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臨界點、極值與鞍點計算器
求函數的臨界點、極值與鞍點
您的輸入
找出並分類$$$f{\left(x,y \right)} = 2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2$$$的臨界點。
解答
第一步是求出所有一階偏導數：
$$$\frac{\partial}{\partial x} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 4 x \left(y - 1\right)$$$ (如需步驟，請參見偏導數計算器).
$$$\frac{\partial}{\partial y} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 2 x^{2} + 3 y^{2} - 4 y$$$ (如需步驟，請參見偏導數計算器).
接下來，解聯立方程組 $$$\begin{cases} \frac{\partial f}{\partial x} = 0 \\ \frac{\partial f}{\partial y} = 0 \end{cases}$$$ 或 $$$\begin{cases} 4 x \left(y - 1\right) = 0 \\ 2 x^{2} + 3 y^{2} - 4 y = 0 \end{cases}$$$。
該系統有以下實數解：$$$\left(x, y\right) = \left(0, 0\right)$$$, $$$\left(x, y\right) = \left(0, \frac{4}{3}\right)$$$, $$$\left(x, y\right) = \left(- \frac{\sqrt{2}}{2}, 1\right)$$$, $$$\left(x, y\right) = \left(\frac{\sqrt{2}}{2}, 1\right)$$$。
現在，讓我們嘗試將它們分類。
求所有二階偏導數：
$$$\frac{\partial^{2}}{\partial x^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 4 y - 4$$$ (如需步驟，請參見偏導數計算器).
$$$\frac{\partial^{2}}{\partial y\partial x} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 4 x$$$ (如需步驟，請參見偏導數計算器).
$$$\frac{\partial^{2}}{\partial y^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 6 y - 4$$$ (如需步驟，請參見偏導數計算器).
定義表達式 $$$D = \frac{\partial ^{2}f}{\partial x^{2}} \frac{\partial ^{2}f}{\partial y^{2}} - \left(\frac{\partial ^{2}f}{\partial y\partial x}\right)^{2} = - 16 x^{2} + 24 y^{2} - 40 y + 16$$$。
由於 $$$D{\left(0,0 \right)} = 16$$$ 大於 $$$0$$$ 且 $$$\frac{\partial^{2}}{\partial x^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right)|_{\left(\left(x, y\right) = \left(0, 0\right)\right)} = -4$$$ 小於 $$$0$$$，可判定 $$$\left(0, 0\right)$$$ 為相對極大值。
由於 $$$D{\left(0,\frac{4}{3} \right)} = \frac{16}{3}$$$ 大於 $$$0$$$ 且 $$$\frac{\partial^{2}}{\partial x^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right)|_{\left(\left(x, y\right) = \left(0, \frac{4}{3}\right)\right)} = \frac{4}{3}$$$ 大於 $$$0$$$，可以斷定 $$$\left(0, \frac{4}{3}\right)$$$ 是相對極小值。
由於 $$$D{\left(- \frac{\sqrt{2}}{2},1 \right)} = -8$$$ 小於 $$$0$$$，可判定 $$$\left(- \frac{\sqrt{2}}{2}, 1\right)$$$ 為鞍點。
由於 $$$D{\left(\frac{\sqrt{2}}{2},1 \right)} = -8$$$ 小於 $$$0$$$，可判定 $$$\left(\frac{\sqrt{2}}{2}, 1\right)$$$ 為鞍點。
答案
相對極大值
$$$\left(x, y\right) = \left(0, 0\right)$$$A, $$$f{\left(0,0 \right)} = 2$$$A
相對極小值
$$$\left(x, y\right) = \left(0, \frac{4}{3}\right)\approx \left(0, 1.333333333333333\right)$$$A, $$$f{\left(0,\frac{4}{3} \right)} = \frac{22}{27}\approx 0.814814814814815$$$A
鞍點
$$$\left(x, y\right) = \left(- \frac{\sqrt{2}}{2}, 1\right)\approx \left(-0.707106781186548, 1\right)$$$A, $$$f{\left(- \frac{\sqrt{2}}{2},1 \right)} = 1$$$A
$$$\left(x, y\right) = \left(\frac{\sqrt{2}}{2}, 1\right)\approx \left(0.707106781186548, 1\right)$$$A, $$$f{\left(\frac{\sqrt{2}}{2},1 \right)} = 1$$$A