Mathematics Question Thread

[jazzycab]

Oh sorry....how stupid of me!  No it must be the product rule not the function of a function......the brackets are multiplied by each other and by loge.....they are not all in one big lot of brackets if that makes sense?

So is it \(\log_e{\left(2x+4\right)}\times\left(3x-1\right)\)?
If so, you'll need to apply the product rule where \(y=\left(3x-1\right)\log_e{\left(2x+4\right)}=u\times v\) where \(u=3x-1\) and \(v=\log_e{\left(2x+4\right)}\).
That is:

[LaraC]

Thanks heaps Jazzycab! Thats great 

[LaraC]

Hello
Umm yeah...sorry...another question from me!!

Can someone help me how to find the point of inflexion on the curve y=xlog_ex-x^2?
Thanks!

[RuiAce]

Hello
Umm yeah...sorry...another question from me!!

Can someone help me how to find the point of inflexion on the curve y=xlog_ex-x^2?
Thanks!

[LaraC]

Thanks RuiAce! You are a legend
I was thinking it was the first derivative =0 at point of inflexion.....but oops! Makes sense now...thanks!

[Jada03]

Hello

I'm struggling a little with the question:

Find the equation of the tangent to the curve y=4^(x+1) (as in 4 to the power of x+1) at the point (0,4).
I'm stuck with how to find the differential of a constant raised to a more complex index such as (x+1) as opposed to just x by itself.

Thanks in advance!! 

[RuiAce]

Hello

I'm struggling a little with the question:

Find the equation of the tangent to the curve y=4^(x+1) (as in 4 to the power of x+1) at the point (0,4).
I'm stuck with how to find the differential of a constant raised to a more complex index such as (x+1) as opposed to just x by itself.

Thanks in advance!! 

(Although we don't quote it - this is how we do it)
\begin{align*}\frac{d}{dx} 4^{x+1} &= \frac{d}{dx} e^{\ln 4^{x+1}}\tag{formula above}\\ &= \frac{d}{dx} e^{(x+1) \ln 4} \tag{log law}\\ &= e^{(x+1)\ln 4}  \ln 4\tag{chain rule}\\ &= 4^{x+1}\ln 4\tag{reversing everything}\end{align*}

[gilliesb18]

Hello, just need help on this question....
Find the area enclosed between the curve y=x^3, the x-axis and the line y=-3x +4.
Me and my friend just can't seem to find where the two points intercept...
Any help??

thanks heaps

[Opengangs]

Hello, just need help on this question....
Find the area enclosed between the curve y=x^3, the x-axis and the line y=-3x +4.
Me and my friend just can't seem to find where the two points intercept...
Any help??

thanks heaps

Just a quick observation:
Try substituting x = 1 and you'll see the point of intersection.

[RuiAce]

Hello, just need help on this question....
Find the area enclosed between the curve y=x^3, the x-axis and the line y=-3x +4.
Me and my friend just can't seem to find where the two points intercept...
Any help??

thanks heaps

Will make a small remark because I do remember seeing this in maths in focus. The fact they're expecting you to solve that question in 2U is unrealistic - this should not be a 2U question. So don't worry too much about it (or just take Opengangs' suggestion for granted to prove that there's an intercept at \(x=1\)).

[gilliesb18]

Yep well we worked it out by graphing it online, but to me thats not the way to do it, cause it would vary for each question... plus i cant stand the method of guess and check!!!!
If you don't think it's necessary for 2U, I will just leave it....
Just btw- do you feel that 'Maths in Focus' covers everything? I just get a bit worried when I look at past papers that it doesn't!

[RuiAce]

It definitely doesn’t. As for textbooks, Cambridge is easily a better suggestion.

Textbooks like Terry Lee’s are also good but involve generally much harder questions.

[skisso]

Hey
I'm trying to do this question in a HSC paper (1994), and I have the answer but I dont get whats happening...

[RuiAce]

Hey
I'm trying to do this question in a HSC paper (1994), and I have the answer but I dont get whats happening...

They simply used the fact that integrals and derivatives undo each other to go from \( xe^{x^2} = \frac12 \, \frac{d}{dx} e^{x^2} \) to \( \int_0^1 xe^{x^2}\,dx = \frac12 \left[e^{x^2} \right]_0^1 \). The whole point is that because \( x e^{x^2} \) is a derivative of \( \frac12 e^{x^2}\), consequently we must have \( \frac12 e^{x^2} \) being an antiderivative of \( x e^{x^2} \)

[whitney.dent]

Hey, just needing some help on this question:
Find the volume of the solid formed when the line x+3y-1=0 is rotated around the x-axis from x=0 and x=8.
Thanks!!