📝 题目
例 4 给定 $f\left( {x,y}\right) = 2{\left( y - {x}^{2}\right) }^{2} - \frac{1}{7}{x}^{7} - {y}^{2}$ .
(1)求 $f\left( {x,y}\right)$ 的极值；
(2)证明:沿(0,0)点的每条直线,(0,0)点都是定义在该直线上的函数 $f\left( {x,y}\right)$ 的极小值点.
💡 答案与解析
解( 1 )由
$$ \left\{ \begin{array}{l} {f}_{x}^{\prime } = - {8x}\left( {y - {x}^{2}}\right) - {x}^{6} = 0, \\ {f}_{y}^{\prime } = 4\left( {y - {x}^{2}}\right) - {2y} = 0 \end{array}\right. $$
解出
$$ \left\{ {\begin{array}{l} {x}_{1} = 0, \\ {y}_{1} = 0 \end{array}\text{ 与 }\left\{ \begin{array}{l} {x}_{2} = - 2, \\ {y}_{2} = 8. \end{array}\right. }\right. $$
又由
$$ {f}_{xx}^{\prime \prime } = - {8y} + {24}{x}^{2} - 6{x}^{5},\;{f}_{xy}^{\prime \prime } = - {8x},\;{f}_{yy}^{\prime \prime } = 2 $$
得
$$ {H}_{f}\left( {-2,8}\right) = \left\lbrack \begin{matrix} {224} & {16} \\ {16} & 2 \end{matrix}\right\rbrack $$
$$ {\Delta }_{1} = {224} > 0,\;{\Delta }_{2} = \left| \begin{matrix} {224} & {16} \\ {16} & 2 \end{matrix}\right| = {192} > 0, $$
所以(-2,8)为函数的极小值点,极小值为 $f\left( {-2,8}\right) = - \frac{352}{7}$ .
对临界点(0,0),函数的海色矩阵不满足极值的充分条件. 但令 $x = 0,f\left( {0,y}\right) = {y}^{2}$ ,这说明原点邻域中 $y$ 轴上的函数值比原点值大. 又令 $y = {x}^{2},f\left( {x,{x}^{2}}\right) = - \frac{1}{7}{x}^{7} - {x}^{4}$ ,这说明原点邻域中抛物线 $y = {x}^{2}$ 上函数值比原点值小,所以原点无极值.
(2)考虑直线 $y = {kx}$ ,
$$ f\left( {x,{kx}}\right) = 2{k}^{2}{x}^{2} - {4k}{x}^{3} + 2{x}^{4} - \frac{1}{7}{x}^{7} - {k}^{2}{x}^{2} $$
$$ = {x}^{2}\left( {{k}^{2} - {4kx} + 2{x}^{2} - \frac{1}{7}{x}^{5}}\right) . $$
$k \neq 0$ 时,可以看出只要 $x$ 充分小有 $f\left( {x,{kx}}\right) \geq 0.k = 0$ 时,
$$ f\left( {x,0}\right) = 2{x}^{4} - \frac{1}{7}{x}^{7} = {x}^{4}\left( {2 - \frac{1}{7}{x}^{3}}\right) , $$
\begin{center} \includegraphics[max width=0.2\textwidth]{images/039.jpg} \end{center} \hspace*{3em}
图 5.4
只要 $x$ 充分小同样有 $f\left( {x,0}\right) \geq 0$ . 这说明限于直线 $y = {kx}$ 上考虑时,函数在(0,0)点有极小值.