Question

已知函数f(x)=x^2+bx+2。若当x属于[-1,4]时，f(x)>=b

f(x)>=b+3恒成立，求f(x)。若函数f(x)的定义域与值域都是[0,2]，求b的值

Answer 1

(1)设g(x)=x^2+bx+2-(b+3)=x^2+bx-b-1. 即x∈[-1,4] g(x)≥0
对称轴 x= -b/2 ① -b/2≥4 b≤-8 且 g(4)≥0 即 b≥-5
② -1＜-b/2＜4 -8＜b＜2 且g(-b/2)≥0 即b=-2
③ -b/2≤-1 b≥2 且g(-1)≥0 即b≤0
综上 b=-2 即 f(x)=x^2-2x+2
(2)由题意f(x)恒过(0,2) ∴ -b/2≥1
f(x)在[0,2]上最小值为0 ① 当1≤-b/2≤2时 -4≤b≤-2 f(-b/2)=0 得b=-2√2
②当-b/2＞2 b＜-4 f(2)=0 得b=-3
综上 b=-2√2