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March 2, 2015

March 2, 2015

Posted by **casey** on Thursday, November 5, 2009 at 4:35pm.

- calc -
**Reiny**, Thursday, November 5, 2009 at 5:42pmg'(s) = -1/(s-2)^2

setting this equal to zero has no solution, so the function has no max/min

and since g'(s) is always negative, the function is decreasing for all values of s except s = 2

so let's look at the endpoints of the interval

g(0) = 1/-2 = - 1/2

g(1) = 1/-1 = -1

so for the given interval, the max is -1/2 and the min is -1

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