Monday

March 30, 2015

March 30, 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?

**Answer this Question**

**Related Questions**

algebra 1 help please - 4) a student score is 83 and 91 on her first two quizzes...

calculus please help me - Let f(x)=9sinx/2sinx+4cosx. Then 'f(x)= ((2sinx+4cosx...

check math - Here are my answers. Can you check if I got the right answers? ...

Precalculus - Determine the solutions to the equation 2sinx − cos2x = ...

Probability - Random variables X and Y are distributed according to the joint ...

probability - Random variables X and Y are distributed according to the joint ...

calculus - im having so much trouble figuring out how to get these answers. i ...

APCalculus - For the function y=2sin(x)-cos^2(x) on [0, 2pi] find the following...

trig - prove that (2cosx+1)(2cosx-1)= 1 +2cos2x

Calculus - Determine the points of inflection of the function. f(x) = x + sin x...