Question

How to find inflection point and concavity? please see picture attached

Answer 1

Differentiate, then differentiate again.

Answer 2

please show ur steps

Answer 3

Inflection points are where f ' = 0 and f '' = 0.

Answer 4

i know these basic definitions

Answer 5

Ok, just have to check.
A lot of people ask questions like these and don't even know the basics.
Did you get -14cos(x)[sin(x)+1] for the first derivative?

Answer 6

I already got 14 sin^2 x + 14sin x - 14 cos^2x = 0 . that's second derivative.
i think that's what u need to find inflection point right?

Answer 7

and yea

Answer 8

i know i need to set them to 0... but then i just couldnt solve it

Answer 9

I see...

Answer 10

then?

Answer 11

I think you can use a double-angle trig identity here.

Answer 12

\[\large cos^2(x)-sin^2(x)=1-2sin^2(x)\]

Answer 13

ok..but dont u need to set it to 0

Answer 14

Yes, make the substitution into the equation you have set equal to zero.

Answer 15

The identity is the identity, you need to plug it in to what you have first.

Answer 16

but it is sin^2x - cos^2x not cos^2x- sin^2x
can u just type out the whole steps..cuz i dont get it ..

Answer 17

a - b = -(b - a)

Answer 18

Something else I noticed . . .
You put DNE for the local maximum value, but you have an interval of increase of (π/2 , 3π/2) and an interval of decrease of (3π/2 , 2π).
Doesn't that mean that there is a local maximum at x=3π/2?

Answer 19

Anyway, when you make the substitution, you get
\[\large 2sin^2(x)+sin(x)-1=0\]
which can be solved as a quadratic form.

Answer 20

oh, ok. i get it now thanks.
it takes me quite a while to understand it. but yea, i get it now. thanks

Answer 21

It's quite the process, yeah - just need to be careful at each step. If you have a graphing calculator to check your work or something like wolframalpha, it helps too.

Answer 22

umm...how do i get the inflection points
i got x = 1/2 and -1..

Answer 23

I got x = 5pi/6, pi/6 and 3pi/2
what are these..

Answer 24

Not sure about the 5π/6, but the solutions to that quadratic were
sin(x) = 1/2 and sin(x) = -1
which means x could be π/6 or 3π/2.

Answer 25

i used the CAST rule
cuz sin = 1/2 can be pi/6 or 5pi/ 6 in the first quadrant and second quadrant

Answer 26

Oh, yes, you are correct. It is both π/6 and 5π/6 because your domain goes all the way to 2π

Answer 27

(what is the CAST rule, by the way, I don't recognize that?)