📝 题目
例 24 设 $U\left( {{x}_{0},\delta }\right) \subset {\mathbf{R}}^{n},f$ 在 $U\left( {{x}_{0},\delta }\right)$ 内连续,在 $U\left( {{x}_{0},\delta }\right) \smallsetminus \left\{ {x}_{0}\right\}$ 内可微. 求证:
(1)如果 $\left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) \nabla f\left( \mathbf{x}\right) < 0\left( {\forall \mathbf{x} \in U\left( {{\mathbf{x}}_{0},\delta }\right) \smallsetminus \left\{ {\mathbf{x}}_{0}\right\} }\right)$ ,那么 ${\mathbf{x}}_{0}$ 是 $f$ 的一个极大值点;
(2)如果 $\left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) \cdot \nabla f\left( \mathbf{x}\right) > 0\left( {\forall \mathbf{x} \in U\left( {{\mathbf{x}}_{0},\delta }\right) \smallsetminus \left\{ {\mathbf{x}}_{0}\right\} }\right)$ ,那么 ${\mathbf{x}}_{0}$ 是 $f$ 的一个极小值点;
(3) 如果 $f\left( {x,y}\right) = {x}^{2} + {2xy} + 3{y}^{2} + {2x} + {10y} + 9$ ,则(1, - 2)是 $f\left( {x,y}\right)$ 的一个极小值点.
💡 答案与解析
证 (1) 设 $F\left( t\right) \overset{\text{ 定义 }}{ = }f\left( {{x}_{0} + t\left( {x - {x}_{0}}\right) }\right) \left( {0 \leq t \leq 1}\right)$ . 显然 $F\left( t\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续,在(0,1)内可微,且
$$ {F}^{\prime }\left( t\right) = \left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) \cdot \nabla f\left( {{\mathbf{x}}_{0} + t\left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) }\right) . $$
根据微分中值定理, $\exists \xi \in \left( {0,1}\right)$ ,使得
$$ f\left( \mathbf{x}\right) - f\left( {\mathbf{x}}_{0}\right) = F\left( 1\right) - F\left( 0\right) = {F}^{\prime }\left( \xi \right) $$
$$ = \left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) \cdot \nabla f\left( {{\mathbf{x}}_{0} + \xi \left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) }\right) . $$
于是,当 $x \in U\left( {{x}_{0},\delta }\right) \smallsetminus \left\{ {x}_{0}\right\}$ 时, ${x}_{0} + \xi \left( {x - {x}_{0}}\right) \in U\left( {{x}_{0},\delta }\right) \smallsetminus \left\{ {x}_{0}\right\}$ . 所以 $\left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) \cdot \nabla f\left( {{\mathbf{x}}_{0} + \xi \left( {\mathbf{x} - {\mathbf{x}}_{0}}\right) }\right) < 0$ ,即有 $f\left( \mathbf{x}\right) < f\left( {\mathbf{x}}_{0}\right)$ .
(2)考虑 ${f}^{ * } = - f$ ,由假设有
$$ \left( {x - {x}_{0}}\right) \cdot \nabla {f}^{ * }\left( x\right) = - \left( {x - {x}_{0}}\right) \cdot \nabla f\left( x\right) < 0 $$
$$ \left( {\forall x \in U\left( {{x}_{0},\delta }\right) \smallsetminus \left\{ {x}_{0}\right\} }\right) . $$
根据 (1) 的结果推出 ${f}^{ * }\left( \mathbf{x}\right) < {f}^{ * }\left( {\mathbf{x}}_{0}\right)$ ,即 $f\left( \mathbf{x}\right) > f\left( {\mathbf{x}}_{0}\right)$ .
(3)令 $u = x - 1,v = y + 2$ ,则
$$ \left( {x - 1,y + 2}\right) \cdot \nabla f\left( {x,y}\right) $$
$$ = \left( {x - 1,y + 2}\right) \cdot \left( {{2x} + {2y} + 2,{2x} + {6y} + {10}}\right) $$
$$ = {2u}\left( {u + v}\right) + {2v}\left( {u + {3v}}\right) = 2\left\lbrack {{\left( u + v\right) }^{2} + 2{v}^{2}}\right\rbrack > 0 $$
$\left( {\forall \left( {u,v}\right) \neq \left( {0,0}\right) }\right)$ . 利用 (2) 的结果,即可肯定(1, - 2)是 $f\left( {x,y}\right)$ 的一个极小值点.