Posted by Sarah on Monday, March 29, 2010 at 2:08am.
Given f(x)=sin(x)2cos(x) on the interval [0,2pi]. Determine where the function is concave up and concave down.

Calculus  Reiny, Monday, March 29, 2010 at 9:00am
f'(x) = cosx + 2sinx
f''(x) = sinx + 2cosx
points of inflection"
f''(x) = 0
sinx + 2cosx = 0
sinx = 2cosx
tanx = 2
x = 63.4 degrees or x = 243.4 degrees
these last two are our important x values, since they are where the curve switches between concave up and concave down
lets's pick values in between
f''(x) = sin0  2cos0 = 2, so at 0 it is concave upwards
f''(90) = sin90  2cos9 = 1, so at 90 degrees it is concave down
f''(270) = sin270  2cos270 = 1 , so at 270 degrees it is concave upwards again.
f''(360) = sin360  2cos360 = 2 , sure enough concave upwards
so concave up: from 0 < x < 63.4
concave down : 63.4 < x < 243.4
concave up : 243.4 < x < 360
Just noticed you probably wanted your answer in radians, my calculator was set to degrees.
No big deal, just repeat my calculations with radian settings
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