## 2017年2月16日木曜日

### 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)

1. $\begin{array}{l}f\text{'}\left(x\right)=3{x}^{2}+2ax+b\\ \frac{D}{4}={a}^{2}-3b\\ {a}^{2}-3b>0\end{array}$

1. $\begin{array}{l}A\left({a}_{0},0\right),B\left(0,{b}_{0}\right),{a}_{0}>0,{b}_{0},>0\\ l:y=-\frac{{b}_{0}}{{a}_{0}}x+{b}_{0}\\ b=-\frac{{b}_{0}}{{a}_{0}}a+{b}_{0}\\ b=\left(1-\frac{a}{{a}_{0}}\right){b}_{0}\\ {b}_{0}=\frac{b}{1-\frac{a}{{a}_{0}}}\\ \frac{1}{2}{a}_{0}{b}_{0}=\frac{1}{2}{a}_{0}·\frac{b}{1-\frac{a}{{a}_{0}}}\\ =\frac{{a}_{0}}{2}·\frac{{a}_{0}b}{{a}_{0}-a}\\ =\frac{1}{2}·\frac{{a}_{0}^{2}b}{{a}_{0}-a}\\ \\ y=\frac{{a}_{0}^{2}b}{{a}_{0}-a}\\ y\text{'}=\frac{2{a}_{0}b\left({a}_{0}-a\right)-{a}_{0}{}^{2}b}{{\left({a}_{0}-a\right)}^{2}}\\ =\frac{{a}_{0}{}^{2}b-2ab{a}_{0}}{{\left({a}_{0}-a\right)}^{2}}\\ =\frac{{a}_{0}\left({a}_{0}b-2ab\right)}{{\left({a}_{0}-a\right)}^{2}}\\ {a}_{0}b=2ab\\ {a}_{0}=2a\\ \frac{1}{2}·\frac{4{a}^{2}b}{a}\\ =2ab\end{array}$

2. $\begin{array}{l}y={a}_{0}{}^{2}+{b}_{0}{}^{2}\\ ={a}_{0}{}^{2}+{\left(\frac{b{a}_{0}}{{a}_{0}-a}\right)}^{2}\\ y\text{'}=2{a}_{0}+2\left(\frac{b{a}_{0}}{{a}_{0}-a}\right)\frac{b\left({a}_{0}-a\right)-b{a}_{0}}{{\left({a}_{0}-a\right)}^{2}}\\ =2{a}_{0}\left(1+\frac{-a{b}^{2}}{{\left({a}_{0}-a\right)}^{3}}\right)\\ =2{a}_{0}\frac{{\left({a}_{0}-a\right)}^{3}-a{b}^{2}}{{\left({a}_{0}-a\right)}^{3}}\\ {\left({a}_{0}-a\right)}^{3}=a{b}^{2}\\ {a}_{0}-a={a}^{\frac{1}{3}}{b}^{\frac{2}{3}}\\ {a}_{0}={a}^{\frac{1}{3}}{b}^{\frac{2}{3}}+a\\ y={\left({a}^{\frac{1}{3}}{b}^{\frac{2}{3}}+a\right)}^{2}+{\left(\frac{b\left({a}^{\frac{1}{3}}{b}^{\frac{2}{3}}+a\right)}{{a}^{\frac{1}{3}}{b}^{\frac{2}{3}}}\right)}^{2}\\ ={\left({a}^{\frac{1}{3}}{b}^{\frac{2}{3}}+a\right)}^{2}+{\left({b}^{\frac{1}{3}}\left({b}^{\frac{2}{3}}+{a}^{\frac{2}{3}}\right)\right)}^{2}\\ ={a}^{\frac{2}{3}}{b}^{\frac{4}{3}}+2{a}^{\frac{4}{3}}{b}^{\frac{2}{3}}+{a}^{2}+{\left(b+{a}^{\frac{2}{3}}{b}^{\frac{1}{3}}\right)}^{2}\\ ={a}^{\frac{2}{3}}{b}^{\frac{4}{3}}+2{a}^{\frac{4}{3}}{b}^{\frac{2}{3}}+{a}^{2}+{b}^{2}+2{a}^{\frac{2}{3}}{b}^{\frac{4}{3}}+{a}^{\frac{4}{3}}{b}^{\frac{2}{3}}\\ \\ ={a}^{2}+3{a}^{\frac{4}{3}}{b}^{\frac{2}{3}}+3{a}^{\frac{2}{3}}{b}^{\frac{4}{3}}+{b}^{2}\\ ={\left({a}^{\frac{2}{3}}+{b}^{\frac{2}{3}}\right)}^{3}\\ {\left({a}^{\frac{2}{3}}+{b}^{\frac{2}{3}}\right)}^{\frac{3}{2}}\end{array}$