posted by Sarah .
Given f(x)=sin(x)-2cos(x) on the interval [0,2pi]. Determine where the function is concave up and concave down.
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