Find the critical numbers?

**mj.alawami** · November 29th 2009, 11:38 AM

How to find the critical number of the given function:
$f(x)= x^3(x-2)^2$

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

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

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

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

Please answer in steps.

Thank you

**hjortur** · November 29th 2009, 11:50 AM

$<br /> f'(x)=(x^3)(2(x-2))+(x-2)^2(3x^2) =x^2\left(2x(x-2)+(x-2)^23\right)$

$<br /> =x^2(x-2)\left(2x+3(x-2)\right)=x^2(x-2)(5x-6)<br />$

**DeMath** · November 29th 2009, 11:57 AM

Originally Posted by mj.alawami

How to find the critical number of the given function:
$f(x)= x^3(x-2)^2$

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

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

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

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

Please answer in steps.

Thank you

Hint:

$f\left( x \right) = {x^3}{\left( {x - 2} \right)^2} = {x^3}\left( {{x^2} - 4x + 4} \right) = {x^5} - 4{x^4} + 4{x^3}.$

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

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