# Understanding concavity and increasing/decreasing of the function

• May 1st 2013, 06:55 AM
ameerulislam
Understanding concavity and increasing/decreasing of the function
I have the function

$\displaystyle f(x)=\frac{x}{x^2+2}$

$\displaystyle f'(x)=\frac{2-x^2}{(x^2+2)^2}$

$\displaystyle f''(x)=\frac{2x(x^2-6)}{(x^2+2)^3}$

finding increasing and decreasing.

I set the first derivative $\displaystyle \frac{2-x^2}{(x^2+2)^2}=0$
and get $\displaystyle +\sqrt{2}, -\sqrt{2}$

They create 3 intervals

1st interval
$\displaystyle -\infty$ to $\displaystyle -\sqrt{2}$

second interval
$\displaystyle -\sqrt{2}$ to $\displaystyle \sqrt{2}$

third interval
$\displaystyle \sqrt{2}$ to $\displaystyle \infty$

I plugin in values from in between these intervals into the first derivative and try to find where it is positive and where it is negative. The positive places are where the function is increasing and negatives are the places where it is decreasing. Right?
My results
$\displaystyle (-\infty,-\sqrt{2})U(\sqrt{2}, \infty)$ are decreasing

$\displaystyle (-\sqrt{2},\sqrt{2})$ are increasing

Next I set the second derivative as zero and undefined.

for this I can see i got a lot of intervals

$\displaystyle x=0$ and $\displaystyle x=+\sqrt{6} and x=-\sqrt{6}$

1st interval
$\displaystyle -\infty$ to $\displaystyle -\sqrt{6}$

second interval
$\displaystyle -\sqrt{6}$ to $\displaystyle 0$

third interval
$\displaystyle 0$ to $\displaystyle \sqrt{6}$

forth interval
$\displaystyle \sqrt{6}$ to $\displaystyle \infty$

Here I will have to do the same thing as I did when finding the increasing and decreasing but in this case negatives are concave down and positives are concave up.
So My results yield

$\displaystyle (-\infty,-\sqrt{6})U(0,\sqrt{6})$ are concave down

$\displaystyle (-\sqrt{6},0)U(\sqrt{6},\infty)$ are concave up (as they are positive)

are my answers are correct? Is there a quick way to understand that from the graph? Does that has be from the original graph?
• May 1st 2013, 07:04 AM
dokrbb
Re: Understanding concavity and increasing/decreasing of the function
All your arguments are correct and calculations too,

what do you mean by a quicker way - you actually can see the concavity on the graph, whether it is Up or Down, and I suppose that you are required to actually use all the data obtained to draw the graph by yourself, without a software, aren't you,

dokrbb
• May 1st 2013, 08:15 AM
ameerulislam
Re: Understanding concavity and increasing/decreasing of the function
I initially had a sense that increasing pretty much means concave up, and decreasing means concave down. But it doesn't seem that simple. How they are related?

By Quicker way I mean by seeing the graph I could guess what is what? Even before figuring out the derivatives. Just to cross check answers...
• May 1st 2013, 08:25 AM
dokrbb
Re: Understanding concavity and increasing/decreasing of the function
Quote:

Originally Posted by ameerulislam
I initially had a sense that increasing pretty much means concave up, and decreasing means concave down. But it doesn't seem that simple. How they are related?

By Quicker way I mean by seeing the graph I could guess what is what? Even before figuring out the derivatives. Just to cross check answers...

It seems that you are confused a little bit with the definition of CU and CD,

you know what it means Concavity - and you can double check it by watching to the graph, whether the concavity is Up or Down oriented, is also clear,

the notion of Increase and decrease is a separate notion - just make them clear separately, OK

dokrbb
• May 1st 2013, 08:26 AM
Plato
Re: Understanding concavity and increasing/decreasing of the function
Quote:

Originally Posted by ameerulislam
I initially had a sense that increasing pretty much means concave up, and decreasing means concave down. But it doesn't seem that simple. How they are related?

If $\displaystyle f'>0~\&~f">0$ the function is increasing and the rate of increase is increasing.

If $\displaystyle f'>0~\&~f"<0$ the function is increasing and the rate of increase is decreasing.

etc.
• May 1st 2013, 08:38 AM
ameerulislam
Re: Understanding concavity and increasing/decreasing of the function
Quote:

Originally Posted by Plato
If $\displaystyle f'>0~\&~f">0$ the function is increasing and the rate of increase is increasing.

If $\displaystyle f'>0~\&~f"<0$ the function is increasing and the rate of increase is decreasing.

etc.

interesting, pretty mind blowing... I'm still trying to understand it though.. but this is interesting indeed.

The inflection points are

points of concave up and down (change)

so $\displaystyle 0$ and $\displaystyle +\sqrt{6}$ and $\displaystyle -\sqrt{6}$

for this math??
• May 1st 2013, 03:30 PM
ameerulislam
Re: Understanding concavity and increasing/decreasing of the function
Quote:

Originally Posted by ameerulislam
interesting, pretty mind blowing... I'm still trying to understand it though.. but this is interesting indeed.

The inflection points are

points of concave up and down (change)

so $\displaystyle 0$ and $\displaystyle +\sqrt{6}$ and $\displaystyle -\sqrt{6}$

for this math??

Bump!
• May 1st 2013, 03:43 PM
Plato
Re: Understanding concavity and increasing/decreasing of the function
Quote:

Originally Posted by ameerulislam
Bump!

Do Not Bump.

Not only is it against forum rules, it is rude, crude, and stupid.
• May 1st 2013, 03:57 PM
ameerulislam
Re: Understanding concavity and increasing/decreasing of the function
okay sorry, I just needed the answer about the inflection points, As I have exam after couple of hours.