Geometrical Applications of Calculus Help... (1 Viewer)

[ScottyTee]

Hi I am stuck on the following question, any help is greatly appreciated... also if demonstrating the answer please do it in as many steps as possible, cheers.

Find the stationary points on the curve y = (3x - 1)(x - 2)^4

I know I have to get dy/dx and that for the stationary points dy/dx should = 0, but the ^4 is the confusing part, cheers.

 

[pLuvia]

With observation, you should realise that the dy/dx can be factorised to (x-2)³(..other..).

And the other equation can be easily simplified and so the x coordinates can be found for the stationary points

 

[ScottyTee]

With observation, you should realise that the dy/dx can be factorised to (x-2)³(..other..).

And the other equation can be easily simplified and so the x coordinates can be found for the stationary points

Sorry, I dont quite understand perhaps you could elaborate to help me understand? Cheers.

 

[ScottyTee]

Scratch that, I got it... cheers.

 

[SoulSearcher]

dy/dx = 3(x-2)⁴ + 3(3x-1)(x-2)³

You can see here that (x-2)³ is a factor of both parts of the derivative, therefore you can simplify it further there.

 

[Starcraftmazter]

Solution Here:
http://starcraftmazter.net/maths/sol1_20.11.png

Hope it's right =/

Edit:
Also, 66kb version for those on dialup
http://starcraftmazter.net/maths/sol1_20.11.jpg

 

[ScottyTee]

Solution Here:
http://starcraftmazter.net/maths/sol1_20.11.png

Hope it's right =/

Edit:
Also, 66kb version for those on dialup
http://starcraftmazter.net/maths/sol1_20.11.jpg

Cheers mate, much appreciated. Good work! it was right.