陰関数の極値

■問題

次の関係で定義される陰関数 ${\displaystyle y=\varphi \left(x\right)}$ の極値を調べよ．

${\displaystyle x^4-16xy+3y^4=0}$

■答

${\displaystyle x=-2}$ のとき，極小値 ${\displaystyle -2}$ をとり， ${\displaystyle x=2}$ のとき，極大値 ${\displaystyle 2}$ をとる．
${\displaystyle \left(0,0\right)}$ は特異点である．

■ヒント

関数の極値の定理２を用いて極大・極小を判断する．

■解説

${\displaystyle f\left(x,y\right)=x^4-16xy+3y^4}$　･･････(1)

とおく．

(1)を偏導関数の定義より， ${\displaystyle y}$ を定数とみなして ${\displaystyle x}$ で微分する．

${\displaystyle f_x=\frac{\partial }{\partial x}f\left(x,y\right)}$ ${\displaystyle =\frac{\partial }{\partial x}\left(x^4-16xy+3y^4\right)}$ ${\displaystyle =4x^3-16y}$

よって，${\displaystyle f_x\left(x,y\right)=0}$となるのは

${\displaystyle 4x^3-16y}$ ${\displaystyle =0}$

${\displaystyle -16y}$ ${\displaystyle =-4x^3}$

${\displaystyle y}$ ${\displaystyle =\frac{1}{4}{x}^3}$　･･････(2)

この関係を与式に代入して

${\displaystyle x^4-16x·\left(\frac{1}{4}{x}^3\right)+3·{\left(\frac{1}{4}{x}^3\right)}^4=0}$

${\displaystyle x^4-4x^4+\frac{3}{4^4}{x}^{12}}$ ${\displaystyle =0}$

${\displaystyle \frac{3}{4^4}{x}^{12}-3x^4}$ ${\displaystyle =0}$

${\displaystyle \frac{1}{4^4}{x}^{12}-x^4}$ ${\displaystyle =0}$

${\displaystyle \frac{1}{4^4}{x}^{12}-x^4}$ ${\displaystyle =0}$

${\displaystyle {\left(\frac{1}{4}{x}^3\right)}^4-x^4}$ ${\displaystyle =0}$

${\displaystyle \left\{{\left(\frac{1}{4}{x}^3\right)}^2+x^2\right\}\left\{{\left(\frac{1}{4}{x}^3\right)}^2-x^2\right\}}$ ${\displaystyle =0}$

${\displaystyle \{\begin{array}{l}{\displaystyle {\left(\frac{1}{4}{x}^3\right)}^2+x^2=0\cdots \cdots \left(3\right)}\\ {\displaystyle \frac{1}{4}{x}^3+x=0\cdots \cdots \left(4\right)}\\ {\displaystyle \frac{1}{4}{x}^3-x=0\cdots \cdots \left(5\right)}\end{array}}$

(3)より

${\displaystyle \frac{1}{16}{x}^6+x^2}$ ${\displaystyle =0}$
${\displaystyle x^2\left(\frac{1}{16}{x}^4+1\right)}$ ${\displaystyle =0}$

${\displaystyle \{\begin{array}{l}{x}^2=0\cdots \cdots \left(6\right)\\ {\displaystyle \frac{1}{16}{x}^4+1=0\cdots \cdots \left(7\right)}\end{array}}$

(6)より

${\displaystyle x=0}$

(7)より

${\displaystyle \frac{1}{16}{x}^4}$ ${\displaystyle =-1}$

${\displaystyle x^4}$ ${\displaystyle =-16}$

となり，これを満たす ${\displaystyle x}$ はない．

次に，(4)より

${\displaystyle \frac{1}{4}{x}^3+x}$ ${\displaystyle =0}$

${\displaystyle x\left(\frac{1}{4}{x}^2+1\right)}$ ${\displaystyle =0}$

${\displaystyle \{\begin{array}{l}x=0\\ {\displaystyle \frac{1}{4}{x}^2+1=0\cdots \cdots \left(8\right)}\end{array}}$

(8)より

${\displaystyle \frac{1}{4}{x}^2}$${\displaystyle =-1}$

となり，これを満たす ${\displaystyle x}$ はない．

最後に(5)から

${\displaystyle x\left(\frac{1}{4}{x}^2-1\right)=0}$

${\displaystyle \{\begin{array}{l}x=0\\ {\displaystyle \frac{1}{4}{x}^2-1=0\cdots \cdots \left(9\right)}\end{array}}$

(9)から

${\displaystyle \frac{1}{4}{x}^2}$ ${\displaystyle =1}$

${\displaystyle x^2}$ ${\displaystyle =4}$

${\displaystyle x}$ ${\displaystyle =\pm 2}$

以上から，${\displaystyle f_x\left(x,y\right)=0}$を満たす$x$ は ${\displaystyle x=0,\pm 2}$ となる．

故に極値をとる候補は，(2) の関係から

${\displaystyle x=0}$ のとき

${\displaystyle y=0}$

${\displaystyle x=2}$ のとき

${\displaystyle y}$ ${\displaystyle =\frac{1}{4}·2^3}$ ${\displaystyle =\frac{1}{4}·8}$ ${\displaystyle =2}$

${\displaystyle x=-2}$ のとき

${\displaystyle y}$ ${\displaystyle =\frac{1}{4}·{\left(-2\right)}^3}$ ${\displaystyle =\frac{1}{4}·\left(-8\right)}$ ${\displaystyle =-2}$

となり， ${\displaystyle \left(0,0\right),\left(2,2\right),\left(-2,-2\right)}$ の3点となる．

次に，上記3点(${\displaystyle y^{\prime }=0}$ ，言い換えると，${\displaystyle f_x\left(x,y\right)=0}$ )における ${\displaystyle y^{″}}$ を求める．この場合

${\displaystyle y^{″}=\frac{d^{2}y}{d{x}^2}=-\frac{f_{xx}}{f_y}}$　･･････(10)

の関係がある．(関数の極値の定理２を参照)

${\displaystyle f_{xx}=\frac{\partial }{\partial x}{f}_x}$ ${\displaystyle =\frac{\partial }{\partial x}\left(4x^3-16y\right)}$ ${\displaystyle =12x^2}$　･･････(11)

${\displaystyle f_y}$ ${\displaystyle =\frac{\partial }{\partial y}\left(x^4-16xy+3y^4\right)}$ ${\displaystyle =-16x+12y^3}$　･･････(12)

(11)，(12)に(10)を代入して

${\displaystyle y^{″}=-\frac{12x^2}{-16x+12y^3}}$ ${\displaystyle =-\frac{3x^2}{-4x+3y^3}}$ ${\displaystyle =\frac{3x^2}{4x-3y^3}}$

${\displaystyle \left(0,0\right)}$ のとき ${\displaystyle y^{″}}$ は

${\displaystyle y^{″}=\frac{3·0^2}{4·0-3·0^3}}$ ${\displaystyle =0}$

よって

${\displaystyle y^{″}=0}$

${\displaystyle \left(2,2\right)}$ のとき ${\displaystyle y^{″}}$ は，

${\displaystyle y^{″}}$ ${\displaystyle =\frac{3·2^2}{4·2-3·2^3}}$ ${\displaystyle =\frac{3·4}{8-3·8}}$ ${\displaystyle =\frac{12}{8-24}}$ ${\displaystyle =-\frac{12}{16}}$ ${\displaystyle =-\frac{3}{4}}$

よって

${\displaystyle y^{″}<0}$

${\displaystyle \left(-2,-2\right)}$ のとき ${\displaystyle y^{″}}$ は

${\displaystyle y^{″}=\frac{3·{\left(-2\right)}^2}{4·\left(-2\right)-3·{\left(-2\right)}^3}}$ ${\displaystyle =\frac{3·4}{-8-3·\left(-8\right)}}$ ${\displaystyle =\frac{12}{-8+24}}$ ${\displaystyle =\frac{12}{16}}$ ${\displaystyle =\frac{3}{4}}$

よって

${\displaystyle y^{″}>0}$

以上から， ${\displaystyle x=-2}$ のとき，極小値 ${\displaystyle -2}$ をとり， ${\displaystyle x=2}$ のとき，極大値 ${\displaystyle 2}$ をとる．
${\displaystyle \left(0,0\right)}$ は特異点である．

　

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最終更新日： 2023年9月18日