Question

以知f(x)＝x²＋bx＋5有最小值1，则f(2)＝？

Answer 1

f(x) = x^2 + bx +3 (1)
x∈[-1,2]
f ' (x) = 2x + b
令：f ' (x) = 0,b
解出：f(x)取最小值的点：x* = - b/2,
将其代入(1) 得到方程：
b^2/4 - b^2/2 +3 = 1,即
b^2 = 8 (2)
b = 土 2√2
舍弃正根（超出x的定义域）,b = - 2√2,x* = - b/2 = √2 ,
于是f(x) 的解析式为：
f(x) = x^2 - 2√2x + 3 (3)
验证：min：f(x*) = f (√2) = √2√2 - 2√2√2 + 3 = 2 - 4 + 3 = 1
符合题设.