Author Topic: Silly Questions Thread (Read 27097 times) Tweet Share

lachymm · « **Reply #270 on:** October 31, 2010, 09:36:18 pm »

Quote from: brightsky on October 31, 2010, 09:33:52 pm

sin(2x) = 0
2x = 0, pi, 2pi, 3pi, etc.
So yes your answer is correct.

but their answer is $\frac{2kpie-pie}{2}<x<pie*k$ ?

m@tty · « **Reply #271 on:** October 31, 2010, 09:43:16 pm »

The derivative is not defined where $\sin(2x)\leq 0$

$\sin(\theta)\leq 0 \text{ where } \pi\leq \theta \leq 2\pi$

$\therefore \sin(2x) \leq 0 \text{ where } \frac{\pi}{2} \leq x \leq \pi$

This is the first solution.

So the general solution is $\frac{\pi(2k-1)}{2} \leq x \leq k\pi \text{ where }k\in\mathbb{Z}$

lachymm · « **Reply #272 on:** October 31, 2010, 09:47:50 pm »

damn your a wizz at maths, thanks again

m@tty · « **Reply #273 on:** October 31, 2010, 10:05:56 pm »

Quote from: brightsky on October 31, 2010, 09:28:33 pm

Quote from: lachymm on October 31, 2010, 09:23:44 pm
When i derive $|5x+3|$ the calculator gives me $5 sign (5x + 3)$ what does this mean?
Or do i need to set the domains of 2 different functions then derive individually?

See signum function. 5sgn(5x + 3) = 5|x|/(x(5x + 3)). For methods, it's probably better to just split it into cases.

I doubt anyone will notice, but on the off chance that this may mislead someone

$5sgn(5x+3) \ne \frac{5|x|}{x(5x+3)}$

Rather, from the definition $x=sgn(x)\cdot |x|$

$5sgn(5x+3)=\frac{5(5x+3)}{|5x+3|}$

The Detective · « **Reply #274 on:** November 01, 2010, 11:11:44 am »

umm... sorry guys but how do you post mathematical symbols? thanks

lachymm · « **Reply #275 on:** November 01, 2010, 11:49:04 am »

Latex there is a little button above the smiley faces called "TEX" then just type ur equations in btw you may need to learn how to do the basic symbols and stuff so go to wikipedia for the commands. Hope i made sense..

m@tty · « **Reply #276 on:** November 01, 2010, 12:23:57 pm »

See here. Mao's beginner's guide.

If you find yourself in need of more obscure symbols, follow the link at the bottom of that page for extensive commands.

lachymm · « **Reply #277 on:** November 01, 2010, 05:36:01 pm »

Alright i need some help with this " $\frac{dy}{dx}=x^3-7x^2=x*e^x-2$ , then the average value over $[\frac{pie}{2},root(7)]$ is ?

I thought you needed to anti differintiate the y function and do that $\frac{1}{a-b}$ times the antiderivitive thingy to get the average value? but answers prove me wrong again...

lachymm · « **Reply #278 on:** November 01, 2010, 05:45:25 pm »

also how would you go about solving $Pr(A)=.6$ and $Pr(B)=.65$ and $Pr(AUB)=.65$ then $Pr(A'|`|B')$ is? BTW that $|`|$ is an upside down U that they sometimes have in probabilty..

Also In the formula $np=mean$ and $np(1-p)=variance$ what does n actually represent? i have forgoten

lachymm · « **Reply #279 on:** November 01, 2010, 05:50:04 pm »

When i find there inverse of $f(x)=x^2-6x+5$ i get $f^{-1}(x)=-sqroot(x+4)+3$ which is correct because you need to choose the negative root in the context of the question however when i went to restrict the domain i said $[-4,oo)$ however the answer is $(-4,oo)$ why is this?

lachymm · « **Reply #280 on:** November 01, 2010, 05:56:36 pm »

How do you use the chain rule for inside functions other than the typical $(x-3)^6$ kind of thing example...
$cos(\frac{1}{x})$ i dont understand how the $u$ and the $v$ thingy works when inside a function

brightsky · « **Reply #281 on:** November 01, 2010, 07:08:36 pm »

Quote from: lachymm on November 01, 2010, 05:56:36 pm

How do you use the chain rule for inside functions other than the typical $(x-3)^6$ kind of thing example...
$cos(\frac{1}{x})$ i dont understand how the $u$ and the $v$ thingy works when inside a function

Let $u = \frac{1}{x}$

Therefore $y = \cos(u)$

$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = -\sin(u) \cdot -x^{-2} = \frac{\sin(u)}{x^2} = \frac{\sin\left(\frac{1}{x}\right)}{x^2}$

Quote from: lachymm on November 01, 2010, 05:50:04 pm

When i find there inverse of $f(x)=x^2-6x+5$ i get $f^{-1}(x)=-sqroot(x+4)+3$ which is correct because you need to choose the negative root in the context of the question however when i went to restrict the domain i said $[-4,oo)$ however the answer is $(-4,oo)$ why is this?

You should be right.

Quote from: lachymm on November 01, 2010, 05:45:25 pm

also how would you go about solving $Pr(A)=.6$ and $Pr(B)=.65$ and $Pr(AUB)=.65$ then $Pr(A'|`|B')$ is? BTW that $|`|$ is an upside down U that they sometimes have in probabilty..

Also In the formula $np=mean$ and $np(1-p)=variance$ what does n actually represent? i have forgoten

Recall that $Pr(A) + Pr(B) - Pr(A \cap B) = Pr (A \cup B)$

Then substituting values we get $0.6 + 0.65 - Pr(A \cap B) = 0.65$

$Pr(A\cap B) = 0.6$

Quote from: lachymm on November 01, 2010, 05:36:01 pm

$\frac{dy}{dx}=x^3-7x^2=x*e^x-2$

I don't understand what you mean by this.

nacho · « **Reply #282 on:** November 01, 2010, 07:21:32 pm »

A pretty standard question, I fear my lack of knowledge may have let me down..
Question:

Let M = [ -2cosx 3 ]
[ 2 cosx ]

Where x is a real number. What is the range of det( M )?

^that is a single square matrice by the way.
My workings lead me to :

cosx - root3 = 0
OR
cosx + root3 = 0

Possibly incorrect, but i'd assume that i'd find what x equals in both scenarios and the answer would be [smaller x value, larger x value]
Just do not know how to get there.

(Sorry, too lazy to learn LaTex

)

lachymm · « **Reply #283 on:** November 02, 2010, 04:40:37 pm »

Thanks bright sky

Alright some more this time from kilabah 2010..

1) How does $\frac{x-2}{x+2}$ become $\frac{4}{x+2}-1$ ?

2) How would you go about solving $m^2-8m+8=0$ by hand?

3) How would you find the inverse of $f(x)=e^{2x}-e^{-2x}$ ? i canot even understand their workings :S

Also is there a good way to remember the binomial formula the one $^3C2*p^2*(1-p)^{1}$

I mean how do you remember the factorial part? which numbers go where on the fraction? I havent really been taught how to do the binomial formula by hand always done it on the calc.

m@tty · « **Reply #284 on:** November 02, 2010, 05:03:38 pm »

1) It doesn't.

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

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

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

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

$(m-4)^2=8$

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

3)

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

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

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

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

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

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

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

By the quadratic formula:

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

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

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

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

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

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