# On what interval is f increasing/decreasing and concave up/down

• October 20th 2009, 11:56 AM
Gitano
On what interval is f increasing/decreasing and concave up/down
f (x)= x - 2sinx, 0<x<2pi (less than or equal to)

Regarding this problem, I think that I have it solved. I just need someone to check my work to make sure I got the correct answer. Here is my answer:

f ' (x)= 1 - 2cosx ; 1-2cosx=0; cosx= 1/2
cosx > 1/2
f is increasing from (0, pi/3) and (5pi/3, 2pi)
f is decreasing from (pi/3, 5pi/3)

f ''(x) = 2sinx; sinx = 0; sinx>0
(0 , 2pi)
concave up: (pi, 2pi)
concave down: (0, pi)

Feedback would be appreciated, thanks.
• October 20th 2009, 11:59 AM
WhoCares357
Quote:

Originally Posted by Gitano
f (x)= x + 2sinx, 0<x<2pi (less than or equal to)

Regarding this problem, I think that I have it solved. I just need someone to check my work to make sure I got the correct answer. Here is my answer:

f ' (x)= 1 - 2cosx ; 1-2cosx=0; cosx= 1/2
cosx > 1/2
f is increasing from (0, pi/3) and (5pi/3, 2pi)
f is decreasing from (pi/3, 5pi/3)

f ''(x) = 2sinx; sinx = 0; sinx>0
(0 , 2pi)
concave up: (pi, 2pi)
concave down: (0, pi)

Feedback would be appreciated, thanks.

You made a mistake with the first derivative. The derivative of sin(x) is cos(x), not -cos(x).
• October 20th 2009, 12:05 PM
Gitano
Quote:

Originally Posted by WhoCares357
You made a mistake with the first derivative. The derivative of sin(x) is cos(x), not -cos(x).

That was a typo, it is actually x - 2sinx.
• October 20th 2009, 12:38 PM
WhoCares357
$f(x)=x-2sin(x)$ given $0
$f'(x)=1-2cos(x) => 1-2cos(x)=0 => \frac{1}{2}=cos(x)$
$cos^{-1}(\frac{1}{2})=x$
$x=\frac{\Pi}{3}$ and $x=\frac{5\Pi}{3}$
<-----a--neg--b--pos--c--neg--d---->
$a = 0;b = \frac{\Pi}{3};c = \frac{5\Pi}{3};d = 2\Pi$

f is decreasing from 0 to $\frac{\Pi}{3}$ and from $\frac{5\Pi}{3}$ to $2\Pi$
if is increasing from $\frac{\Pi}{3}$ to $\frac{5\Pi}{3}$

$f''(x)=2sin(x)$
[tex]0=2sin(x) => sin(x)=0 => $sin^{-1}(0)=x$
$x=\Pi$ and $x=2\Pi$
<---a--pos--b--neg--c--->
$a = 0;b = \Pi;c = 2\Pi$
f is concaved up from 0 to $\Pi$
f is concaved down from $\Pi$ to $2\Pi$

Edit: You can see the graph here:
http://www.wolframalpha.com/input/?i=plot+x-2sin%28x%29

Look at it and you will see that my claims are verified.
• October 20th 2009, 01:12 PM
Gitano
Thanks man, that was a very nice and thorough explanation.