第1章 1.3

(1)

$y = x^3 + 3x^2 - 9x$

$y' = 3x^2 + 6x - 9$

$y'' = 6x + 6$

$y' = 0$ のとき

$3x^2 + 6x - 9 = 0$

$(x+3)(x-1) = 0 \qquad \therefore x = -3, 1$

$x = -3$ のとき $y'' = -18 + 6 = -12 < 0$ （極大）

$x = 1$ のとき $y'' = 6 + 6 = 12 > 0$ （極小）

$y'' = 0$ のとき

$6x + 6 = 0 \qquad \therefore x = -1$

(2)

$y = 2 + x - \frac{x^3}{3}$

$y' = 1 - x^2$

$y'' = -2x$

$y' = 0$ のとき

$1 - x^2 = 0 \qquad \therefore x = \pm 1$

$x = -1$ のとき $y'' = 2 > 0$ （極大）

$x = 1$ のとき $y'' = -2 < 0$ （極小）

(3)

$y = x^4 + 2x^3$

$y' = 4x^3 + 6x^2$

$y'' = 12x^2 + 12x$

$y' = 0$ のとき

$4x^3 + 6x^2 = 0$

$2x^2(2x+3) = 0 \qquad \therefore x = 0, -\frac{3}{2}$

$y'' = 0$ のとき

$12x^2 + 12x = 0$

$12x(x+1) = 0 \qquad \therefore x = 0, -1$

$x = 0$ のとき $y'' = 0$

$x = -\dfrac{3}{2}$ のとき $y'' = 27 - 18 = 9 > 0$ （極小）

(4)

$y = \frac{1}{x^3+1} \qquad (\text{定義域}\ x \neq -1)$

$y' = \frac{-3x^2}{(x^3+1)^2}$

$y'' = \frac{-6x(x^3+1)^2+3x^2\{2(x^3+1)\cdot 3x^2\}}{(x^3+1)^4} = \frac{-6x(x^3+1)^2+18x^4(x^3+1)}{(x^3+1)^4}$

$y'=0$ のとき

$\frac{-3x^2}{(x^3+1)^2} = 0 \qquad \therefore x=0$

$x=0$ のとき $y''=0$

$y''=0$ のとき

$\frac{-6x(x^3+1)^2+18x^4(x^3+1)}{(x^3+1)^4} = 0$

$-6x(x^3+1)^2+18x^4(x^3+1) = 0$

$-x(x^3+1)+3x^4 = -x^4-x+3x^4 = 2x^4-x = x(2x^3-1) = 0 \qquad \therefore x=0,\ \frac{1}{\sqrt[3]{2}}$

(5)

$y = (x-3)\sqrt{x} \quad （\text{定義域} x \ge 0）$

$y' = \sqrt{x} + (x-3)\frac{1}{2\sqrt{x}}$

$y'' = \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{x}} + (x-3)\left(-\frac{1}{2\sqrt{x^3}} \cdot \frac{1}{2}\right) = \frac{1}{\sqrt{x}} + (x-3)\left(-\frac{1}{4\sqrt{x^3}}\right)$

$y' = 0$ のとき

$\sqrt{x} + (x-3)\frac{1}{2\sqrt{x}} = 0$

$x + (x-3)\frac{1}{2} = 0$

$2x + (x-3) = 3x - 3 = 0 \qquad \therefore x = 1$

$x = 1 \qquad y'' = 1 + (1-3)\left(-\frac{1}{4}\right) = 1 + \frac{1}{2} = \frac{3}{2} > 0 \qquad （\text{極小}）$

$y'' = 0$ のとき

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

$\frac{4x - (x-3)}{4\sqrt{x^3}} = 0$

$4x - (x-3) = 0$

$3x + 3 = 0 \qquad \therefore x = -1$

(6)

$y = \frac{1}{3}x\sqrt{18-x^2} \qquad (\text{定義域} -3\sqrt{2} \le x \le 3\sqrt{2})$

$y' = \frac{1}{3}\left\{\sqrt{18-x^2} + x \cdot \frac{-2x}{2\sqrt{18-x^2}}\right\} = \frac{1}{3}\left\{\sqrt{18-x^2} - \frac{x^2}{\sqrt{18-x^2}}\right\}$

$y'' = \frac{1}{3}\left\{\frac{-2x}{2\sqrt{18-x^2}} - \frac{2x\sqrt{18-x^2} - x^2 \cdot \dfrac{-2x}{2\sqrt{18-x^2}}}{(\sqrt{18-x^2})^2}\right\} = \frac{1}{3}\left\{\frac{-x}{\sqrt{18-x^2}} - \frac{2x\sqrt{18-x^2} + \dfrac{x^3}{\sqrt{18-x^2}}}{18-x^2}\right\}$

$y'=0$ のとき

$\frac{1}{3}\left\{\sqrt{18-x^2} - \frac{x^2}{\sqrt{18-x^2}}\right\} = 0$

$\frac{1}{3} \cdot \frac{18-x^2-x^2}{\sqrt{18-x^2}} = 0$

$18-2x^2=0 \qquad \therefore x = \pm 3$

$x=-3$ のとき $y'' = \frac{1}{3}\left\{\frac{-3}{\sqrt{18-9}} - \frac{-6\sqrt{18-9}+\dfrac{-27}{\sqrt{18-9}}}{18-9}\right\} = \frac{2}{3} > 0 \qquad (\text{極小})$

$x=3$ のとき $y'' = \frac{1}{3}\left\{\frac{3}{\sqrt{18-9}} - \frac{6\sqrt{18-9}+\dfrac{27}{\sqrt{18-9}}}{18-9}\right\} = -\frac{2}{3} > 0 \qquad (\text{極大})$

$y''=0$ のとき

$\frac{1}{3}\left\{\frac{-x}{\sqrt{18-x^2}} - \frac{2x\sqrt{18-x^2}+\dfrac{x^3}{\sqrt{18-x^2}}}{18-x^2}\right\} = 0$

$x\sqrt{18-x^2} + \frac{x^3}{\sqrt{18-x^2}} = 0$

$\frac{x(18-x^2)+x^3}{\sqrt{18-x^2}} = 0$

$2x^3 + 18x = 0$

$2x(x^2+9) = 0 \qquad \therefore x = 0$