Author Topic: Dekoyl's question thread (Read 24254 times) Tweet Share

Collin Li · « **Reply #15 on:** October 13, 2008, 12:02:40 am »

Question 2

$f(x) = \log_2x$

Let $\log_2x = p \implies 2^p = x$

$\implies p \log_e2 = \log_ex$

$\therefore\; f(x) = p = \log_2x = \frac{\log_ex}{\log_e2}$

$f'(x) = \frac{1}{x\log_e2}$

Now, since $f(x+h) \approx f(x) + hf'(x)$

Choosing $x = 16$ and $h = 4$ yields: $f(19) \approx f(16) + 4f'(16)$

$\implies \log_2(19) \approx \log_2(16) + \frac{4}{16\log_e2}$

$\implies \log_2(19) \approx 4 + \frac{1}{4\log_e2}$

Since $4\log_e2 = 4\times 0.69 = (0.36 + 2.40) = 2.76$ (using $\log_e2 = 0.69$ )

Use long-division to obtain the first 2 decimal places of $\frac{1}{2.76} = \frac{100}{276} = \frac{25}{69}$

This will require multiplication of single digit numbers with two digit numbers along the way...

...nothing too difficult to get: $\frac{25}{69} = 0.362\ldots \approx 0.36$

Therefore, $\log_2(19) \approx 4.36$ (to 2 d.p.)

N.B.: I agree this is a ridiculous question, but it's also possible.

shinny · « **Reply #16 on:** October 13, 2008, 12:07:33 am »

Quote from: dekoyl on October 13, 2008, 12:01:52 am

Nope there were no domain restrictions.

If so, am I right in saying that technically, $f^{-1}(x)$ doesn't actually exist at all since $f(x)$ isn't a one-to-one function? o_O

EDIT: Wow go coblin. But yeh, to anyone who's freaked out about that question...don't worry. I HIGHLY doubt VCAA would ever give a question like that which is just more of a test of arithmetic rather than actual mathematics -_-"

Collin Li · « **Reply #17 on:** October 13, 2008, 12:12:30 am »

Quote from: shinjitsuzx on October 13, 2008, 12:07:33 am

EDIT: Wow go coblin. But yeh, to anyone who's freaked out about that question...don't worry. I HIGHLY doubt VCAA would ever give a question like that which is just more of a test of arithmetic rather than actual mathematics -_-"

I agree, but the first part (how to differentiate a non-natural logarithm) is useful.

dekoyl · « **Reply #18 on:** October 13, 2008, 12:17:05 am »

Quote from: coblin on October 13, 2008, 12:02:40 am

Therefore, $\log_2(19) \approx 4.36$ (to 2 d.p.)

I thank thou again. What a question. $:-\$

Mao · « **Reply #19 on:** October 13, 2008, 09:08:44 pm »

erm, coblin: is that "h=4" meant to be "h=3"?

Collin Li · « **Reply #20 on:** October 13, 2008, 10:06:56 pm »

Quote from: Mao on October 13, 2008, 09:08:44 pm

erm, coblin: is that "h=4" meant to be "h=3"?

Ah yep.

I wouldn't bother with the question with $h=3$ . The arithmetic is unbearable - although the fraction becomes $\frac{3}{11.04} \approx \frac{3}{11}$ . You can't really be sure it's accurate to 2 d.p. though...

Mao · « **Reply #21 on:** October 14, 2008, 02:44:12 pm »

3.000 = 0.2 * 11.04 + 0.792 = 0.2 * 11.04 + 0.07 * 11.04 + 0.0192 = 0.2 * 11.04 + 0.07 * 11.04 + 0.01 * 11.04 + ...

correct to 2 decimals: 0.27

dekoyl · « **Reply #22 on:** October 15, 2008, 12:38:41 am »

^ Heh thanks for the continuous help

I have another one. However, I don't agree with the answers.

Given $f(x) = -x(2x+a)^{3}(x-2b)^{2}$ and $a, b \epsilon R^{-}$ , some x-intercepts of the graph $y = f'(x)$ are?

The answer is $\frac{-a}{2} , 2b$

My answer is $0, \frac{a}{2}, -2b$

Thanks again.

Collin Li · « **Reply #23 on:** October 15, 2008, 12:49:23 am »

It says $f'(x)$

You should be able to sketch the polynomial $f(x)$ , and see that there is a stationary point at $2x+a=0$ and $x-2b = 0$ . However, $x=0$ is merely an x-intercept, and not a stationary point.

dekoyl · « **Reply #24 on:** October 15, 2008, 12:50:50 am »

What the.. Oh this is stupid. I must have read the question 10 times. I even typed it out correctly and I didn't even pick up the $f'(x)$ .

Please excuse my foolishness.

Collin Li · « **Reply #25 on:** October 15, 2008, 12:53:08 am »

Haha

I still don't understand why your x-intercepts are negative to what they should be though. The negative out the front shouldn't affect it (it's a reflection in the x-axis, so the x-intercepts stay the same, since $x=0$ reflects to the same point).

dekoyl · « **Reply #26 on:** October 15, 2008, 12:59:22 am »

I thought $a, b \epsilon R^{-} \implies a, b$ are negative numbers.
This made me remember the common "trick" question whereby a graph has intercepts in the $x < 0$ region at $a$ and it asks you which is a possible equation to the graph to which people would often choose $f(x) = (x + a)...$
Gah I'm just spewing crap.
Don't worry about it if I don't make sense. Slightly tired. :uglystupid2:

Collin Li · « **Reply #27 on:** October 15, 2008, 01:39:12 am »

If $a=-2$ is an x-intercept, for example, then the factor would be $[x - (-2)] = (x+2)$

Don't you agree that's still $(x-a)$ , regardless of whether $a$ is positive or negative?

dekoyl · « **Reply #28 on:** October 15, 2008, 01:50:03 am »

Quote from: coblin on October 15, 2008, 01:39:12 am

If $a=-2$ is an x-intercept, for example, then the factor would be $[x - (-2)] = (x+2)$

Don't you agree that's still $(x-a)$ , regardless of whether $a$ is positive or negative?

Yeah I had to think for a while

Gotcha now. Thanks again.

dekoyl · « **Reply #29 on:** October 18, 2008, 05:50:50 pm »

I came across this question where it asks for the "point of inflection" on a cos graph. Is there one?

What is the point of inflection of the graph $y = 1 + cos(\frac{x}{2})$ for $x \in [0,2\pi]$ ?

I'm required to find the tangent to the graph at that point of inflection. :S

Thanks.

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