2. 考研
  3. 已知f(x)二阶可导，且f(x)＞0，f(x)f’’(x)－[f’(x)]2≥0(x∈R)． (1)证明f(x1)－f(x2)≥(x1，x2∈R)； (2)若f(0)=1，证明f(x)≥ef(0)x(x∈R)．

已知f(x)二阶可导，且f(x)＞0，f(x)f’’(x)－[f’(x)]2≥0(x∈R)． (1)证明f(x1)－f(x2)≥(x1，x2∈R)； (2)若f(0)=1，证明f(x)≥ef(0)x(x∈R)．

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问题 已知f(x)二阶可导，且f(x)＞0，f(x)f’’(x)－[f’(x)]2≥0(x∈R)．
(1)证明f(x1)－f(x2)≥(x1，x2∈R)；
(2)若f(0)=1，证明f(x)≥ef(0)x(x∈R)．
选项
答案(1)记g(x)=lnf(x)，则 [*] (2) [*] 即f(x)≥ef’(0)x．
解析
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考研数学一

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