Author Topic: VCE Methods Question Thread! (Read 5705607 times) Tweet Share

soNasty · « **Reply #3990 on:** February 14, 2014, 06:41:39 pm »

yeah, that's what i keep getting.
the back of the book says it's:

$(5x^2+4x)/\sqrt{2x+2}$

ealam2 · « **Reply #3991 on:** February 14, 2014, 06:47:52 pm »

Quote from: alchemy on February 13, 2014, 09:54:13 pm

Hehe, fair enough. On another note: has anyone found the textbook helpful during an exam (if they brought it in for reference)?

I think it's generally better if you make your own bound reference as it serves for your own revision and you actually know what's in it and can paste in cut-outs and practice SACs if you want with extra examples.

lzxnl · « **Reply #3992 on:** February 14, 2014, 07:23:14 pm »

Quote from: lzxnl on February 14, 2014, 06:34:24 pm

$\frac{d}{dx}(x^2\sqrt{2x+2})$
$=\sqrt{2x+2}\frac{d}{dx}(x^2) + x^2\frac{d}{dx}(\sqrt{2x+2})$

Now, the first derivative is just 2x. As for the derivative of $\sqrt{2x+2}=(2x+2)^{1/2}$ , a chain rule application yields $\frac{2}{2\sqrt{2x+2}} = \frac{1}{\sqrt{2x+2}}$

Putting this altogether, you should get $\frac{d}{dx}(x^2\sqrt{2x+2}) = \sqrt{2x+2}\frac{d}{dx}(x^2) + x^2\frac{d}{dx}(\sqrt{2x+2}) = 2x\sqrt{2x+2} + \frac{x^2}{\sqrt{2x+2}}$

As mentioned above.

So that was my working above. Now, moving on from the final result by putting over a common denominator (notice how the square root disappears):

$2x\sqrt{2x+2} + \frac{x^2}{\sqrt{2x+2}} = \frac{2x(2x+2)+x^2}{\sqrt{2x+2}} = \frac{5x^2+4x}{\sqrt{2x+2}}$

Anchy · « **Reply #3993 on:** February 14, 2014, 07:46:35 pm »

Hi can I pls have some help with this:

Transform 1/x

- Translate 6 units right
- Translate 1 unit up
- Reflection in yaxis
- Dilation factor of 2 from x axis
( in that order)

The textbook answer is -2/(x+6) + 2 and I can't get the correct answer. Thank you

abcdqd · « **Reply #3994 on:** February 14, 2014, 08:02:13 pm »

Quote from: Anchy on February 14, 2014, 07:46:35 pm

Hi can I pls have some help with this:

Transform 1/x

- Translate 6 units right
- Translate 1 unit up
- Reflection in yaxis
- Dilation factor of 2 from x axis
( in that order)

The textbook answer is -2/(x+6) + 2 and I can't get the correct answer. Thank you

Let's do this step by step:

- Translate 6 units right $\frac{1}{x}$ becomes $\frac{1}{x-6}$
- Translate 1 unit up $\frac{1}{x-6}$ becomes $\frac{1}{x-6}+1$

Reflections in the y-axis are done by changing $f(x)$ to $f(-x)$ . Note that you only change the $x$ term, i.e. $x$ becomes $-x$
- Reflection in y-axis $\frac{1}{x-6}+1$ becomes $\frac{1}{-x-6}+1$

Dilations in the x-axis are done by multiplying the entire function by the dilation factor.

- Dilation factor of 2 from x axis $\frac{1}{-x-6}+1$ becomes $2 \left(\frac{1}{-x-6}+1\right)$

When you expand this and take out a factor of -1 from your denominator it is equal to the answer provided in the book.

soNasty · « **Reply #3995 on:** February 14, 2014, 08:13:47 pm »

Quote from: lzxnl on February 14, 2014, 07:23:14 pm

So that was my working above. Now, moving on from the final result by putting over a common denominator (notice how the square root disappears):

$2x\sqrt{2x+2} + \frac{x^2}{\sqrt{2x+2}} = \frac{2x(2x+2)+x^2}{\sqrt{2x+2}} = \frac{5x^2+4x}{\sqrt{2x+2}}$

could you please show me in a bit more detail how you reached that final result? its confusing me, the whole denominator thing

IndefatigableLover · « **Reply #3996 on:** February 14, 2014, 08:23:42 pm »

Quote from: andrew2910 on February 14, 2014, 08:13:47 pm

could you please show me in a bit more detail how you reached that final result? its confusing me, the whole denominator thing

Alright so $2x\sqrt{2x+2}$ is the same as $\frac{2x\sqrt{2x+2}}{1}$ . When we want to add fractions together we need to make the denominator the same for both (so in this case we will multiply the denominator of $\frac{2x\sqrt{2x+2}}{1}$ with $2x\sqrt{2x+2}$ . And whatever we do with the bottom we have to do to the top. Funnily when we multiply $\frac{2x\sqrt{2x+2}}{1}$ with $2x\sqrt{2x+2}$ , the square roots cancel out.

In the end when you multiply the numerator and the denominator with $2x\sqrt{2x+2}$ you'll end up with $\frac{2x(2x+2)}{\sqrt{2x+2}}$ and then you add the $x^2$ to it to give you $\frac{2x(2x+2)+x^2}{\sqrt{2x+2}}$ .

Professor Polonsky · « **Reply #3997 on:** February 14, 2014, 08:25:58 pm »

Quote from: andrew2910 on February 14, 2014, 08:13:47 pm

could you please show me in a bit more detail how you reached that final result? its confusing me, the whole denominator thing

The main step there was to multiply both the numerator and the denominator of the first fraction by $\sqrt{2x+2}$ . You then have a common denominator.

$2x\sqrt{2x+2} + \frac{x^2}{\sqrt{2x+2}}$

$=\frac{2x \times \sqrt{2x+2} \times \sqrt{2x+2}}{\sqrt{2x+2}} + \frac{x^2}{\sqrt{2x+2}}$

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

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

Anchy · « **Reply #3998 on:** February 14, 2014, 08:37:36 pm »

Quote from: abcdqd on February 14, 2014, 08:02:13 pm

Let's do this step by step:

- Translate 6 units right $\frac{1}{x}$ becomes $\frac{1}{x-6}$
- Translate 1 unit up $\frac{1}{x-6}$ becomes $\frac{1}{x-6}+1$

Reflections in the y-axis are done by changing $f(x)$ to $f(-x)$ . Note that you only change the $x$ term, i.e. $x$ becomes $-x$
- Reflection in y-axis $\frac{1}{x-6}+1$ becomes $\frac{1}{-x-6}+1$

Dilations in the x-axis are done by multiplying the entire function by the dilation factor.

- Dilation factor of 2 from x axis $\frac{1}{-x-6}+1$ becomes $2 \left(\frac{1}{-x-6}+1\right)$

When you expand this and take out a factor of -1 from your denominator it is equal to the answer provided in the book.

Thank you so much for that!

I also have another question. How would you dilate 1/(x+1) -2 by a factor of 1/3?

soNasty · « **Reply #3999 on:** February 14, 2014, 08:42:56 pm »

Amazing thank you both

RKTR · « **Reply #4000 on:** February 14, 2014, 08:55:00 pm »

Quote from: Anchy on February 14, 2014, 08:37:36 pm

Thank you so much for that!

I also have another question. How would you dilate 1/(x+1) -2 by a factor of 1/3?

From which axis?

Anchy · « **Reply #4001 on:** February 14, 2014, 08:56:34 pm »

Quote from: RKTR on February 14, 2014, 08:55:00 pm

From which axis?

Oh sorry, from y-axis

RKTR · « **Reply #4002 on:** February 14, 2014, 09:23:20 pm »

1/(3x+1) -2?

alchemy · « **Reply #4003 on:** February 14, 2014, 09:26:01 pm »

Quote from: Anchy on February 14, 2014, 08:37:36 pm

Thank you so much for that!

I also have another question. How would you dilate 1/(x+1) -2 by a factor of 1/3 from the y-axis?

Using the dash method:

y=1/(x+1) - 2
(x',y')->(1/3x,y)
x'=x/3
x=3x', y=y'
y=1/(3x'+1) - 2
Hence, the dilated graph has the equation of 1/(3x+1) - 2

IndefatigableLover · « **Reply #4004 on:** February 14, 2014, 09:26:44 pm »

Quote from: Anchy on February 14, 2014, 08:37:36 pm

Thank you so much for that!

I also have another question. How would you dilate 1/(x+1) -2 by a factor of 1/3?

Quote from: Anchy on February 14, 2014, 08:56:34 pm

Oh sorry, from y-axis

So when you're dilating from the y-axis, it'll be: $y=f(x)$ to $y=f(\frac{x}{k})$
In this case, 'k' is 3 so you just put a '3' in front of every 'x' you see in the equation

So you're final equation should be:

$\frac{1}{3x+1}-2$

EDIT: Beaten by RKTR & alchemy (but it's good to see the Dash Method being used too)!
Mod Edit: Fixed your latex tags - Phy124

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