娘娘HW2 - 幻想境界「醉生夢死」 - 兔 @ ウィキ - atwiki（アットウィキ）

幻想境界「醉生夢死」 - 兔 @ ウィキ

幻想境界「醉生夢死」 - 兔 @ ウィキ

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娘娘HW2

1.求f在所予閉區間上的絕對極大值與絕對極小值

$f(x) = x^{3} - 6x^{2} + 9x + 2; \verb|[0,2]|$

sol:

$f'(x)= 3x^{2} - 12x + 9$

$\verb|Set | f'(x) = 0 \verb| => | 3(x-1)(x-3)=0$

Thus we have a maximum at x=1, and a minimum at x = 3.

Because the slope of this function is positive in [0,1) and negative in (1,2], f(1) = 6 is the absolute max in [0,2].

f(0) = 2, f(2) = 4, so f(0) = 2 is the absolute min in [0,2]

2.求函數的遞增區間與遞減區間

$f(x) = x^{4} - 4x^{3} - 8x^{2} +3$

sol:

$f'(x) = 4x^{3} - 12x^{2} - 16x = 4x(x-4)(x+1)$

$\verb|Ths slope of f(x) (namely f'(x)) is negative in |$
$\verb|(|\infty\verb|,-1) and (0,4), positive in (-1,0) and|$
$\verb|(4,|\infty\verb|), so it's decreasing in |$
$\verb|(|\infty\verb|,-1) and (0,4), increasing in (-1,0) and |$
$\verb|(4, |\infty\verb|).|$

3.討論函數圖形的凹性並找出反曲點

$f(x) = \frac{1}{x^{2}+1}$

sol:

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

$f''(x) = \frac{-2(x^{2}+1)^{2} + 4x(x^{2}+1)(2x)}{(x^{2}+1)^{4}}$

$= \frac{6x^{2}-2}{(x^{2}+1)^{3}}$

$\verb|Turning points at |x = \pm\frac{1}{\sqrt{3}}$

開口向上 - 反曲點 - 開口向下 - 反曲點 - 開口向上

4.試作函數的圖形

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

sol:

$f'(x) = \frac{4x(x^{2}-1) - 2x^{2}(2x)}{(x^{2}-1)^{2}}$

$= \frac{-4x}{(x^{2}-1)^{2}}$

$f''(x) = \frac{-4(x^{2}-1)^{2} + 8x(x^{2}-1)(2x)}{(x^{2}-1)^{4}}$

$= \frac{12x^{2} + 4}{(x^{2}-1)^{3}}$

f(x) has a local maximum at x=0, and two singularities at x = +1, -1.

Note that f(x) gets closer and closer to (yet still larger than) y=2 while x goes to infinity at the two ends of the x-axis.

Plot the function according to the conditions above and watch out for the singularities.

To better draw the graph, you may start with drawing the asymptotes(漸進線) first: y=2, x=1, x=-1.

5.求

$\lim_{x \to 1} \frac{sin(x-1)}{x^{2}+x-2}$

sol:

$\verb|By L'Hospital's Rule, |$

$\lim_{x \to 1} \frac{sin(x-1)}{x^{2}+x-2}$

$= \lim_{x \to 1} \frac{cos(x-1)}{2x+1}$

$= \frac{1}{3}$

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最終更新：2010年07月21日 17:54

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