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# 示例：二阶偏导数测试

## 二阶偏导数测试的解法(参考)

$\begin{array}{r}{f}_{x}\left({x}_{0},{y}_{0}\right)=0\\ \\ {f}_{y}\left({x}_{0},{y}_{0}\right)=0\end{array}$

$H={f}_{xx}\left({x}_{0},{y}_{0}\right){f}_{yy}\left({x}_{0},{y}_{0}\right)-\left({f}_{xy}\left({x}_{0},{y}_{0}\right){\right)}^{2}$
${f}_{xx}$, ${f}_{yy}$${f}_{xy}$均为$f$的二阶偏导数。

• 如果 ${f}_{xx}\left({x}_{0},{y}_{0}\right)>0$$f$ 具有局部最小值。
• 如果 ${f}_{xx}\left({x}_{0},{y}_{0}\right)<0$$f$ 具有局部最大值。

## 例 1: 全部的驻点!

$\begin{array}{r}{x}^{4}-4{x}^{2}+{y}^{2}\end{array}$

## 步骤 1:找到所有驻点

${f}_{x}\left(x,y\right)=$
${f}_{y}\left(x,y\right)=$

$\begin{array}{rl}{f}_{x}\left({x}_{0},{y}_{0}\right)& =0\\ \\ {f}_{y}\left({x}_{0},{y}_{0}\right)& =0\end{array}$

## 步骤 2; 应用二阶导数测试

${f}_{xx}\left(x,y\right)=$
${f}_{yy}\left(x,y\right)=$
${f}_{xy}\left(x,y\right)=$

${f}_{xx}\left(x,y\right){f}_{yy}\left(x,y\right)-\left({f}_{xy}\left(x,y\right){\right)}^{2}$

${f}_{xx}\left(x,y\right){f}_{yy}\left(x,y\right)-\left({f}_{xy}\left(x,y\right){\right)}^{2}=$

• 驻点 1:
$\left(x,y\right)=\left(0,0\right)$，表达式变为
$\begin{array}{r}\phantom{\rule{1em}{0ex}}24{x}^{2}-16=24\left(0{\right)}^{2}-16=-16\end{array}$

• 驻点 2: 在$\left({x}_{0},{y}_{0}\right)=\left(\sqrt{2},0\right)$, 表达式变为
$\begin{array}{rl}24{x}^{2}-16& =24\left(\sqrt{2}{\right)}^{2}-16\\ \\ & =48-16\\ \\ & =32\end{array}$

$\begin{array}{rl}{f}_{xx}\left(\sqrt{2},0\right)& =12\left(\sqrt{2}{\right)}^{2}-8\\ \\ & =24-8\\ \\ & =16\end{array}$

• 驻点 3: 我们可以代入点$\left(-\sqrt{2},0\right)$，但我们也可以发现函数$f\left(x,y\right)={x}^{4}-4{x}^{2}+{y}^{2}$是对称的，所以我们只需在表达式里用$-x$代替$x$:
$\left(-x{\right)}^{4}-4\left(-x{\right)}^{2}+{y}^{2}={x}^{4}-4{x}^{2}+{y}^{2}$

## 例 2: 深入探究

$\begin{array}{r}f\left(x,y\right)={x}^{2}y-{y}^{2}x-{x}^{2}-{y}^{2}\end{array}$

## 步骤 1:找到所有驻点

${f}_{x}\left(x,y\right)=$
${f}_{y}\left(x,y\right)=$

$\begin{array}{rl}2xy-{y}^{2}-2x& =0\\ \\ {x}^{2}-2xy-2y& =0\end{array}$

• 解一个等式，用$x$来表示$y$
• 将其代入另一个表达式并得到一个只包含$x$的等式。
• $x$
• $x$的解代入两个等式中并解$y$
• 检查哪一对$\left(x,y\right)$真正是表达式的解。

$\begin{array}{rl}\phantom{\rule{1em}{0ex}}2xy-{y}^{2}-2x& =0\\ \\ +\phantom{\rule{1em}{0ex}}{x}^{2}-2xy-2y& =0\\ \\ \\ \\ {x}^{2}-{y}^{2}-2\left(x+y\right)& =0\\ \\ \left(x+y\right)\left(x-y\right)-2\left(x+y\right)& =0\\ \\ \left(x+y\right)\left(x-y-2\right)& =0\end{array}$

Either

$\begin{array}{rl}& \left(x,y\right)=\overline{)\left(0,0\right)}\\ \\ & \left(x,y\right)=\overline{)\left(-\frac{2}{3},\frac{2}{3}\right)}\end{array}$

$\begin{array}{r}x=2-1+\sqrt{5}=1+\sqrt{5}\\ \\ x=2-1-\sqrt{5}=1-\sqrt{5}\end{array}$

$\begin{array}{r}\left(x,y\right)=\overline{)\left(1+\sqrt{5},-1+\sqrt{5}\right)}\\ \\ \left(x,y\right)=\overline{)\left(1-\sqrt{5},-1-\sqrt{5}\right)}\end{array}$

## 步骤 2; 应用二阶导数测试

$\begin{array}{r}f\left(x,y\right)={x}^{2}y-{y}^{2}x-{x}^{2}-{y}^{2}\end{array}$
${f}_{xx}\left(x,y\right)=$
${f}_{yy}\left(x,y\right)=$
${f}_{xy}\left(x,y\right)=$

$\begin{array}{r}{f}_{xx}\left(x,y\right){f}_{yy}\left(x,y\right)-\left({f}_{xy}\left(x,y\right){\right)}^{2}\end{array}$

$\left(y-1\right)\left(-x-1\right)-\left(x-y{\right)}^{2}\phantom{\rule{1em}{0ex}}←\text{关键表达式}$

• 驻点 $\left(0,0\right)$

• 驻点 $\left(-\frac{2}{3},\frac{2}{3}\right)$

• 驻点 $\left(1+\sqrt{5},-1+\sqrt{5}\right)$

• 驻点 $\left(1-\sqrt{5},-1-\sqrt{5}\right)$: