Question

Find the Concavity:

f(x) = x - 2cosx on the closed interval [0, 2pi]

Answer 1

I know I need the second derivative

f'(x) = 1+2sinx
f''(x) = 2cosx

Where do I go from there?

Answer 2

ok... so look at the stationary points

let

f'(x) = 0 and solve.....

there are 2 stationary points

1 + 2sin(x) = 0

the solutions are exact values in 3rd and 4th quadrants...

test them in the 2nd derivative...

you'll also need to solve the 2nd derivative as the curve will have a point of inflection, given it has 2 stationary points...

hope it helps

Answer 3

Sorry I'm here

Answer 4

I already found that Min = -2 and Max = 2pi-2

Is that what you mean?

Answer 5

no you need to find f''(x) = 0

Answer 6

Well cos = 0 at pi/2 and 3pi/2 within 0 to 2pi
Is that it?

Answer 7

well not quite you are solving the 1st derivative

\[0 = 1 + 2\sin(x)\]

now solve for x..... this will give the the x values of any stationary points....

and there are 2 of those...

Answer 8

there are also 2 points of inflection in the domain, so there are 2 concavities

Answer 9

-1 = 2sin(x)
-1/2 = sinx

But that is not within [0,2pi]

Answer 10

cos(x) = 0
x = pi/2 or 3pi/3

so yes. Now you need to find the concavity on
(0,pi/2) , (pi/2,3pi/2) , (3pi/2, 2pi)

Answer 11

thats correct... so the angles that give than values are in the 3rd and 4th quadrants...

so what are the angles...?

Answer 12

Okay so my points of inflection are pi/2, 3pi/2 and 2pi.

Answer 13

2 of them are.... 2 pi isn't a point of inflection

Answer 14

yes, so what are the interval you need to check the concavity on?

Answer 15

Oh okay thank you.

Answer 16

I wish I could give you a medal too sourwing

Answer 17

the points of inflection show where the concavity changes... so you well need to look at the intervals left and right of the points of concavity... and the interval between them...

Answer 18

the easiest way to do this question is to graph it using some sort of graphing software

Answer 19

I got -infinity,pi/2 is concave down

pi/2, 3pi/2 is concave up

and 3pi/2 +infinity is concave down

Answer 20

well thats interesting as by looking at the 1st derivative

0 = 1 + 2sin(x)

solutions are

\[x = \frac{7\pi}{6}...and..... \frac{11\pi}{6}\]

and then you use the 2nd derivative test

\[f"(\frac{7\pi}{6}) = 2\cos(\frac{7\pi}{6})\]

then

\[f"(\frac{7\pi}{6}) < 0... a... maximum\]

so you have the concavity wrong... you have the curve concave down between the
points of inflection.

Answer 21

oops you should have the curve concave down between the points of inflection...

Answer 22

if you want to use the 2nd derivative... you need to select points either side of the points of inflection and substitute them into the 2nd derivative to test for concavity...


hope it helps

Answer 23

(0,pi/2) concave up
(pi/2,3pi/2) concave down
(pi/2,2pi) concave up

Answer 24

here is the graph of your curve over the domain,

finding max and min values from the 1st derivative helps to identify the concavity

Answer 25

Wow that is very detailed. Thank you for putting the time into it. Okay I see my mistake thank you. I have to go. @campbell_st