Question

已知函數(shù)f(x)=,x∈［1,+∞).

(1)當(dāng)a=時(shí),判斷證明f(x)的單調(diào)性并求f(x)的最小值;

(2)若對(duì)任意x∈［1,+∞),f(x)＞1恒成立,試求實(shí)數(shù)a的取值范圍.

Answer 1

解:(1)當(dāng)a=時(shí),f(x)=x++2,在［1,+∞)上任取x₁＜x₂,

f(x₁)-f(x₂)=(x₁-x₂)(1)＞0,所以f(x)在［1,+∞)為單調(diào)遞增,

f(x)的最小值應(yīng)為當(dāng)x=1時(shí)取到f(1)=;

(2)在區(qū)間［1,+∞)上,f(x)=＞1等價(jià)于x²+x+a＞0,

而g(x)=x²+x+a=(x+)²+a在［1,+∞)上遞增,

所以當(dāng)x=1時(shí),g(x)_min=2+a,當(dāng)且僅當(dāng)g(x)_min=2+a＞0時(shí),恒有f(x)＞1,

即實(shí)數(shù)a的取值范圍為a＞-2.