2023-M2-09

另外，

$\begin{array}{rcl}{f}^{\prime }(x)& =& (1-2x^2)e^{-x^2}\\ f”(x)& =& (1-2x^2)e^{-x^2}\times (-2x)+e^{-x^2}\times (-4x)\\ f”(x)& =& (-2x+4x^3-4x)e^{-x^2}\\ f”(x)& =& 2x(2x^2-3)e^{-x^2}\end{array}$

$\begin{array}{rcl}{f}^{\prime }(x)& =& 0\\ (1-2x^2)e^{-x^2}& =& 0\\ 2x^2& =& 1\\ x^2& =& {\displaystyle \frac{1}{2}}\\ x& =& \pm {\displaystyle \frac{\sqrt{2}}{2}}\end{array}$

$\begin{array}{|lccccc|}\hline x& x<{\displaystyle \frac{-\sqrt{2}}{2}}& x={\displaystyle \frac{-\sqrt{2}}{2}}& {\displaystyle \frac{-\sqrt{2}}{2}}<x<{\displaystyle \frac{\sqrt{2}}{2}}& x={\displaystyle \frac{\sqrt{2}}{2}}& x>{\displaystyle \frac{\sqrt{2}}{2}}\\ f^{\prime }(x)& -ve& 0& +ve& 0& -ve\\ f(x)& \text{遞減}& \text{極小點}& \text{遞增}& \text{極大點}& \text{遞減}\\ \hline\end{array}$

所以，$G$ 的極大點為 $({\displaystyle \frac{\sqrt{2}}{2}},{\displaystyle \frac{\sqrt{2}}{2}}{e}^{\frac{-1}{2}})$。

$G$ 的極小點為 $({\displaystyle \frac{-\sqrt{2}}{2}},{\displaystyle \frac{-\sqrt{2}}{2}}{e}^{\frac{-1}{2}})$。

1. 留意 $f^{\prime }(1)=-e^{-1}$。利用點斜式，$L$ 的方程為

   $\begin{array}{rcl}{\displaystyle \frac{y-e^{-1}}{x-1}}& =& -e^{-1}\\ y-e^{-1}& =& e^{-1}x+e^{-1}\\ y& =& {\displaystyle \frac{1}{e}}x+{\displaystyle \frac{2}{e}}\end{array}$

2. 留意 $L$ 為 $G$ 於 $(1,{\displaystyle \frac{1}{e}})$ 的切線。

   對於 $x\in (0,1)$, $f”(x)=2x(2x^2-3)e^{-x^2}<0$。由此，$y=f(x)$ 的圖像在區間 $(0,1)$ 向下凹。

   所以，$G$ 於區間 $(0,1)$ 在 $L$ 之下。

3. 所求的面積

   $\begin{array}{cl}=& {\displaystyle {\int }_0^1({\displaystyle \frac{-1}{e}}x+{\displaystyle \frac{2}{e}}-x{e}^{-x^2})dx}\\ =& {\displaystyle {\int }_0^1({\displaystyle \frac{-1}{e}}x+{\displaystyle \frac{2}{e}})dx-{\displaystyle {\int }_0^{1}x{e}^{-x^2}dx}}\\ =& {[{\displaystyle \frac{-1}{2e}}{x}^2+{\displaystyle \frac{2}{e}}x]}_0^1+{\displaystyle \frac{1}{2}}{\displaystyle {\int }_0^1{e}^{-x^2}d(-x^2)}\\ =& {\displaystyle \frac{-1}{2e}}+{\displaystyle \frac{2}{e}}+{\displaystyle \frac{1}{2}}{\left[e^{-x^2}\right]}_0^1\\ =& {\displaystyle \frac{3}{2e}}+{\displaystyle \frac{1}{2}}(e^{-1}-1)\\ =& {\displaystyle \frac{2}{e}}-{\displaystyle \frac{1}{2}}\end{array}$