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September 21, 2014

September 21, 2014

Posted by **mary** on Thursday, March 15, 2012 at 3:07pm.

Thank you for your time.

- calculus-derivatives -
**Reiny**, Thursday, March 15, 2012 at 3:48pmtried x = ± 1 , no good

tried ± 1/2 and ahhh! x = 1/2 is a root

so (2x-1) is a factor

long division:

4x^3 -6x^2 +1 = (2x-1)((2x^2 - 2x-1)

2x^2 - 2x-1 = 0 gave me

x = (1 ± √3)/2

so there are 3 critical points to worry about.

take it from here.

- calculus-derivatives -
**Steve**, Thursday, March 15, 2012 at 3:51pmthis factors into

2(2x-1)(x^2 - 2x - 1)

x = 1/2 or (1±√3)/2

Having that, you can find where f'<0 or f'>0

- calculus-derivatives -
**mary**, Thursday, March 15, 2012 at 4:50pmOooooh! Thank you two so much :) I'm more confident in regards to my test now, thank you

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