Finding Critical Numbers

**break** · November 21st 2009, 03:06 PM

Find the critical numbers of the function and describe the behavior of f at these numbers.

$f(x) = x^6(x - 2)^5$

I took the derivative using product rule

$f'(x) = 6x^5 (x-2)^5 + x^6 5(x-2)^4$

Then i simplified and set $f'(x) = 0$

$f'(x) = x^5(x-2)^4 [6(x-2) + 5x] = 0$
$f'(x) = x^5(x-2)^4 [6x-12 + 5x] = 0$
$f'(x) = x^5(x-2)^4 (x-12) = 0$

and the three points i got were

$x=0$ $x=2$ and $x=12$

i know 0 is right, but 2 and 12 is wrong.

what am i doing wrong?

**makaan** · November 21st 2009, 03:29 PM

When finding f'(x)=0,
6x - 12 + 5x is actually 11x - 12. This root gives you x = 12/11.

The critical points are x = {0, 12/11, 2}.

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