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January 28, 2015

January 28, 2015

Posted by **Janet** on Monday, April 18, 2011 at 11:52pm.

This is what I did so far.

f'(x)=3x^2-4=0

3x^2=4

x= 2/sqrt3 The -2/sqrt3 is outside of interval so can't use.

x=2/sqrt3

f(0)=5

lim x-->5- is 110. Can't use this though because 5 is not included in the interval.

Since this is the largest of the three numbers, 2/sqrt3, 5, 110, there is no absolute max.

The local minimum is at x=2/sqrt3.

I don't know if this is right or if there are any local max or absolute minimum.

Please help.

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