Author Topic: Silly Questions Thread (Read 34642 times) Tweet Share

lachymm · « **Reply #285 on:** November 02, 2010, 05:34:47 pm »

Quote from: m@tty on November 02, 2010, 05:03:38 pm

1) It doesn't.

$\frac{x-2}{x+2}=\frac{x+2-4}{x+2}=\frac{x+2}{x+2}-\frac{4}{x+2}=1-\frac{4}{x+2}$

Thanks

2)
Quadratic formula is the easiest, or you could complete the square.

$m^2-8m+8=0$

$(m-4)^2-16+8=0$

$(m-4)^2=8$

$m=\pm 2 \sqrt{2}+4$

I dont understand this last step how $(m-4)^2=8$ means that $m=\pm 2 \sqrt{2}+4$

3)

$f(x)=e^{2x}-e^{-2x}$

$\text{Let }f(x)=y$

$f(y)=x\text{ for inverse}$

$x=e^{2y}-2^{-2y}$

$xe^{2y}=e^{4y}-1$

I also dont understand how you got $xe^{2y}=e^{4y}-1$ from $x=e^{2y}-2^{-2y}$

$\text{Let }A=e^{2y}$

$\implies 0=A^2-xA-1$

By the quadratic formula:

$A=\frac{x \pm \sqrt{x^2+4}}{2}$

$\text{Sub }e^{2y}=A$

$\therefore e^{2y}=\frac{x+\sqrt{x^2+4}}{2}\text{ Reject -'ve root as }e^{2y}>0 \forall x\in\mathbb{R}$

$\therefore 2y=\log_e\left(\frac{x+\sqrt{x^2+4}}{2}\right)$

$y=\frac{1}{2} \log_e\left(\frac{x+\sqrt{x^2+4}}{2}\right)$

Thanks m@tty for your time

Declanator · « **Reply #286 on:** November 02, 2010, 07:31:05 pm »

hey can you guys help me with this one?
its one of the examples for the new exam one content
thanks

$f (x) = x^3 + (k + 1)x^2 + kx$
Solve f(x)=0 for x

fady_22 · « **Reply #287 on:** November 02, 2010, 07:46:29 pm »

The Detective · « **Reply #288 on:** November 03, 2010, 09:31:56 am »

alright thanks guys

lachymm · « **Reply #289 on:** November 03, 2010, 10:04:35 am »

what does "tbh" mean i see it all the time on vcenotes

fazida · « **Reply #290 on:** November 03, 2010, 10:42:21 am »

Quote from: lachymm on November 03, 2010, 10:04:35 am

what does "tbh" mean i see it all the time on vcenotes

it means 'to be happy'

Whatlol · « **Reply #291 on:** November 03, 2010, 10:51:40 am »

Quote from: fazida on November 03, 2010, 10:42:21 am

Quote from: lachymm on November 03, 2010, 10:04:35 am
what does "tbh" mean i see it all the time on vcenotes

it means 'to be happy'

... it means to be honest

d-ea-6 · « **Reply #292 on:** November 03, 2010, 10:58:21 am »

Quote from: Whatlol on November 03, 2010, 10:51:40 am

Quote from: fazida on November 03, 2010, 10:42:21 am
Quote from: lachymm on November 03, 2010, 10:04:35 am
what does "tbh" mean i see it all the time on vcenotes

it means 'to be happy'
... it means to be honest

Yeah, I hope that other guy was just joking...

lovingit · « **Reply #293 on:** November 03, 2010, 12:13:19 pm »

Can someone please explain to me how you know when a connecting point in a hybrid graph is diffentiable
How do you determine if it is continous or not?

m@tty · « **Reply #294 on:** November 03, 2010, 02:13:42 pm »

Quote from: lachymm on November 02, 2010, 05:34:47 pm

Quote
$(m-4)^2=8$

$m=\pm 2 \sqrt{2}+4$

I dont understand this last step how $(m-4)^2=8$ means that $m=\pm 2 \sqrt{2}+4$

3)

$x=e^{2y}-2^{-2y}$

$xe^{2y}=e^{4y}-1$

I also dont understand how you got $xe^{2y}=e^{4y}-1$ from $x=e^{2y}-2^{-2y}$

Thanks m@tty for your time

$(m-4)^2=8$

$m-4=\pm\sqrt{8}$

$m-4=\pm 2\sqrt{2}$

$\therefore m=\pm2\sqrt{2}+4$

That second one was a typo. It was meant to read:

$x=e^{2y}-e^{-2y}$

Multiply both sides by $e^{2y}$ .

$xe^{2y}=e^{4y}-1$

lachymm · « **Reply #295 on:** November 03, 2010, 03:49:56 pm »

Quote from: lovingit on November 03, 2010, 12:13:19 pm

Can someone please explain to me how you know when a connecting point in a hybrid graph is diffentiable
How do you determine if it is continous or not?

If a 2 functions are apart of a hybrid function and they both have the same y value when there domains intersect (They continue where the other left off) while having the same gradient at this point they will be continuous thus not requiring a open circle..Finally i am able to help someone else out on here, feels good

doter · « **Reply #296 on:** November 03, 2010, 04:17:06 pm »

hey how does this this equation simplify into the one below it
thanks in advance!!

m@tty · « **Reply #297 on:** November 03, 2010, 05:22:36 pm »

$\tan^2(x)+(1-\sqrt{3})\tan(x)-\sqrt{3}=0$

$\tan^2(x)+\tan(x)-\sqrt{3}\tan(x)-\sqrt{3}=0$

$\tan(x)\left(\tan(x)+1\right)-\sqrt{3}\left(\tan(x)+1\right)=0$

Factor out $\tan(x)+1$

$(\tan(x)+1)(\tan(x)-\sqrt{3})=0$

doter · « **Reply #298 on:** November 03, 2010, 06:44:51 pm »

thanks matty
understand it now

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