数学 - 解析学 - 微分法 - 平均値の定理(3次関数、極値、三角形の面積と線分の長さ、最大値・最小値)

解析入門〈1〉

学習環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Google Chrome...)用JavaScript Library: MathJax

解析入門〈1〉(松坂 和夫(著)、岩波書店)の第4章(微分法)、4.2(平均値の定理)、問題4.2-2、3.を取り組んでみる。

2. 
   $\begin{array}{l}f\text{'}\left(x\right)=3x^2+2ax+b\\ \frac{D}{4}=a^2-3b\\ a^2-3b>0\end{array}$
3. 1. 
      $\begin{array}{l}A\left(a_0,0\right),B\left(0,b_0\right),a_0>0,b_0,>0\\ l:y=-\frac{b_0}{a_0}x+b_0\\ b=-\frac{b_0}{a_0}a+b_0\\ b=\left(1-\frac{a}{a_0}\right)b_0\\ b_0=\frac{b}{1-\frac{a}{a_0}}\\ \frac{1}{2}{a}_0{b}_0=\frac{1}{2}{a}_0·\frac{b}{1-\frac{a}{a_0}}\\ =\frac{a_0}{2}·\frac{a_{0}b}{a_0-a}\\ =\frac{1}{2}·\frac{a_0^{2}b}{a_0-a}\\ \\ y=\frac{a_0^{2}b}{a_0-a}\\ y\text{'}=\frac{2a_{0}b\left(a_0-a\right)-a_0{}^{2}b}{{\left(a_0-a\right)}^2}\\ =\frac{a_0{}^{2}b-2ab{a}_0}{{\left(a_0-a\right)}^2}\\ =\frac{a_0\left(a_{0}b-2ab\right)}{{\left(a_0-a\right)}^2}\\ a_{0}b=2ab\\ a_0=2a\\ \frac{1}{2}·\frac{4a^{2}b}{a}\\ =2ab\end{array}$
   2. 
      $\begin{array}{l}y=a_0{}^2+b_0{}^2\\ =a_0{}^2+{\left(\frac{b{a}_0}{a_0-a}\right)}^2\\ y\text{'}=2a_0+2\left(\frac{b{a}_0}{a_0-a}\right)\frac{b\left(a_0-a\right)-b{a}_0}{{\left(a_0-a\right)}^2}\\ =2a_0\left(1+\frac{-a{b}^2}{{\left(a_0-a\right)}^3}\right)\\ =2a_0\frac{{\left(a_0-a\right)}^3-a{b}^2}{{\left(a_0-a\right)}^3}\\ {\left(a_0-a\right)}^3=a{b}^2\\ a_0-a=a^{\frac{1}{3}}{b}^{\frac{2}{3}}\\ a_0=a^{\frac{1}{3}}{b}^{\frac{2}{3}}+a\\ y={\left(a^{\frac{1}{3}}{b}^{\frac{2}{3}}+a\right)}^2+{\left(\frac{b\left(a^{\frac{1}{3}}{b}^{\frac{2}{3}}+a\right)}{a^{\frac{1}{3}}{b}^{\frac{2}{3}}}\right)}^2\\ ={\left(a^{\frac{1}{3}}{b}^{\frac{2}{3}}+a\right)}^2+{\left(b^{\frac{1}{3}}\left(b^{\frac{2}{3}}+a^{\frac{2}{3}}\right)\right)}^2\\ =a^{\frac{2}{3}}{b}^{\frac{4}{3}}+2a^{\frac{4}{3}}{b}^{\frac{2}{3}}+a^2+{\left(b+a^{\frac{2}{3}}{b}^{\frac{1}{3}}\right)}^2\\ =a^{\frac{2}{3}}{b}^{\frac{4}{3}}+2a^{\frac{4}{3}}{b}^{\frac{2}{3}}+a^2+b^2+2a^{\frac{2}{3}}{b}^{\frac{4}{3}}+a^{\frac{4}{3}}{b}^{\frac{2}{3}}\\ \\ =a^2+3a^{\frac{4}{3}}{b}^{\frac{2}{3}}+3a^{\frac{2}{3}}{b}^{\frac{4}{3}}+b^2\\ ={\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)}^3\\ {\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)}^{\frac{3}{2}}\end{array}$