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October 31, 2014

October 31, 2014

Posted by **jeffery** on Wednesday, March 17, 2010 at 5:22am.

- college calculus 1 -
**Damon**, Wednesday, March 17, 2010 at 6:34amf(x)=x^3-6x^2+9x+1, [2,4]

f' = 3 x^2 -12 x + 9

look for zeros in domain

x^2 -4x + 3 = 0

(x-3)(x-1) = 0

x = 1 or x = 3

3 is in our piece of domain between 2 and 4

What is function at x = 3 ? Calculate it.

Then also calculate the function at x = 2 and x = 4

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