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December 18, 2014

December 18, 2014

Posted by **Cam** on Wednesday, March 6, 2013 at 10:06pm.

- calculus -
**Reiny**, Thursday, March 7, 2013 at 9:11amf ' (x) = 3ax^2 + 12x + b

given:

f ' (-1) = 0 ----> 3a -12 + b = 0 or 3a + b = 12

f ' (2) = 0 ----> 12a + 24 + b = 0 or 12a + b = -24

subtract them:

9a = -36

a = -4

back in 3a + b = 12

-12 + b = 12

b = 24

a = -4, b = 24

check:

f ' (x) = -12x^2 + 12x + 24

= 0 for a max/min

x^2 - x - 2 = 0

(x-2)(x+1)=0

x = 2 or x = -1 , yeahhhh!

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