第2章 2.1

(1)

$f(x) = x^2 - 2x$    $[-1, 3]$

$f' = 2x - 2$

$\frac{(3^2 - 2\cdot 3) - \{(-1)^2 - 2\cdot(-1)\}}{3 - (-1)} = 2C - 2$

$\frac{3-3}{4} = 2C - 2 \qquad \therefore C = 1$

(2)

$f(x) = x^3 - 5x^2 - 3x$    $[-2, 6]$

$f' = 3x^2 - 10x - 3$

$\frac{6^3 - 5\cdot 6^2 - 3\cdot 6 - \{(-2)^3 - 5\cdot(-2)^2 - 3\cdot(-2)\}}{6 - (-2)} = 3C^2 - 10C - 3$

$\frac{40}{8} = 3C^2 - 10C - 3 \qquad 3C^2 - 10C - 8 = 0 \qquad (3C+2)(C-4) = 0 \qquad \therefore C = -\frac{2}{3}, 4$

(3)

$f(x) = \dfrac{x-1}{x+1}$    $[0, 3]$

$f' = \dfrac{x+1-(x-1)}{(x+1)^2} = \dfrac{2}{(x+1)^2}$

$\dfrac{\dfrac{3-1}{3+1} - \dfrac{-1}{1}}{3-0} = \frac{2}{(x+1)^2} \qquad \frac{1}{2} = \frac{2}{(C-1)^2} \qquad (C+1)^2 = 4 \qquad C^2 + 2C + 1 - 4 = 0$

$C^2 + 2C - 3 = 0 \qquad (C-1)(C+3) = 0 \qquad C = 1, -3 \qquad 0 \le x \le 3 \text{から} \qquad \therefore C = 1$

(4)

$f_{(x)} = \sqrt{25-x^2} \qquad [-3,4]$

$f' = \frac{-2x}{2\sqrt{25-x^2}} = \frac{-x}{\sqrt{25-x^2}}$

$\frac{\sqrt{25-16}-\sqrt{25-9}}{4-(-3)} = \frac{-C}{\sqrt{25-C^2}} \qquad \frac{3-4}{7} = \frac{-C}{\sqrt{25-C^2}} \qquad \sqrt{25-C^2} = 7C$

$25-C^2 = 49C^2 \qquad 50C^2 = 25 \qquad C^2 = \frac{1}{2} \qquad C = \pm\frac{1}{\sqrt{2}}$

ここで $7C \geq 0$ となる必要があるので $\therefore C = \dfrac{1}{\sqrt{2}}$