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December 20, 2014

December 20, 2014

Posted by **sh** on Sunday, December 5, 2010 at 4:15pm.

f(x)=sin(x+(pi/2)) for 0≤x≤2pi

so my derivative is f'(x)=cos(x+(pi/2))?

and my critical number is -pi/2?

- Calculus -
**Damon**, Sunday, December 5, 2010 at 5:04pmcosine is positive for x +pi/2 = 0 to pi/2 and from 3 pi/2 to 2 pi

when x+pi/2 = 0 then x = -pi/2 or 3pi/2

when x + pi/2 = pi/2, x = 0

so one interval is x = 3pi/2 to 0

when x+pi/2 = 3 pi/2 then x = pi

when x+pi/2 = 2 pi then x = 3 pi/2

so second interval is frm x = pi to 3 pi/2

so finally the entire interval from x = pi to x = 2 pi (which is 0) has a positive derivative, function increasing

- Calculus -
**sh**, Sunday, December 5, 2010 at 5:19pmWow, so confusing. Thanks!

- Calculus -
**sh**, Sunday, December 5, 2010 at 5:24pmSo the whole interval is increasing? The function of cosx has a section of decreasing though.

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