2.5 Answer

2.5

1.

$\displaystyle{f(x) = x - \tan{x}}$ より，

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

Now， $\cos^{2}{x} \leq 1$ and the equality holds at $x = 0$

$x = 0$

．Thus， $f'(x) \leq 0$ . Thus, $f(x)$

$f(x)$

is strictly monotonically decreasing on $(-\frac{\pi}{2},\frac{\pi}{2})$ ．

2.

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

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

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

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

Next we let $\displaystyle{g(x) = \tan^{-1}{x} - \frac{x}{1+x^2}}$ . Then $f(0) = 0$

$f(0) = 0$

. Now for

$x > 0$

, we show

$f'(x) > 0$

．

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

(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

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

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 $\displaystyle{\frac{4}{e^2}}$ at $x = 2$ .

$\begin{figure}\begin{center} \includegraphics[width=8cm]{CALCFIG/Fig9-2-6-3.eps} \end{center}\end{figure}$