📝 题目
例 1 设函数 $f$ 在 $\mathbb{R}$ 上有二阶导数. 如果
$$ {f}^{\prime \prime }\left( x\right) = 0,\;\forall x \in \mathbb{R}, $$
那么
$$ f\left( x\right) \equiv {C}_{0}x + {C}_{1}. $$
💡 答案与解析
证明 因为
$$ {\left( {f}^{\prime }\right) }^{\prime }\left( x\right) = {f}^{\prime \prime }\left( x\right) = 0,\;\forall x \in \mathbb{R}, $$
所以
$$ {f}^{\prime }\left( x\right) \equiv {C}_{0}\text{ (常数). } $$
记
$$ \varphi \left( x\right) = f\left( x\right) - {C}_{0}x, $$
则有
$$ {\varphi }^{\prime }\left( x\right) = {f}^{\prime }\left( x\right) - {C}_{0} = 0,\;\forall x \in \mathbb{R}, $$
因而
$$ \varphi \left( x\right) \equiv {C}_{1}\text{ (常数). } $$
即
$$ f\left( x\right) - {C}_{0}x \equiv {C}_{1}. $$
由此得到
$$ f\left( x\right) \equiv {C}_{0}x + {C}_{1}. $$