Wednesday

March 4, 2015

March 4, 2015

Posted by **sh** on Wednesday, March 24, 2010 at 2:23am.

y=2cosx + sin2x 0≤x≤2pi

y'=-2sinx + 2cos2x

y"=-2cosx-4sinx

How do I find the IP(s)?

- Calculus -
**Reiny**, Wednesday, March 24, 2010 at 9:09amYour 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 9:23amfrom cosx + 4sinxcosx = 0

it should have been

cosx(1 + 4sinx) = 0

cosx = 0 or sinx = -1/4

- Calculus -
**sh**, Wednesday, March 24, 2010 at 10:32amI got-4sin2x for the second term, I made a typo.

How do you translate sinx=-1/4 to decimal numbers?

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