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November 28, 2014

November 28, 2014

Posted by **chrissy** on Monday, November 26, 2012 at 1:53pm.

- calc -
**Steve**, Monday, November 26, 2012 at 2:44pmf' = 1 + cos(x)

f'=0 when cos(x) = -1, or x=pi

however,

f''(x) = -sin(x) which is also 0 at x=pi.

So, (pi,pi) is an inflection point, not a max or min.

f(2pi) = 2pi

so that is the absolute max on [0,2pi]

Note that f'(x) >= 0 for all x, so there is no max or min, which require changing direction.

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