2.4 Answer

2.4

1.

(a)

$${f'(x) = 3x^2 - 2x = \frac{f(1) - f(-1)}{1 - (-1)} = 1}$$ より

$$3x^2 - 2x - 1 = 0$$

Then

$$(3x + 1)(x - 1) = 0$$

$$x = -\frac{1}{3}, 1$$

Now $x$ must be in the interval $(-1,1)$. Then

$$x = -\frac{1}{3}$$

(b)

$${f'(x) = \frac{1}{\sqrt{1 - x^2}} = \frac{f(1) - f(0)}{1} = \frac{\pi}{2}}$$ implies

$$\sqrt{1 - x^2} = \frac{2}{\pi}$$

Square both sides. Then

$$1 - x^2 = (\frac{2}{\pi})^2$$

Solve this for $x$. We have

$$x = \pm \sqrt{ 1 - (\frac{2}{\pi})^{2}}$$

Now $x$ must be in the interval $(0,1)$. Then

$$x = \sqrt{ 1 - (\frac{2}{\pi})^{2}}$$

(c)

$${f'(x) = \frac{1}{x} = \frac{f(e) - f(1)}{e - 1} = \frac{1}{e - 1}}$$ implies that

$$x = e - 1$$

Now $x$ is in $(1,e)$. Then

$$x = e - 1$$