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

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

I differentiated it using the udv + vdu, then i tried to work it out and i got no were close to the answer i cant figure how to do my working out properly.

back of book says it (2,0) and (2/3, 3/13/81)

can sum1 plz show me how they would do working out on this question.

thx

The derivatives are:



And:



So:



Setting f'(x) = 0, we get:



Now we can't simply divide by (x-2), or we will be missing out on a solution, so instead we factor out:









Then, after substituting these values back into the original equation, we get the stationary points as:

2,0),\left(\frac{2}{3},\frac{256}{81}\right)\rbrac e}" alt="\lbrace{(x,y)2,0),\left(\frac{2}{3},\frac{256}{81}\right)\rbrac e}" />

thanx a lot

Hello, KavX!

Find the stationary points on the curve: .

Advice: After doing a Product Rule, think of factoring.

We have: .

Simplify: .

Factor: . .

Simplify: .


To solve , we have: .

Got it?

aahh yer thanx for that