第7章 7.1

この問の図は変換原稿に含まれていません。

(1)

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

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

$y'=0$ のとき $4x^3 - 6x^2 + 2x = 0$ $2x(2x^2 - 3x + 1) = 2x(2x-1)(x-1) = 0$ $x = 0, \dfrac{1}{2}, 1$

(2)

$y = \dfrac{x^2 - 1}{x^2 + 1}$

$y' = \dfrac{2x(x^2+1) - (x^2-1)2x}{(x^2+1)^2} = \dfrac{2x^3 + 2x - 2x^3 + 2x}{(x^2+1)^2} = \dfrac{4x}{(x^2+1)^2}$

$y'=0$ のとき $4x = 0$ $x = 0$

(3)

$y = \dfrac{x^2 - 4x}{x^2 + 2}$

$y' = \dfrac{(2x-4)(x^2+2) - (x^2-4x)2x}{(x^2+2)^2} = \dfrac{2x^3 - 4x^2 + 4x - 8 - 2x^3 + 8x^2}{(x^2+2)^2} = \dfrac{4x^2 + 4x - 8}{(x^2+2)^2} = \dfrac{4(x+2)(x-1)}{(x^2+2)^2}$

$y'=0$ のとき $4(x+2)(x-1) = 0$ $x = -2, 1$

(4)

$y = \sqrt{x^2 - x + 1}$

$y' = \dfrac{2x - 1}{2\sqrt{x^2 - x + 1}}$

$y' = 0$ のとき $2x - 1 = 0$ $x = \dfrac{1}{2}$

(5)

$y = \sqrt{3 + 2\cos x}$

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

$y' = 0$ のとき $\sin x = 0$ $x = 0, \pi, 2\pi$