2.4 Properties of functions

1.

(a) $${-\frac{1}{3}}$$(b) $${\sqrt{1 - \left(\frac{2}{\pi}\right)^2}}$$(c) $e - 1$

2.

$f'(x) = 1 - \sec^{2}{x} \leq 0$ and the equality holds only for $x = 0$． 3.

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

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

Next, let $${g(x) = \tan^{-1}{x} - \frac{x}{1+x^2}}$$. Then，$f(0) = 0$. Thus for $x > 0$, we need to 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 to both sides，show $\pi > e\log{\pi}$．Let $f(x) = x - e\log{x}$. Then $f(e) = 0$．For $x > e$

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

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

4.

(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 $x = 2$