Find the critical numbers?

• Nov 29th 2009, 11:38 AM
mj.alawami
Find the critical numbers?
How to find the critical number of the given function:
$\displaystyle f(x)= x^3(x-2)^2$

Attempt:
$\displaystyle f'(x)=(x^3)(2[x-2][1])+(x-2)^2(3x^2)$

Now we have to equate f'(x) to zero

$\displaystyle f'(x)=(x^3)(2[x-2])+(x-2)^2(3x^2) =0$

After this step I dont know how to simplify to get the x co-ordinates or critical numbers

Thank you
• Nov 29th 2009, 11:50 AM
hjortur
$\displaystyle f'(x)=(x^3)(2(x-2))+(x-2)^2(3x^2) =x^2\left(2x(x-2)+(x-2)^23\right)$

$\displaystyle =x^2(x-2)\left(2x+3(x-2)\right)=x^2(x-2)(5x-6)$
• Nov 29th 2009, 11:57 AM
DeMath
Quote:

Originally Posted by mj.alawami
How to find the critical number of the given function:
$\displaystyle f(x)= x^3(x-2)^2$

Attempt:
$\displaystyle f'(x)=(x^3)(2[x-2][1])+(x-2)^2(3x^2)$

Now we have to equate f'(x) to zero

$\displaystyle f'(x)=(x^3)(2[x-2])+(x-2)^2(3x^2) =0$

After this step I dont know how to simplify to get the x co-ordinates or critical numbers

$\displaystyle f\left( x \right) = {x^3}{\left( {x - 2} \right)^2} = {x^3}\left( {{x^2} - 4x + 4} \right) = {x^5} - 4{x^4} + 4{x^3}.$
$\displaystyle f'\left( x \right) = 5{x^4} - 16{x^3} + 12{x^2} = {x^2}\left( {5{x^2} - 16x + 12} \right) = {x^2}\left( {5x - 6} \right)\left( {x - 2} \right).$