Differentiation help please! Second Derivative (stationary points) question...

Hi, I have this question that I can't quite figure out (I think I'm just doing something stupid! )...

Find any stationary points on the curve of the following function and identify their nature:

$y= x^{3}+\frac{12}{x}$

I differentiated, then rewrote the equation:

$\frac{\mathrm{d}{y}}{\mathrm{d} x} = 3x^{2}-\frac{12}{x^{2}} = 3x^{2}-12x^{-2} [br]$

From there I was unable to factorise and so find the stationary points

Thanks for your help!

Hi, I have this question that I can't quite figure out (I think I'm just doing something stupid! )...

Find any stationary points on the curve of the following function and identify their nature:

$x^{2}$

I differentiated, then rewrote the equation:

$\frac{\mathrm{d}{y}}{\mathrm{d} x} = 3x^{2}-\frac{12}{x^{2}} = 0[br]$

From there I was unable to factorise and so find the stationary points

Thanks for your help!

to solve

$\frac{\mathrm{d}{y}}{\mathrm{d} x} = 3x^{2}-\frac{12}{x^{2}} = 0 [br]$

you could start by multiplying each term by $x^{2}$

Would that mean the equation becomes
$3x^{4}-12=0$?

I'm a little stuck in factorising this equation ...

this is what you should get....

$3x^{4}-12=0$

Sorry, I get it now! To find the stationary points then, I would factorise, resulting in $3(x^{4}-4)$
Then $\frac{\mathrm{dy} }{\mathrm{d} x}=0$ when $x^{4}-4 \Rightarrow x^{4}=4 \Rightarrow x=\sqrt[4]{4}$

Is this correct?
The answer in my book lists the stationary points as the x values being

$\sqrt{2}$ and $-\sqrt{2}$ respectively.

Would the x values obtained from the equation then have to equal these for my working to be correct?

You need both to get full marks. Do you see where they come from?

No - that's just where I can't work the problem further... I don't see how they got their answers from where I was with my partial answer of $x= \sqrt[4]{4}$ ??

If x^4=4, then $x= \pm \sqrt[4]{4}$

Use laws of indices to progress further. Perhaps rewrite it as (2^n)^m