(x+1),

即x＝√2-1 (只取正值)

(2)当0