Posted by sh on Wednesday, March 24, 2010 at 2:23am.
Determine the concavity and find the points of inflection.
y=2cosx + sin2x 0≤x≤2pi
y'=2sinx + 2cos2x
y"=2cosx4sinx
How do I find the IP(s)?

Calculus  Reiny, Wednesday, March 24, 2010 at 9:09am
Your second derivative's last term should be 4sin2x
set y'' = 0 for points of inflection
2cosx  4sin2x = 0
cosx + 2sin2x = 0
cosx + 4sinxcosx = 0
cosx(1 + 4cosx) = 0
cosx = 0 or cosx = 1/4
You seem to know what you are doing, and can probably take it from here.
Concavity in short:
If for some f(x), the second derivative exists,
the curve is concave upwards if f''(x) > 0
the curve is concave downwards if f''(x) < 0
if f''(x) = 0 it is neither and you have a point of inflection as seen above.

correction  Calculus  Reiny, Wednesday, March 24, 2010 at 9:23am
from cosx + 4sinxcosx = 0
it should have been
cosx(1 + 4sinx) = 0
cosx = 0 or sinx = 1/4

Calculus  sh, Wednesday, March 24, 2010 at 10:32am
I got4sin2x for the second term, I made a typo.
How do you translate sinx=1/4 to decimal numbers?
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