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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