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December 19, 2014

December 19, 2014

Posted by **Britney** on Wednesday, November 7, 2012 at 4:59pm.

f(x)=x^3+3x^2-1 on [-3,1]

- math -
**Steve**, Wednesday, November 7, 2012 at 5:46pmf(-3) = -1

f(1) = 3

f'(x) = 3x^2 + 6x = 3x(x+2)

so, there are relative max/min at x= -2,0

f(-2) = 3

f(0) = -1

So, on [-3,1] we have

min = -1

max = 3

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