If anyone else is looking at this, please don’t assume that all you have to do, is check all the integers in the interval. Just find the max by taking the derivative again, setting it equal to 0 and solving for t:

a(t) = 3t^2 – 6t + 12; a'(t) = 6t – 6 = 0; 6t = 6; t = 1.

Now, just check the endpoints and that value, so: t=0, t=1, t=3. ]]>

A good friend mailed this link the other day and I’m excitedly waiting your next post. Keep on on the incredible work.

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