第3章 3.9

(1)

$f = \sqrt{x} \qquad (x = 4)$

$f' = \frac{1}{2\sqrt{x}} \qquad x = 4 \text{のとき} \qquad f' = \frac{1}{4}$

$f'' = \frac{-1}{4\sqrt{x^3}} \qquad x = 4 \text{のとき} \qquad f'' = \frac{-1}{32}$

$f''' = \frac{3}{8\sqrt{x^5}} \qquad x = 4 \text{のとき} \qquad f''' = \frac{3}{256}$

これから

$f \cong 2 + \frac{1}{4}(x-4) + \frac{\frac{-1}{32}}{2!}(x-4)^2 + \frac{\frac{3}{256}}{3!}(x-4)^3$

$= 2 + \frac{1}{4}(x-4) - \frac{1}{64}(x-4)^2 + \frac{1}{512}(x-4)^3$

(2)

$f = \frac{1}{x} \qquad (x = 1)$

$f' = -\frac{1}{x^2} \qquad x = 1 \text{のとき} \qquad f' = -1$

$f'' = \frac{2}{x^3} \qquad x = 1 \text{のとき} \qquad f'' = 2 = 2!$

$f''' = -\frac{6}{x^4} \qquad x = 1 \text{のとき} \qquad f''' = -6 = -3!$

これから

$f \cong 1 - (x-1) + (x-1)^2 - (x-1)^3$