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

1. ## 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.

2. 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).

3. 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.

4. $\displaystyle f(x)=x-2sin(x)$ given $\displaystyle 0<x<2\Pi$
$\displaystyle f'(x)=1-2cos(x) => 1-2cos(x)=0 => \frac{1}{2}=cos(x)$
$\displaystyle cos^{-1}(\frac{1}{2})=x$
$\displaystyle x=\frac{\Pi}{3}$ and $\displaystyle x=\frac{5\Pi}{3}$
<-----a--neg--b--pos--c--neg--d---->
$\displaystyle a = 0;b = \frac{\Pi}{3};c = \frac{5\Pi}{3};d = 2\Pi$

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

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

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.

5. Thanks man, that was a very nice and thorough explanation.

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# (pi, 2pi) increasing

Click on a term to search for related topics.