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

• Nov 3rd 2007, 01:34 AM
KavX
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
• Nov 3rd 2007, 01:49 AM
DivideBy0
$f(x) = (3x-1)(x-2)^4$

$f'(x) = (3x-1)\cdot \frac{d}{\,dx}((x-2)^4) + (x-2)^4 \cdot \frac{d}{\,dx}(3x-1)$

The derivatives are:

$\frac{d}{\,dx} ((x-2)^4) = 4(x-2)^3$

And:

$\frac{d}{\,dx} (3x-1) = 3$

So:

$\Rightarrow f'(x) = 4(3x-1)(x-2)^3 + 3(x-2)^4$

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

$4(3x-1)(x-2)^3 + 3(x-2)^4 =0$

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

$(x-2)^3(4(3x-1)+3(x-2))=0$

$(x-2)^3(12x-4+3x-6)=0$

$(x-2)^3(15x-10)=0$

$\mbox{\Rightarrow x = 2 or x=\frac{2}{3}}\\$

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

$\lbrace{(x,y):(2,0),\left(\frac{2}{3},\frac{256}{8 1}\right)\rbrace}$
• Nov 3rd 2007, 01:53 AM
KavX
thanx a lot
• Nov 3rd 2007, 11:08 AM
Soroban
Hello, KavX!

Quote:

Find the stationary points on the curve: . $y\:=\:(3x-1)(x-2)^4$
Advice: After doing a Product Rule, think of factoring.

We have: . $y' \;=\;(3x-1)\cdot4(x-2)^3\cdot 1 \:+\: 3\cdot(x-2)^4$

Simplify: . $y' \;=\;4(x-2)^3(3x-1) + 3(x-2)^4$

Factor: . . $y' \;=\;(x-2)^3\left[4(3x-1) + 3(x-2)\right]$

Simplify: . $y' \;=\;(x-2)^3(15x-10) \;=\;5(x-2)^3(3x-2)$

To solve $y' = 0$, we have: . $5(x-2)^3(3x-2) \;=\;0$

Got it?

• Nov 3rd 2007, 02:52 PM
KavX
aahh yer thanx for that