2.5 Answer

2.5

1.

$${f(x) = x - \tan{x}}$$より，

$$f'(x) = 1 - \sec^{2}{x} = 1 - \frac{1}{\cos^{2}{x}} = \frac{\cos^{2}{x} - 1}{\cos^{2}{x}}.$$

Now， $\cos^{2}{x} \leq 1$ and the equality holds at $x = 0$．Thus， $f'(x) \leq 0$. Thus, $f(x)$ is strictly monotonically decreasing on $(-\frac{\pi}{2},\frac{\pi}{2})$．

2.

(a) Let $${f(x) = \log{(1 + x)} - \frac{x}{1+x}}$$. Then $f(0) = 0$. Now for $x > 0$, we show $f'(x) > 0$．

$$f'(x) = \frac{1}{1+x} - \frac{1}{(1+x)^2} = \frac{x}{(1+x)^2} > 0$$

(b) Let $${f(x) = x - \tan^{-1}{x}}$$. Then $f(0) = 0$. Now for $x > 0$, we show $f'(x) > 0$．

$$f'(x) = 1 - \frac{1}{1+x^2} = \frac{x^2}{1+x^2} > 0$$

Next we let $${g(x) = \tan^{-1}{x} - \frac{x}{1+x^2}}$$. Then $f(0) = 0$. Now for $x > 0$, we show $f'(x) > 0$．

$$f'(x) = \frac{1}{1+x^2} - \frac{1 - x^2}{(1+x^2)^2} = \frac{2x^2}{(1+x^2)^2} > 0$$

(c) Take logarithm on both sides，we show $\pi > e\log{\pi}$． Let $f(x) = x - e\log{x}$. Then $f(e) = 0$．Also, for $x > e$, we have

$$f'(x) = 1 - \frac{e}{x} = \frac{x - e}{x} > 0$$

Thus, $f(\pi) = \pi - e\log{\pi} > 0$

3.

(a) Local maximum 7 at $x = 1$，local minimum 3 at $x = 3$.

(b) Local minimum 0 at $x = 0$， local maximum $${\frac{4}{e^2}}$$ at $x = 2$.