Author Topic: Some differentiation questions (Read 6297 times) Tweet Share

brightsky · « **Reply #1 on:** March 29, 2010, 05:51:55 pm »

4. a)

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

$= \frac{x^n + 1 - 1}{1+x^n}$

$= \frac{x^n + 1}{1 + x^n} - \frac{1}{1 + x^n}$

$= 1 - \frac{1}{1+x^n}$

the.watchman · « **Reply #2 on:** March 29, 2010, 05:56:25 pm »

3. (a) $f(x) = (x-a)^n \cdot (x-b)^m = 0$ , $x=a,b$

(b) $f'(x) = (x-a)^n \cdot [(x-b)^m]' + [(x-a)^n]' \cdot (x-b)^m$

$f'(x) = (x-a)^n \cdot m(x-b)^{m-1} + n(x-a)^{n-1} \cdot (x-b)^m$

$f'(x) = (x-a)^{n-1}[(x-a) \cdot m(x-b)^{m-1} + n \cdot (x-b)^m]$

$f'(x) = (x-a)^{n-1} \cdot (x-b)^{m-1}[(x-a) \cdot m + n \cdot (x-b)]$

the.watchman · « **Reply #3 on:** March 29, 2010, 06:08:23 pm »

3. (c) $f'(x) = (x-a)^{n-1} \cdot (x-b)^{m-1}[(x-a) \cdot m + n \cdot (x-b)] = 0$

So $x=a,b$

OR $(x-a) \cdot m + n \cdot (x-b)=0$

$x=\frac{am+bn}{m+n}$

brightsky · « **Reply #4 on:** March 29, 2010, 06:14:31 pm »

4. c)

$f(x) = 1 - \frac{1}{x^n + 1}$ where n is a positive even integer.

$\because$ n is a positive even integer

$\therefore x^n \ge 0$

So $x^n + 1 \ge 1$

Hence $0 < \frac{1}{x^n+1} \le 1$

$\therefore 0 \le 1 - \frac{1}{x^n + 1} < 1$

m@tty · « **Reply #5 on:** March 29, 2010, 06:17:17 pm »

Quote from: brightsky on March 29, 2010, 06:14:31 pm

$\therefore 0 \ge 1 - \frac{1}{x^n + 1} > 1$

$0>1$ ?

EDIT: Haha, look at my post count and average per day. 1333 and 3.333 per day

EDIT OF EDIT: ... Not anymore

the.watchman · « **Reply #6 on:** March 29, 2010, 06:18:56 pm »

You forgot to swap the signs when multiplying all sides by -1

brightsky · « **Reply #7 on:** March 29, 2010, 06:25:35 pm »

4. b)

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

Let $g(x) = x^n$ and $h(x) = x^n + 1$

$f'(x) = \frac{g'(x)h(x) - g(x) h'(x)}{h(x)^2}$

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

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

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

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

brightsky · « **Reply #8 on:** March 29, 2010, 06:26:09 pm »

Quote from: the.watchman on March 29, 2010, 06:18:56 pm

You forgot to swap the signs when multiplying all sides by -1

Woops :p Edited.

vea · « **Reply #9 on:** March 29, 2010, 06:28:52 pm »

These are pretty good questions, which book are they from?

m@tty · « **Reply #10 on:** March 29, 2010, 06:30:59 pm »

Looks like Essentials typeset.

the.watchman · « **Reply #11 on:** March 29, 2010, 06:52:45 pm »

Quote from: m@tty on March 29, 2010, 06:30:59 pm

Looks like Essentials typeset.

Second that

Except I can't find it in the textbook, mind telling me where it is someone?

superflya · « **Reply #12 on:** March 29, 2010, 06:57:12 pm »

Quote from: the.watchman on March 29, 2010, 06:52:45 pm

Quote from: m@tty on March 29, 2010, 06:30:59 pm
Looks like Essentials typeset.

Second that
Except I can't find it in the textbook, mind telling me where it is someone?

chapter 9 review, extended response questions. Q3

the.watchman · « **Reply #13 on:** March 29, 2010, 06:59:33 pm »

Quote from: superflya on March 29, 2010, 06:57:12 pm

Quote from: the.watchman on March 29, 2010, 06:52:45 pm
Quote from: m@tty on March 29, 2010, 06:30:59 pm
Looks like Essentials typeset.

Second that
Except I can't find it in the textbook, mind telling me where it is someone?

chapter 9 review, extended response questions. Q3

Ok, ta

brightsky · « **Reply #14 on:** March 29, 2010, 07:12:35 pm »

4. f)

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

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

$\because$ n is a positive even integer

$\therefore f(-x) = \frac{x^n}{x^n + 1}$

Hence $f(x) = f(-x)$ , and so $f$ is an even function.

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