Question

已知函數(shù)f(x)＝ex－e－x(x∈R且e為自然對(duì)數(shù)的底數(shù))．

(1)判斷函數(shù)f(x)的奇偶性與單調(diào)性；

(2)是否存在實(shí)數(shù)t，使不等式f(x－t)＋f(x2－t2)≥0對(duì)一切x都成立？若存在，求出t；若不存在，請(qǐng)說明理由．

 

Answer 1

(1) f(x)是奇函數(shù) (2) 存在實(shí)數(shù)t＝－，使不等式f(x－t)＋f(x2－t2)≥0對(duì)一切x都成立

【解析】(1)∵f(x)＝ex－x，且y＝ex是增函數(shù)，

y＝－x是增函數(shù)，∴f(x)是增函數(shù)．

由于f(x)的定義域?yàn)?/span>R，且f(－x)＝e－x－ex＝－f(x)，

∴f(x)是奇函數(shù)．

(2)由(1)知f(x)是增函數(shù)和奇函數(shù)，

∴f(x－t)＋f(x2－t2)≥0對(duì)一切x∈R恒成立

?f(x2－t2)≥f(t－x)對(duì)一切x∈R恒成立

?x2－t2≥t－x對(duì)一切x∈R恒成立

?t2＋t≤x2＋x對(duì)一切x∈R恒成立

?2≤對(duì)一切x∈R恒成立

?2≤0?t＝－.

即存在實(shí)數(shù)t＝－，使不等式f(x－t)＋f(x2－t2)≥0對(duì)一切x都成立．